A) [tex]3.6\cdot 10^5 J[/tex]
The power used by an object is defined as
[tex]P=\frac{E}{t}[/tex]
where
E is the energy used
t is the time elapsed
In this problem, we have
P = 100 W is the power of the light bulb
t = 1 h = 3600 s is the time elapsed
Solving for E, we find the amount of energy used by the light bulb:
[tex]E=Pt = (100 W)(3600 s)=3.6\cdot 10^5 J[/tex]
B) [tex]1.1 \cdot 10^2 m/s[/tex]
The kinetic energy of an object is given by
[tex]K=\frac{1}{2}mv^2[/tex]
where
m is the mass
v is the speed of the object
In this problem, we have a man of mass
m = 65 kg
we want its kinetic energy to be
[tex]E=3.6\cdot 10^5 J[/tex]
Therefore, we can calculate its speed from the previous formula:
[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.6\cdot 10^5 J)}{65 kg}}=105.2 m/s = 1.1 \cdot 10^2 m/s[/tex]
A 100-watt light bulb uses 360,000 joules of energy per hour. A 65 kg person would need to run at approximately 105 m/s to have the same amount of kinetic energy.
Energy = Power × Time = 100 W × 3600 s = 360000 J
Therefore, a 100-watt light bulb uses 360,000 joules of energy per hour.
KE = ½ mv²
Where m is mass in kilograms and v is velocity in meters per second. Plugging in the values, we can solve for v:
360,000 J = ½ × 65 kg × v²
v² = (2 × 360,000 J) / 65 kg = 11076.92 m²/s²
v = √11076.92 m²/s²
v ≈ 105 m/s
Hence, a 65 kg person would need to run at approximately 105 meters per second to have 360,000 joules of kinetic energy, which is equivalent to the energy used by a 100-watt light bulb in one hour.
Wonder Woman and Superman fly to an altitude of 1690 km , carrying between them a chest full of jewels that they intend to put into orbit around Earth. They want to make this tempting treasure inaccessible to their evil enemies who are trying to gain possession of it, yet keep it available for themselves for future use when they retire and settle down. But perhaps the time to retire is now! They accidentally drop the chest, which leaves their weary hands at rest, and discover that they are no longer capable of catching it as it falls into the Pacific Ocean. At what speed does the chest impact the surface of the water? Ignore air resistance, although in the real world it would make a world of difference. The radius and mass of Earth are 6370 km and 5.98×1024 kg , respectively.
Answer:
5120 m/s
Explanation:
The acceleration due to gravity is:
g = MG / r²
where M is the mass of the earth, G is the universal constant of gravitation, and r is the distance from the earth's center to the object's center.
Here, r = h + R, where h is the height of the chest above the surface and R is the radius of the earth.
g = MG / (h + R)²
Acceleration is the derivative of velocity:
dv/dt = MG / (h + R)²
Using chain rule, we can say:
(dv/dh) (dh/dt) = MG / (h + R)²
(dv/dh) v = MG / (h + R)²
Separate the variables:
v dv = MG / (h + R)² dh
Integrating:
∫₀ᵛ v dv = MG ∫₀ʰ dh / (h + R)²
½ v² |₀ᵛ = -MG / (h + R) |₀ʰ
½ (v² − 0²) = -MG / (h + R) − -MG / (0 + R)
½ v² = -MG / (h + R) + MG / R
½ v² = MGh / (R(h + R))
v² = 2MGh / (R(h + R))
Given:
M = 5.98×10²⁴ kg
R = 6.37×10⁶ m
h = 1.69×10⁶ m
G = 6.67×10⁻¹¹ m³/kg/s²
Plugging in:
v² = 2 (5.98×10²⁴) (6.67×10⁻¹¹) (1.69×10⁶) / ((6.37×10⁶) (1.69×10⁶ + 6.37×10⁶))
v² = 2 (5.98) (6.67) (1.69) / ((6.37) (1.69 + 6.37)) × 10⁷
v ≈ 5120 m/s
Notice that if we had approximated g as a constant 9.8 m/s², we would have gotten an answer of:
v² = v₀² + 2a(x - x₀)
v² = (0 m/s)² + 2 (9.8 m/s²) (1.69×10⁶ m - 0 m)
v ≈ 5760 m/s
So we know that our calculated velocity of 5120 m/s is a reasonable answer.
The Newton's second law and the law of universal gravitation allows to find the result for the speed of the chest when reaching the ocean is:
The velocity is v = 5120 m / s
The law of universal gravitation is stable that the gravitational force between bodies is attractive and is proportional to the mass of the bodies and inversely proportional to the square of the distance.
[tex]F= - G \frac{Mm}{r^2}[/tex]
Where M and m are the mass of the two bodies and r is the distance.
Indicate the height from where the chest falls h= 1690 km = 1,690 10⁶ m, they also give the radius and the mass of the earth.
Newton's second law establishes a relationship between force, mass and the acceleration of bodies.
F = m a
[tex]- G \frac{Mm}{r^2} = m a \\a= - G \frac{M}{r^2}[/tex]
The distance of the body from the center of the planet is
r = R + h
The acceleration is defined as the variation of velocity with time.
[tex]a= \frac{dv}{dt} \\ \frac{dv}{dt} = - G \frac{M}{(h+R)^2}[/tex]
Let's use the chain rule
[tex]\frac{dv}{dh}\ \frac{dh}{dt} = - GM \frac{1}{(h+R)^2 }[/tex]
The velocity is the derivative of the position with respect to the time.
[tex]v=\frac{dh}{dt} \\ v \ \frac{dv}{dh} = - GM \ \frac{1}{(h+R)^2}[/tex]
To solve we use the method of separation of variables and we integrate.
[tex]\int\limits^v_0 {v} \, dv = -GM \int\limits^0_h {\frac{1}{(h+R)^2} } \, dh \\\frac{1}{2} ( v^2 - 0) = -GM (-1) [ \frac{1}{(0+R)} - \frac{1}{(h+R)}] \\\frac{1}{2} v^2 = GM \ \frac{h}{(h+R)R}[/tex]
[tex]v^2 = 2GM \ \frac{h}{(h+R) R }[/tex]
Let's calculate
v² = [tex]2 \ 6.67 \ 10^{-11} \ 5.98 \ 10^{24}} \ \frac{1.690 }{(1.609 + 6.370) 6.370} \ 10^{-6}[/tex]
v = 5120 m/s
In conclusion we use Newton's second law and the universal gravitation's law we can find the result for the speed of the chest when reaching the ocean is;
The velocity is v = 5120 m / s
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The siren on an ambulance is emitting a sound whose frequency is 2250 Hz. The speed of sound is 343 m/s. (a) If the ambulance is stationary and you (the "observer") are sitting in a parked car, what are the wavelength and the frequency of the sound you hear? (b) Suppose that the ambulance is moving toward you at a speed of 26.6 m/s. Determine the wavelength and the frequency of the sound you hear. (c) If the ambulance is moving toward you at a speed of 26.6 m/s and you are moving toward it at a speed of 11.0 m/s, find the wavelength and frequency of the sound you hear.
(a) 2250 Hz, 0.152 m
In this situation, both the ambulance and observer are stationary.
This means that there is no shift in frequency/wavelength due to the Doppler effect. So, the frequency heard by the observer is exactly identical to the frequency emitted by the ambulance:
f = 2250 Hz
While the wavelength is given by the formula:
[tex]\lambda=\frac{v}{f}[/tex]
where
v = 343 m/s is the speed of sound
f = 2250 Hz is the frequency of the sound
Substituting, we find
[tex]\lambda=\frac{343 m/s}{2250 Hz}=0.152 m[/tex]
(b) 2439.2 Hz, 0.141 m
The Doppler effect formula for a moving source is
[tex]f'=(\frac{v}{v+v_s})f[/tex]
where
f' is the apparent frequency
f is the original frequency
v is the speed of sound
[tex]v_s[/tex] is the velocity of the source (the ambulance), which is positive if the source is moving away from the observer, negative otherwise
Here the ambulance is moving toward the observer, so
[tex]v_s = -26.6 m/s[/tex]
Substituting into the formula, we find the frequency heard by the observer:
[tex]f'=(\frac{343 m/s}{343 m/s-26.6 m/s})(2250 Hz)=2439.2 Hz[/tex]
while the wavelength seen by the observer will be:
[tex]\lambda' = \frac{v}{f'}=\frac{343 m/s}{2439.2 Hz}=0.141 m[/tex]
(c) 2517.4 Hz, 0.136 m
In this situation, we must use the most general formula for the Doppler effect, which is
[tex]f'=(\frac{v+v_r}{v+v_s})f[/tex]
where
[tex]v_r[/tex] is the velocity of the observer, which is positive if the observer is moving toward the source, negative otherwise
[tex]v_s[/tex] is the velocity of the source (the ambulance), which is positive if the source is moving away from the observer, negative otherwise
In this situation,
[tex]v_s = -26.6 m/s[/tex]
[tex]v_r = +11.0 m/s[/tex]
Therefore, the frequency heard by the observer is
[tex]f'=(\frac{343 m/s+11.0 m/s}{343 m/s-26.6 m/s})(2250 Hz)=2517.4 Hz[/tex]
while the wavelength seen by the observer will be:
[tex]\lambda' = \frac{v}{f'}=\frac{343 m/s}{2517.4 Hz}=0.136 m[/tex]
a)Find the minimum magnetic field needed to exert a 5.7 fN force on an electron moving at 23Mm/s .b)Find the field strength required if the field were at 45 degrees to the electron's velocity.
a) [tex]1.55\cdot 10^{-3} T[/tex]
The magnetic force exerted on a charged particle in motion is given by:
[tex]F=qvB sin \theta[/tex]
where
q is the charge of the particle
v is the velocity of the particle
B is the magnetic field strength
[tex]\theta[/tex] is the angle between the direction of v and B
The expression can be rewritten as
[tex]B=\frac{F}{qv sin \theta}[/tex]
We see that the minimum magnetic field needed is the one for which [tex]sin \theta=1[/tex], so with [tex]\theta=90^{\circ}[/tex]. In this problem, we have:
[tex]q=1.6\cdot 10^{-19} C[/tex] (charge of the electron)
[tex]f=5.7 fN=5.7\cdot 10^{-15}N[/tex] is the force
[tex]v=23 Mm/s = 23\cdot 10^6 m/s[/tex] is the electron's velocity
Substituting, we find
[tex]B=\frac{5.7\cdot 10^{-15}N}{(1.6\cdot 10^{-19} C)(23\cdot 10^6 m/s) sin 90^{\circ}}=1.55\cdot 10^{-3} T[/tex]
b) [tex]2.19\cdot 10^{-3} T[/tex]
In this case, the field is at 45 degrees to the electron's velocity, so we have
[tex]\theta=45^{\circ}[/tex]
Therefore, the field strength required to obtain a force of
[tex]f=5.7 fN=5.7\cdot 10^{-15}N[/tex] is the force
will be equal to
[tex]B=\frac{F}{qv sin \theta}=\frac{5.7\cdot 10^{-15}N}{(1.6\cdot 10^{-19} C)(23\cdot 10^6 m/s) sin 45^{\circ}}=2.19\cdot 10^{-3} T[/tex]
The minimum magnetic field needed to exert the given force is [tex]1.55 \times 10^{-3} \ T[/tex]
When the field is 45 degrees to the electron's speed, the magnetic field strength is [tex]2.19\times 10^{-3} \ T[/tex]
The given parameters;
force exerted on the electron, F = 5.7 Nspeed of the electron, v = 23 M m/sThe minimum magnetic field is calculated as follows;
[tex]F = qvB \\\\B = \frac{F}{qv} \\\\B = \frac{5.7\times 10^{-15}}{(1.602\times 10^{-19})(23\times 10^6)} \\\\B = 1.55 \times 10^{-3} \ T[/tex]
When the field is 45 degrees to the electron's speed, the magnetic field strength is calculated as follows;
[tex]B = \frac{F}{q\times v \times sin(45)} = \frac{5.7 \times 10^{-15}}{1.602 \times 10^{-19} \times 23 \times 10^6 \times sin(45)} = 2.19 \times 10^{-3} \ T[/tex]
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which statement is a characteristic of a concave lens
"A concave lens is thinner at the center than it is at the edges."
If this isn't on the list of choices, that's tough. We can't help you choose the best one if we don't know what any of them is.
Predict the products of the combustion of methanol, CH3OH(l).
Answer : The products of the combustion of methanol are, carbon dioxide [tex](CO_2)[/tex] and water [tex](H_2O)[/tex]
Explanation :
Combustion reaction : It is a type of reaction in which the hydrocarbon react with the oxygen gas to give carbon dioxide and water as products.
The balanced chemical reaction of combustion of methanol [tex](CH_3OH)[/tex] is :
[tex]2CH_3OH(l)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(g)[/tex]
By the stoichiometry we can say that, 2 mole of methanol react with 3 moles of oxygen gas to give 2 mole of carbon dioxide and 4 moles of water as products.
In this reaction, methanol and oxygen gas are the reactants and carbon dioxide and water are the products.
Therefore, the products of the combustion of methanol are, carbon dioxide [tex](CO_2)[/tex] and water [tex](H_2O)[/tex]
The combustion of methanol produces carbon dioxide and water in the presence of oxygen, as represented by the reaction equation CH3OH(l) + 1.5 O2(g) -> CO2(g) + 2 H2O(l).
Explanation:In a combustion reaction, a substance combines with oxygen to produce heat and light. For methanol, CH3OH(l), the reactants are methanol and oxygen (O2), and the products are carbon dioxide (CO2) and water (H2O). The balanced chemical equation for this reaction is CH3OH(l) + 1.5 O2(g) -> CO2(g) + 2 H2O(l). This indicates that one molecule of methanol reacts with 1.5 molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water.
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Can someone help me??!!!!!
Answer:
III quadrant
Explanation:
First of all, let's write the components of each vector along the two directions. We have:
[tex]B_x = B cos 33^\circ = (5.6) cos 33^\circ=-4.7\\B_y = B sin 33^\circ = (5.6) sin 33^\circ=-3.0[/tex]
where we put a negative sign on both components, since they are in the negative x- and y- direction.
Similarly,
[tex]C_x = C sin 22^\circ = (4.8) sin 22^\circ=1.8\\C_y = C cos 22^\circ = (4.8) cos 22^\circ=-4.5[/tex]
where we put a negative sign on the y-direction, since it is along the negative direction.
Now we sum the components along each axis:
[tex]B_x + C_x = -4.7+1.8=-2.9\\B_y + C_y = -3.0 + (-4.5)=-7.5[/tex]
We see that both components of (B+C) are negative, so this vector must be in the 3rd quadrant.
Electromagnetic radiation behaves both as particles (called photons) and as waves. Wavelength (λ) and frequency (ν) are related according to the equation c=λ×ν where c is the speed of light (3.00×108 m/s). The energy (E in joules) contained in one quantum of electromagnetic radiation is described by the equation E=h×ν where h is Planck's constant (6.626×10−34 J⋅s). Note that frequency has units of inverse seconds (s−1), which are more commonly expressed as hertz (Hz). A microwave oven operates at 2.20 GHz . What is the wavelength of the radiation produced by this appliance?
Explanation:
We can find the wavelength of the radiation produced by the microwave oven by using the following given equation:
[tex]c=\lambda.\nu[/tex] (1)
Clearing [tex]\lambda[/tex] :
[tex]\lambda=\frac{c}{\nu}[/tex] (2)
Knowing [tex]\nu=2.20 GHz=2.20(10)^{6}Hz=2.20(10)^{6}s^{-1}[/tex]
[tex]\lambda=\frac{3(10)^{8}m/s}{2.20(10)^{6}s^{-1}}[/tex] (3)
[tex]\lambda=136.363m[/tex] This is the wavelength of the radiation produced by the microwave
The wavelength of the radiation produced by a microwave oven operating at 2.20 GHz is approximately 13.6 cm.
Explanation:Calculating the Wavelength of Microwave Oven Radiation
To find the wavelength of the radiation produced by a microwave oven that operates at 2.20 GHz, we use the formula c = λν, where c is the speed of light (3.00 × 10¸ m/s), λ is the wavelength in meters, and ν is the frequency in hertz (Hz).
First, we convert the frequency from gigahertz (GHz) to hertz (Hz) by multiplying it by 10¹:
2.20 GHz × 10¹ = 2.20 × 10¹ Hz.
Next, we rearrange the formula to solve for λ:
λ = c / ν
Now, we plug in the values:
λ = (3.00 × 10¸ m/s) / (2.20 × 10¹ Hz)
Calculating this gives:
λ = (3.00 × 10¸) / (2.20 × 10¹)
λ = 1.36 × 10² m
Therefore, the wavelength of the radiation emitted by the microwave oven is approximately 13.6 cm.
Which of the following most accurately describes the development of the antibiotic penicillin?
A.) One scientist worked tirelessly for years to create a drug that could prevent death from infections
B.) It was developed by Ernest Chain, who knew about the Penicillium mold’s antibacterial properties from his own life experiences
C.) Fleming shows that the Penicillium mood could kill bacteria and observed that it was very effective when taken orally or intravenously
D.) It was developed over many years through work of multiple scientists, some of whom collaborated and some of whom didn’t know about the work of the others
Answer:
C.) Fleming shows that the Penicillium mood could kill bacteria and observed that it was very effective when taken orally or intravenously
Out of the given options, “Fleming shows that the Penicillium mold could kill bacteria and observed that it was very effective when taken orally or intravenously” is most accurately describes the development of the antibiotic penicillin
Answer: Option C
Explanation:
It was in 1928, when Alexander Fleming discovered the most effective antibacterial drug to knock down severe bacteria and hence, saved lives. It all began with the development of mold on a staphylococcus culture plate while Fleming was experimenting with the influenza virus.
Penicillin has helped numerous people to get aided from severe illness in the 20th century and especially during the World War II. The use of penicillin has helped in treating bacterial endocarditis, pneumococcal pneumonia, bacterial meningitis etc.
A circular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux through the loop is a maximum. At this position, the electric flux is 7.50 × 105 N⋅m2/C. Determine the magnitude of the electric field. A) 8.88 × 105 N/C B) 1.07 × 106 N/C C) 2.44 × 106 N/C D) 4.24 × 106 N/C E) 6.00 × 106 N/C
Answer:
C) 2.44 × 106 N/C
Explanation:
The electric flux through a circular loop of wire is given by
[tex]\Phi = EA cos \theta[/tex]
where
E is the electric field
A is the cross-sectional area
[tex]\theta[/tex] is the angle between the direction of the electric field and the normal to A
The flux is maximum when [tex]\theta=0^{\circ}[/tex], so we are in this situation and therefore [tex]cos \theta =1[/tex], so we can write
[tex]\Phi = EA[/tex]
Here we have:
[tex]\Phi = 7.50\cdot 10^5 N/m^2 C[/tex] is the flux
d = 0.626 m is the diameter of the coil, so the radius is
r = 0.313 m
and so the area is
[tex]A=\pi r^2 = \pi (0.313 m)^2=0.308 m^2[/tex]
And so, we can find the magnitude of the electric field:
[tex]E=\frac{\Phi}{A}=\frac{7.50\cdot 10^5 Nm^2/C}{0.308 m^2}=2.44\cdot 10^6 N/C[/tex]
In a double-slit experiment, it is observed that the distance between adjacent maxima on a remote screen is 1.0 cm. What happens to the distance between adjacent maxima when the slit separation is cut in half? 13) ______ A) It decreases to 0.25 cm. B) It decreases to 0.50 cm. C) It increases to 2.0 cm. D) It increases to 4.0 cm. E) None of these choices are correct.
Answer:
C) It increases to 2.0 cm
Explanation:
In a double-slit diffraction experiment, the distance on the screen between two adjacent maxima is given by
[tex]\Delta y = \frac{\lambda D}{d}[/tex]
where
[tex]\lambda[/tex] is the wavelength of the wave
D is the distance of the screen from the slits
d is the separation between the slits
In this problem, the initial distance between adjacent maxima is 1.0 cm. Later, the slit separation is cut in a half, which means that the new slit separation is
[tex]d'=\frac{d}{2}[/tex]
Substituting into the equation, we find that the new separation between the maxima is
[tex]\Delta y' = \frac{\lambda D}{d/2}=2(\frac{\lambda D}{d})=2\Delta y[/tex]
So, the distance increases by a factor 2: therefore, the new separation between the maxima will be 2.0 cm.
In a double-slit experiment, when the slit separation is halved, the distance between adjacent maxima will double. Therefore, the correct answer is C) It increases to 2.0 cm.
Explanation:In a double-slit experiment, the distance between adjacent maxima on a remote screen is determined by the separation of the slits and the wavelength of the light used. According to the formula for double-slit interference d sin θ = mλ (d - distance between slits, m - order of maxima, λ - wavelength of the light), when the slit separation (d) is cut in half, the distance between adjacent maxima will increase. Therefore, between the given options, C) It increases to 2.0 cm is the correct answer. This is because the maxima result from constructive interference which occurs when the path difference is an integral multiple of the wavelength.
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Bob has just finished climbing a sheer cliff above a beach, and wants to figure out how high he climbed. All he has to use, however, is a baseball, a stopwatch, and a friend on the beach below with a long measuring tape. Bob is a pitcher and he knows that the fastest he can throw the ball is about ????0=34.1 m/s.v0=34.1 m/s. Bob starts the stopwatch as he throws the ball (with no way to measure the ball's initial trajectory), and watches carefully. The ball rises and then falls, and after ????1=0.510 st1=0.510 s the ball is once again level with Bob. Bob cannot see well enough to time when the ball hits the ground. Bob's friend then measures that the ball landed ????=126 mx=126 m from the base of the cliff. How high up is Bob, if the ball started exactly 2 m above the edge of the cliff?
Answer:
56.0 m
Explanation:
We know that after 0.510 s, the ball is level with Bob again. We can use this to find the vertical component of the initial velocity.
y = y₀ + v₀ᵧ t + ½ gt²
h+2 = h+2 + v₀ᵧ (0.510) + ½ (-9.8) (0.510)²
v₀ᵧ = 2.50 m/s
Since the magnitude is 34.1 m/s, we can now find the horizontal component:
v₀² = v₀ₓ² + v₀ᵧ²
(34.1)² = v₀ₓ² + (2.50)²
v₀ₓ = 34.0 m/s
And since we know the ball lands 126 m from the base of the cliff, we can find the time it takes to land:
x = x₀ + v₀ₓ t + ½ at²
126 = 0 + (34.0) t + ½ (0) t²
t = 3.71 s
Finally, we can now find the height of the cliff:
y = y₀ + v₀ᵧ t + ½ gt²
0 = h+2 + (2.50) (3.71) + ½ (-9.8) (3.71)²
h = 56.0 m
Bob climbed approximately 4.4 meters.
Explanation:To determine the height that Bob climbed, we can use the kinematic equation for vertical motion:
Y = Yo + Voy*t -1/2gt^2
where:
Y = final height (unknown)
Yo = initial height (2m)
Voy = initial vertical velocity (unknown)
t = time taken to reach the final height (0.510s)
g = acceleration due to gravity (9.8m/s^2)
We need to find the initial vertical velocity. When the ball reaches the final height, it has the same vertical velocity as it did when it was thrown. Using the equation for vertical velocity:
Vy = Voy - gt
where:
Vy = vertical velocity (0 m/s)
Substituting the known values:
0 = Voy - (9.8)(0.510)
Solving for Voy, we get:
Voy ≈ 5.0 m/s
Now we can calculate the final height:
Y = Yo + Voy*t - 1/2gt^2
Y = 2 + (5.0)(0.510) - 1/2(9.8)(0.510)^2
Y ≈ 2 + 2.55 - 0.126
Y ≈ 4.425m
Therefore, Bob climbed approximately 4.4 meters.
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In deep space, sphere A of mass 94 kg is located at the origin of an x axis and sphere B of mass 100 kg is located on the axis at x = 1.8 m. Sphere B is released from rest while sphere A is held at the origin. (a) What is the gravitational potential energy of the two-sphere system just as B is released? (b) What is the kinetic energy of B when it has moved 0.60 m toward A?
(a) [tex]-3.48\cdot 10^{-7} J[/tex]
The gravitational potential energy of the two-sphere system is given by
[tex]U=-\frac{Gm_A m_B}{r}[/tex] (1)
where
G is the gravitational constant
[tex]m_A = 94 kg[/tex] is the mass of sphere A
[tex]m_B = 100 kg[/tex] is the mass of sphere B
r = 1.8 m is the distance between the two spheres
Substitutign data in the formula, we find
[tex]U=-\frac{(6.67\cdot 10^{-11})(94 kg)(100 kg)}{1.8 m}=-3.48\cdot 10^{-7} J[/tex]
and the sign is negative since gravity is an attractive force.
(b) [tex]1.74\cdot 10^{-7}J[/tex]
According to the law of conservation of energy, the kinetic energy gained by sphere B will be equal to the change in gravitational potential energy of the system:
[tex]K_f = U_i - U_f[/tex] (2)
where
[tex]U_i=-3.48\cdot 10^{-7} J[/tex] is the initial potential energy
The final potential energy can be found by substituting
r = 1.80 m -0.60 m=1.20 m
inside the equation (1):
U=-\frac{(6.67\cdot 10^{-11})(94 kg)(100 kg)}{1.2 m}=-5.22\cdot 10^{-7} J
So now we can use eq.(2) to find the kinetic energy of sphere B:
[tex]K_f = -3.48\cdot 10^{-7}J-(-5.22\cdot 10^{-7} J)=1.74\cdot 10^{-7}J[/tex]
If a 4.0Ω resistor, a 6.0Ω resistor, and an 8.0Ω resistor are connected in parallel across a 12 volt battery, what is the total current of the circuit?
6.7 amps
18 amps
1.8 amps
3.6 amps
Answer:
Explanation:
6.7 amps
The total current of the circuit is 6.7 amps.
Explanation:In a parallel circuit, the total current is the sum of the individual currents flowing through each resistor. To find the total current, we can use Ohm's law, which states that I = V/R, where I is the current, V is the voltage, and R is the resistance.
For the given circuit, the total resistance can be calculated by adding the reciprocals of the individual resistances: 1/Rt = 1/R1 + 1/R2 + 1/R3. Plugging in the resistance values, we get 1/Rt = 1/4.0 + 1/6.0 + 1/8.0.
After calculating the reciprocal of the sum, we find that Rt = 1.714 ohms. Finally, using Ohm's law, we can find the total current: I = V/Rt = 12 volts / 1.714 ohms = 6.7 amps.
Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudinal load of 46500 N (10400 lbf). Assume Vf = 0.3, Vm = 0.7, Ef = 131 GPa and Em = 2.4 GPa. (a) Calculate the fiber-matrix load ratio. (b) Calculate the actual load carried by fiber phase. (c) Calculate the actual load carried by matrix phase. (d) Compute the magnitude of the stress on the fiber phase. (e) Compute the magnitude of the stress on the matrix phase. (f) What strain is expected by the composite?
(a) 23.4
The fiber-to-matrix load ratio is given by
[tex]\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}[/tex]
where
[tex]E_f = 131 GPa[/tex] is the fiber elasticity module
[tex]E_m = 2.4 GPa[/tex] is the matrix elasticity module
[tex]V_f=0.3[/tex] is the fraction of volume of the fiber
[tex]V_m=0.7[/tex] is the fraction of volume of the matrix
Substituting,
[tex]\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4[/tex] (1)
(b) 44,594 N
The longitudinal load is
F = 46500 N
And it is sum of the loads carried by the fiber phase and the matrix phase:
[tex]F=F_f + F_m[/tex] (2)
We can rewrite (1) as
[tex]F_m = \frac{F_f}{23.4}[/tex]
And inserting this into (2):
[tex]F=F_f + \frac{F_f}{23.4}[/tex]
Solving the equation, we find the actual load carried by the fiber phase:
[tex]F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N[/tex]
(c) 1,906 N
Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:
[tex]F=F_f + F_m[/tex] (2)
Using
F = 46500 N
[tex]F_f = 44594 N[/tex]
We can immediately find the actual load carried by the matrix phase:
[tex]F_m = F-F_f = 46,500 N - 44,594 N=1,906 N[/tex]
(d) 437 MPa
The cross-sectional area of the fiber phase is
[tex]A_f = A V_f[/tex]
where
[tex]A=340 mm^2=340\cdot 10^{-6}m^2[/tex] is the total cross-sectional area
Substituting [tex]V_f=0.3[/tex], we have
[tex]A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2[/tex]
And the magnitude of the stress on the fiber phase is
[tex]\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa[/tex]
(e) 8.0 MPa
The cross-sectional area of the matrix phase is
[tex]A_m = A V_m[/tex]
where
[tex]A=340 mm^2=340\cdot 10^{-6}m^2[/tex] is the total cross-sectional area
Substituting [tex]V_m=0.7[/tex], we have
[tex]A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2[/tex]
And the magnitude of the stress on the matrix phase is
[tex]\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa[/tex]
(f) [tex]3.34\cdot 10^{-3}[/tex]
The longitudinal modulus of elasticity is
[tex]E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa[/tex]
While the total stress experienced by the composite is
[tex]\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa[/tex]
So, the strain experienced by the composite is
[tex]\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}[/tex]
The fiber-matrix load ratio, actual load carried by the fiber and matrix phase, stress on the fiber and matrix phase, and the strain in the composite can be calculated using the given volume fractions and elastic moduli.
Explanation:SolutionGiven the load P = 46500N, volume fraction of fiber (Vf) = 0.3, volume fraction of matrix (Vm) = 0.7, elastic modulus of fiber (Ef) = 131 GPa = 131 * 10⁹ Pa, and elastic modulus of matrix (Em) = 2.4 Gpa = 2.4 * 10⁹ Pa, the fiber-matrix load ratio is given by the ratio of the product of the volume fraction and elastic modulus of fiber to the product of the volume fraction and elastic modulus of matrix, i.e., (Vf * Ef)/(Vm * Em). (b) The actual load carried by the fiber phase can be calculated by multiplying load P with Vf and the ratio of Ef to the sum of Vf * Ef and Vm * Em. (c) The actual load carried by the matrix phase can be calculated by subtracting the load carried by fiber from total load P. (d) The stress on the fiber phase can be calculated as the load carried by the fiber phase divided by the cross-sectional area. (e) The stress on the matrix phase can be calculated as the load carried by the matrix phase divided by the cross-sectional area. (f) The strain in the composite can be calculated by dividing the total applied stress by the equivalent stiffness of the composite material (i.e., (Vf * Ef + Vm * Em)).
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Can someone help me?!!!
The answer is:
The time took for the rock to reach its maximum height is 0.110 seconds.
Why?In order to calculate the time needed for the rock to reach its maximum height, we need to calculate the initial vertical speed.
From the statement we know that the rock was launched at an initial speed of 2.1 m/s at an angle of 30° above the horizontal, so, calculating we have:
[tex]V_x=2.1\frac{m}{s} *cos(30\°)=1.82\frac{m}{s} \\\\V_y=2.1\frac{m}{s} *sin(30\°)=1.05\frac{m}{s}[/tex]
Also, we know that at the maximum height, the speeds tends to 0. So, using the following equation and substituting "v" equal to 0, we have:
[tex]V=V_o-g*t\\\\0=V_o-g*t\\\\g*t=V_o\\\\t=\frac{V_o}{g}[/tex]
Where,
t is the time in seconds.
V is the initial speed
g is the gravity acceleration.
Using gravity acceleration equal to [tex]9.81\frac{m}{s^{2} }[/tex] we have:
[tex]V=V_o-g*t\\\\0=V_o-g*t\\\\g*t=V_o\\\\t=\frac{V_o}{g}[/tex]
[tex]t=\frac{V_o}{g}[/tex]
[tex]t=\frac{1.05\frac{m}{s}}{9.81\frac{m}{s^{2}}}=0.1070seconds=0.110seconds[/tex]
Hence, the correct option is the last option, the time took for the rock to reach its maximum height is 0.110 seconds.
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An air bubble at the bottom of a lake 40.5 m deep has a volume of 1.00 cm3. Part A If the temperature at the bottom is 2.3 ∘C and at the top 28.1 ∘C, what is the radius of the bubble just before it reaches the surface? Express your answer to two significant figures and include the appropriate units.
Answer:
5.4 cm³
Explanation:
Ideal gas law:
PV = nRT
where P is absolute pressure, V is volume, n is number of moles, R is ideal gas constant, and T is absolute temperature.
Since n and R are constant, we can say:
PV / T = constant
At the bottom of the lake, the pressure is:
P = ρgh + Patm
P = (1000 kg/m³) (9.8 m/s²) (40.5 m) + 101,325 Pa
P = 498,225 Pa
And the temperature is:
T = 2.3 + 273.15 K
T = 275.45 K
At the top of the lake, the pressure is:
P = Patm
P = 101,325 Pa
And the temperature is:
T = 28.1 + 273.15 K
T = 301.25 K
Therefore:
PV / T = PV / T
(498225 Pa) (1.00 cm³) / (275.45 K) = (101325 Pa) V / (301.25 K)
V = 5.4 cm³
Final answer:
The radius of the air bubble just before it reaches the surface is calculated using Charles's Law and the volume of a sphere. It is approximately 6.20 mm once we consider the change in temperature from the bottom to the top of the lake.
Explanation:
To determine the radius of the air bubble just before it reaches the surface of the lake, we must consider the effects of pressure and temperature changes on the volume of the bubble. Due to the nature of the question involving thermal expansion and the principles of gas laws, Physics is the subject, and the content is suitable for High School level.
First, we acknowledge some basic relations for the behavior of gases. Assuming that the amount of gas and the pressure remain constant (since it's mentioned that we can ignore the pressure change), we can use Charles's Law (V1/T1 = V2/T2), where V1 and T1 are the original volume and temperature, and V2 and T2 are the final volume and temperature respectively. Temperatures should be in Kelvin for gas laws, where Kelvin is the Celsius temperature plus 273.15.
V1 = 1.00 cm3 = 1.00 x 10⁻⁶ m³
T1 = 2.3 °C + 273.15 = 275.45 K
T2 = 28.1 °C + 273.15 = 301.25 K
Applying Charles's Law:
V2 = V1 * (T2/T1) = (1.00 x 10⁻⁶ m³) * (301.25 K / 275.45 K)
Now let's calculate V2.
V2 = (1.00 x 10⁻⁶ m³) * (301.25 / 275.45)
V2 ≈ 1.0937 x 10⁻⁶ m3
The volume of a sphere is given by (4/3)πr³. To find the radius, we solve for r:
V = (4/3)πr³
r = ∛(3V / 4π)
Substituting the value of V2:
r ≈ ∛(3 * 1.0937 x 10⁻⁶ m3) / (4 * π)
r ≈ 6.20 x 10⁻³ m
So, the radius of the bubble just before it reaches the surface is approximately 6.20 mm.
A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on the horizontal section of a track, as shown. It rolls around the inside of a vertical circular loop 90.0 cm in diameter and finally leaves the track at a point 20.0 cm below the horizontal section. (a) Find the speed of the ball at the top of the loop. Demonstrate that it will not fall from the track. (b) Find its speed as it leaves the track. (c) Suppose that static friction between the ball and the track was negligible, so that the ball slide instead of rolling. Would its speed then be higher, lower, or the same at the top of the loop
Answer:
2.38 m/s, 4.31 m/s, lower
Explanation:
a)
Initial energy = final energy
½ m v₀² + ½ I ω₀² = mgh + ½ m v₁² + ½ I ω₁²
Since the ball is rolling without slipping, ω = v / r.
For a hollow sphere, I = ⅔ m r².
½ m v₀² + ½ (⅔ m r²) (v₀ / r)² = mgh + ½ m v₁² + ½ (⅔ m r²) (v₁ / r)²
½ m v₀² + ⅓ m v₀² = mgh + ½ m v₁² + ⅓ m v₁²
⅚ m v₀² = mgh + ⅚ m v₁²
⅚ v₀² = gh + ⅚ v₁²
v₀² = 1.2gh + v₁²
v₁ = √(v₀² − 1.2gh)
Given v₀ = 4.03 m/s, g = 9.80 m/s, h = 0.900 m:
v₁ = √((4.03)² − 1.2 (9.80) (0.900))
v₁ ≈ 2.38 m/s
At the top of the loop, the sum of the forces in the radial direction is:
∑F = ma
W + N = m v² / R
N = m v² / R - mg
N = m (v² / R - g)
Given v = 2.38 m/s, R = 0.450 m, and g = 9.80 m/s²:
N = m ((2.38)² / 0.450 - 9.80)
N = 2.77m
N ≥ 0, so the ball stays on the track.
b)
Initial energy = final energy
Borrowing from part a):
v₂ = √(v₀² − 1.2gh)
This time, h = -0.200 m:
v₂ = √((4.03)² − 1.2 (9.80) (-0.200))
v₂ ≈ 4.31 m/s
c)
Without the rotational energy:
½ m v₀² = mgh + ½ m v₁²
½ v₀² = gh + ½ v₁²
v₀² = 2gh + v₁²
v₁ = √(v₀² - 2gh)
This is less than v₁ we calculated earlier.
Refer the below Solution for better understanding.
Given :
Speed = 4.03 m/sec
Vertical circular loop of 90 cm diameter.
Solution :
a)
Initial energy = Final Energy
[tex]\rm \dfrac{1}{2}mv_0^2 + \dfrac{1}{2}I\omega_0^2 = mgh + \dfrac{1}{2}mv_1^2 + \dfrac{1}{2}I\omega_1^2[/tex]
here,
[tex]\rm \omega = \dfrac{v}{r}[/tex]
for a hollow sphere,
[tex]\rm I = \dfrac{2}{3}mr^2[/tex]
[tex]\rm \dfrac{1}{2}mv_0^2 + \dfrac{1}{2}\times\dfrac{2}{3}mr^2(\dfrac{v_0}{r})^2 = mgh + \dfrac{1}{2}mv_1^2 + \dfrac{1}{2}\times\dfrac{2}{3}mr^2(\dfrac{v_1}{r})^2[/tex]
by further solving above equation,
[tex]\rm v_1=\sqrt{v_0^2-1.2gh}[/tex] --- (1)
Now put the values of [tex]\rm v_0 , \;g\;and\;h[/tex] in equation (1),
[tex]\rm v_1 = \sqrt{4.03^2-1.2(9.8)(0.9)}[/tex]
[tex]\rm v_1 = 2.38 \; m/sec[/tex]
Now,
F = ma
[tex]\rm mg + N = \dfrac{mv_1^2}{R}[/tex]
[tex]\rm N = m(\dfrac{v_1^2}{R}-g)[/tex] --- (2)
Now put the values of R, g, m and [tex]\rm v_1[/tex] in equation (2) we get,
N = 2.77m
[tex]\rm N\geq 0[/tex]
ball stays on the track.
b) To find the speed of the ball as it leaves the track,
[tex]\rm v_2=\sqrt{v_0^2-1.2gh}[/tex] ---- (3)
put h = -0.2m in equation (3)
[tex]\rm v_2=\sqrt{4.03^2-1.2(9.8)(-0.2)}[/tex]
[tex]\rm v_2=4.31\;m/sec[/tex]
c) Again, but without rotational energy
Initial energy = Final energy
[tex]\rm \dfrac{1}{2}mv_0^2 = mgh + \dfrac{1}{2}mv_1^2[/tex]
by further solving the above equation we get,
[tex]\rm v_1 = \sqrt{v_0^2-2gh}[/tex] and this is less than [tex]\rm v_1[/tex] we calculated earlier.
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Calculate the moment of inertia for a solid cylinder with a mass of 100g and a radius of 4.0 cm
Solution here,
mass of cylinder(m)=100 g
radius of cylinder(r)= 4 cm
momoent of inertia(I)=?
we have,
I=mr^2=100×4^2=1600 g/cm^2
Which one of the following statements concerning waves is false? A transverse wave is one in which the disturbance is parallel to the direction of travel. A wave can have both transverse and longitudinal components. A wave carries energy from one place to another. A wave does not result in the bulk flow of the material of its medium. A wave is a traveling disturbance.
Answer:
A transverse wave is one in which the disturbance is parallel to the direction of travel
Explanation:
There are two types of waves:
- Transverse waves: in transverse waves, the disturbance (oscillation) occurs in a plane perpendicular to the direction of propagation of the wave
- Longitudinal waves: in longitudinal waves, the disturbance (oscillation) occurs parallel to the direction of propagation of the wave
Therefore, we immediately see that the statement:
"A transverse wave is one in which the disturbance is parallel to the direction of travel"
is wrong, because it is actually the opposite: in a transverse wave, the disturbance is perpendicular to the direction of travel.
The false statement is that a transverse wave has a disturbance parallel to its direction of travel. A transverse wave has the disturbance perpendicular to its direction of travel, not parallel. An example of this would be the waves on a stringed instrument, contrasted with sound waves which are longitudinal (disturbance and wave travel in the same direction).
Explanation:The false statement among the given options is: "A transverse wave is one in which the disturbance is parallel to the direction of travel." This is incorrect because in a transverse wave, the disturbance or oscillation is perpendicular to the direction of wave propagation. For example, the movement of a stringed instrument creates transverse waves, where the strings vibrate upwards and downwards while the wave travels horizontally along the string.
On the other hand, in a longitudinal wave, the disturbance is parallel to the direction of wave propagation. An example would be sound waves, where the air particles vibrate back and forth, in the same direction that the wave propagates.
It's also correct that waves can have both transverse and longitudinal components, such as seismic waves from earthquakes. Furthermore, a wave does carry energy from one location to another but does not result in the bulk movement of the material medium.
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can someone help me?!!
The answer is:
The first option, the force tending to lift Rover is equal to 14.5 N.
Why?To calculate the force that is tending to lift Rover vertically, we need to calculate the vertical component force.
Since we know that the angle between the force and the ground is 29°, we can calculate the vertical component of the force using the following formula:
[tex]F_y=Force*Sin(29\°)[/tex]
We are given that the force is equal to 30.0 N, so, calculating we have:
[tex]F_y=Force*Sin(29\°)[/tex]
[tex]F_y=30N*Sin(29\°)=14.5N[/tex]
Also, we can calculate the horizontal component of the force using the following formula:
[tex]F_x=Force*Cos(29\°)[/tex]
[tex]F_x=30N*Cos(29\°)=26.24N[/tex]
Hence, we have that the correct option is the first option, the force tending to lift Rover is equal to 14.5 N.
Have a nice day!
Amy is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward; and Carlos is driving his car at 15 m/s westward. How fast does Bill see Carlos moving and in what direction?
Answer:
20 m/s westward
Explanation:
Taking eastward as positive direction, we have:
[tex]v_B = +5 m/s[/tex] is the velocity of Bill with respect to Amy (which is stationary)
[tex]v_C = -15 m/s[/tex] is the velocity of Carlos with respect to Amy
Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is
[tex]v_B = +5 m/s[/tex]
Therefore, Carlos velocity in Bill's reference frame will be
[tex]v_C' = v_C - v_B = -15 m/s - (+5 m/s) = -20 m/s[/tex]
and the direction will be westward (negative sign).
A hard-boiled egg of mass 48.0 g moves on the end of a spring with force constant 25.0 N/m . The egg is released from rest at an initial displacement of 0.305 m . A damping force Fx=−bvx acts on the egg, and the amplitude of the motion decreases to 0.113 m in a time of 4.70 s . Calculate the magnitude of the damping constant b.
b = kg/s
Answer:
they did it
Explanation:
If a 4Ω resistor, an 8Ω resistor, and a 12Ω resistor are connected in series, which resistor has the most current in it?
4
8
12
They all have the same current.
Answer: Last option
They all have the same current.
Explanation:
A connection of three elements in series is represented as follows:
--------[4Ω]-------[8Ω]--------[12Ω]----------
→ I
Note that the three elements share the same current line I .
By definition when the resistors or other electrical components are connected in series then the same current passes through them. Therefore in this case the magnitude of the resistance does not influence the magnitude of the current.
The answer is the last option
can someone help me?!!!!!
Answer:
7 units, north
Explanation:
This is a problem of vector subtraction. We have:
- Vector A: magnitude 2 units, direction to the north
- Vector B: magnitude 5 units, direction to the south
If we take the north as positive direction, we can write
[tex]A = +2\\B=-5[/tex]
Since we want to find [tex]A-B[/tex] (vector subtraction), we have to change the sign of B, so we find:
[tex]A-B=+2-(-5))=+2+5=+7[/tex]
And the positive sign means the direction is north.
can someone help me?!!!!!
The answer is:
The third option, the approximate magnitude of the given vector is 9.9 units.
[tex]|A|=9.9units[/tex]
Why?To calculate the magnitude (length) of a vector, we need to apply the following formula:
[tex]|A|=\sqrt{(A_x)^{2} +(A_y)^{2} }[/tex]
So, we are given the vector:
[tex]A=(7.6,-6.4)[/tex]
Then, substituting and calculating the magnitude of the vector, we have:
[tex]|A|=\sqrt{(7.6)^{2} +(-6.4)^{2}}=\sqrt{57.76+40.96}=\sqrt{98.72}\\\\|A|=\sqrt{98.72}=9.9units[/tex]
Hence, we have that the correct option is the third option, the approximate magnitude of the given vector is 9.9 units.
[tex]|A|=9.9units[/tex]
Have a nice day!
can someone help me?!!!!!
Answer:
Explanation:
The first one is false. A vector has a magnitude and a direction; a scalar only has a magnitude. The two cannot be added together.
The second one is false. The magnitude of a vector is found using Pythagorean theorem: c² = a² + b². The only way the magnitude of a vector (c) can be 0 is if both components are 0 (a=0 and b=0).
The third one is true. A vector in Quadrant III will have negative components but can still have a positive magnitude. For example, a vector with magnitude 1 and direction 225° has a positive magnitude and negative components.
The fourth one is false. Rotating a vector will change it.
The fifth one is false. A vector sum can only be 0 if the two vectors have equal magnitudes and opposite directions.
Nail tips exert tremendous pressures when they are hit by hammers because they exert a large force over a small area. What force (in N) must be exerted on a nail with a circular tip of 1.15 mm diameter to create a pressure of 2.63 ✕ 109 N/m2? (This high pressure is possible because the hammer striking the nail is brought to rest in such a short distance.)
Answer:
2780 N
Explanation:
Pressure is defined as the ratio between the force applied and the area of the surface:
[tex]p=\frac{F}{A}[/tex]
Here we know the pressure:
[tex]p=2.63 \cdot 10^9 N/m^2[/tex]
we also know the diameter of the tip, d = 1.15 mm, so we can calculate the radius
[tex]r=\frac{1.15 mm}{2}=0.58 mm = 5.8\cdot 10^{-4} m[/tex]
and so the area
[tex]A=\pi r^2 = \pi (5.8\cdot 10^{-4} m)^2=1.057\cdot 10^{-6} m^2[/tex]
And so we can re-arrange the equation to find the force:
[tex]F=pA=(2.63\cdot 10^9 N/m^2)(1.057\cdot 10^{-6} m^2)=2780 N[/tex]
We can calculate the necessary force to drive the nail home by first calculating the area of the nail tip using its given diameter then using the given pressure and the formula for pressure (P = F/A) to solve for the force.
Explanation:To calculate the force required to drive the nail we need to use the equation for pressure which is pressure = force/area. Pressure (P) is given as 2.63 x 109 N/m2. The area (A) can be calculated using the formula for the area of a circle which is A = πr2. The radius (r) of the nail tip can be calculated from its diameter (1.15 mm divided by 2) making sure to convert the units to meters.
After you derive the area, you will use it to calculate the force (F). Rearranging the formula for pressure to solve for force gives F = P * A. This will give the force necessary to drive the nail home exerting the stated pressure.
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A mass suspended from a spring is oscillating up and down as indicated. Consider the following possibilities. A At some point during the oscillation the mass has zero velocity but its acceleration is non-zero (can be either positive or negative); B At some point during the oscillation the mass has zero velocity and zero acceleration; C At some point during the oscillation the mass has non-zero velocity (can be either positive or negative) but has zero acceleration; D At all points during the oscillation the mass has non-zero velocity and has nonzero acceleration (either can be positive or negative). Which possibility or possibilities occur?
A mass suspended from a spring is oscillating up and down, (as stated but not indicated).
A). At some point during the oscillation the mass has zero velocity but its acceleration is non-zero (can be either positive or negative). Yes. This statement is true at the top and bottom ends of the motion.
B). At some point during the oscillation the mass has zero velocity and zero acceleration. No. If the mass is bouncing, this is never true. It only happens if the mass is hanging motionless on the spring.
C). At some point during the oscillation the mass has non-zero velocity (can be either positive or negative) but has zero acceleration. Yes. This is true as the bouncing mass passes through the "zero point" ... the point where the upward force of the stretched spring is equal to the weight of the mass. At that instant, the vertical forces on the mass are balanced, and the net vertical force is zero ... so there's no acceleration at that instant, because (as Newton informed us), A = F/m .
D). At all points during the oscillation the mass has non-zero velocity and has nonzero acceleration (either can be positive or negative). No. This can only happen if the mass is hanging lifeless from the spring. If it's bouncing, then It has zero velocity at the top and bottom extremes ... where acceleration is maximum ... and maximum velocity at the center of the swing ... where acceleration is zero.
In a bouncing spring, the mass will have zero velocity and non-zero acceleration at the top and bottom end of motion.
What is an Acceleration?It is defined as the rate of change in velocity or change speed/direction of the object.
An object can have zero velocity but non-zero acceleration. It can be understood by 2 examples,
In a horizontal direction, if a forward-moving object is forced to accelerate in the opposite direction then it will slow down and suddenly stop at a point and start moving in the opposite direction. In a vertical direction, an upward-moving object stops where its velocity became zero but at the same time, it experiences gravitational acceleration in a downward (opposite direction) and instantly starts falling down.Therefore, in a bouncing spring, the mass will have zero velocity and non-zero acceleration at the top and bottom end of motion.
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A power lifter performs a dead lift, raising a barbell with a mass of 305 kg to a height of 0.42 m above the ground, giving the barbell 1256.66 J of potential energy. The power lifter then releases the barbell, letting it drop towards the ground. Determine the magnitude of the vertical velocity of the barbell when it reaches a height of 0.21 m (on the way down) using a mechanical energy approach. Then calculate the velocity right as it reaches the ground using a mechanical energy approach. Confirm your answer for the velocity of the barbell right before it hits the ground by also calculating this velocity using a projectile motion approach.
Answer:
Explanation:
Before it hits the ground:
The initial potential energy = the final potential energy + the kinetic energy
mgH = mgh + 1/2 mv²
gH = gh + 1/2 v²
v = √(2g (H - h))
v = √(2 * 9.81 m/s² * (0.42 m - 0.21 m))
v ≈ 2.0 m/s
When it hits the ground:
Initial potential energy = final kinetic energy
mgH = 1/2 mv²
v = √(2gH)
v = √(2 * 9.81 m/s² * 0.42 m)
v ≈ 2.9 m/s
Using a kinematic equation to check our answer:
v² = v₀² + 2a(x - x₀)
v² = (0 m/s)² + 2(9.8 m/s²)(0.42 m)
v ≈ 2.9 m/s
In conventional television, signals are broadcast from towers to home receivers. Even when a receiver is not in direct view of a tower because of a hill or building, it can still intercept a signal if the signal diffracts enough around the obstacle, into the obstacle’s “shadow region.” Current television signals have a wavelength of about 54 cm, but future digital television signals that are to be transmitted from towers will have a wavelength of about 13 mm. (a) Did this change in wavelength increase or decrease the diffraction of the signals into the shadow regions of obstacles? Assume that a signal passes through an opening of 5.7 m width between two adjacent buildings. What is the angular spread of the central diffraction maximum (out to the first minima) for wavelengths of (b) 54 cm and (c) 13 mm?
(a) The diffraction decreases
The formula for the diffraction pattern from a single slit is given by:
[tex]sin \theta = \frac{n \lambda}{a}[/tex]
where
[tex]\theta[/tex] is the angle corresponding to nth-minimum in the diffraction pattern, measured from the centre of the pattern
n is the order of the minimum
[tex]\lambda[/tex] is the wavelength
a is the width of the opening
As we see from the formula, the longer the wavelength, the larger the diffraction pattern (because [tex]\theta[/tex] increases). In this problem, since the wavelength of the signal has been decreased from 54 cm to 13 mm, the diffraction of the signal has decreased.
(b) [tex]10.8^{\circ}[/tex]
The angular spread of the central diffraction maximum is equal to twice the distance between the centre of the pattern and the first minimum, with n=1. Therefore:
[tex]sin \theta = \frac{(1) \lambda}{a}[/tex]
in this case we have
[tex]\lambda=54 cm = 0.54 m[/tex] is the wavelength
[tex]a=5.7 m[/tex] is the width of the opening
Solving the equation, we find
[tex]\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.54 m}{5.7 m})=5.4^{\circ}[/tex]
So the angular spread of the central diffraction maximum is twice this angle:
[tex]\theta = 2 \cdot 5.4^{\circ}=10.8^{\circ}[/tex]
(c) [tex]0.26^{\circ}[/tex]
Here we can apply the same formula used before, but this time the wavelength of the signal is
[tex]\lambda=13 mm=0.013 m[/tex]
so the angle corresponding to the first minimum is
[tex]\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.013 m}{5.7 m})=0.13^{\circ}[/tex]
So the angular spread of the central diffraction maximum is twice this angle:
[tex]\theta = 2 \cdot 0.13^{\circ}=0.26^{\circ}[/tex]
The change from a 54 cm wavelength to a 13 mm wavelength will decrease the diffraction of signals. The angular spread for a 54 cm wavelength is 0.1154 degrees and for a 13 mm wavelength, it is 0.00279 degrees.
Explanation:The shift from conventional television signals with a wavelength of about 54 cm to future digital television signals with a wavelength of about 13 mm will decrease the diffraction of the signals into shadow regions of obstacles. This is because diffraction is inversely proportional to wavelength.
The angular spread of the central diffraction maximum can be calculated using diffraction spreading for a single slit given by the equation θ = 1.22 λ / D, where θ is the angle, λ is the wavelength, and D is the slit width. For a wavelength (λ) of 54 cm and a slit width (D) of 5.7 m, the angular spread θ is:
θ = 1.22 * (0.54 m) / 5.7 m = 0.1154 °
For a wavelength of 13 mm, the angular spread θ is:
θ = 1.22 * (0.013 m) / 5.7 m = 0.00279 °