Answer: Third option
More than w and equal to F
Explanation:
If the horse presses the floor with a force F, then it follows that by Newton's third law, the floor makes a force -F on the horse.
Then if this force F is greater than the weight w of the horse, then the horse accelerates upwards and rises from the ground.
We know that after pressing the floor with a force F, the horse gets up off the floor. Then it is true that:
The force exerted by the floor on the horse is equal to F and greater than w
The answer is the third option more than w and equal to F
To become airborne, the force exerted by the ground on the horse must be greater than both the horse's weight and the force applied by the horse.
When a horse becomes airborne after pressing down on the ground, the magnitude of the force that the ground exerted on the horse must have been more than w (weight of the horse) and more than F (force applied by the horse on the ground). This is because for the horse to become airborne, the ground must exert a force greater than the horse's weight, which is the normal reaction force, to overcome gravitational pull. Additionally, the applied force F would be a combination of the horse's weight (mg) and the additional force caused by the horse's muscles to initiate the jump.
Tina is driving her sports car down US1, traveling 27m/s. She sees her friend Rita up ahead, waiting to cross the street at the next intersection. Tina quickly slams down on the horn. If the horn has a frequency of 400Hz, which of the following is the only possible frequency heard by Rita, as Tina drives toward Rita?400Hz, 434Hz or 371Hz
Answer:
434 Hz
Explanation:
According to the Doppler effect, when a source of a wave is moving towards an observer at rest, then the observer will observe an apparent frequency which is higher than the original frequency of the source.
In this situation, Tina is driving towards Rita. Tina is the source of the sound wave (the horn), while RIta is the observer. Since the original frequency of the sound is 400 Hz, Rita will hear a sound with a frequency higher than this value.
The only choice which is higher than 400 Hz is 434 Hz, so this is the frequency that Rita will hear.
Answer:b
Explanation:test
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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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]
An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 9.0 μC flows to the positive plate. The 4.0 V battery is then disconnected and replaced with a 5.0 V battery, with the positive and negative terminals connected in the same manner as before. Part A How much additional charge flows to the positive plate? Express your answer in microcoulombs.
Answer:
[tex]2.25\mu C[/tex]
Explanation:
At the beginning, we have:
V = 4.0 V potential difference across the capacitor
[tex]Q=9.0 \mu C=9.0\cdot 10^{-6}C[/tex] charge stored on the capacitor
Therefore, we can calculate the capacitance of the capacitor:
[tex]C=\frac{Q}{V}=\frac{9.0 \cdot 10^{-6} C}{4.0 V}=2.25\cdot 10^{-6} F[/tex]
Later, the battery is replaced with another battery whose voltage is
V = 5.0 V
Since the capacitance of the capacitor does not change, we can calculate the new charge stored:
[tex]Q=CV=(2.25\cdot 10^{-6} F)(5.0 V)=11.25 \cdot 10^{-6} C=11.25 \mu C[/tex]
Since the capacitor has been connected exactly as before, we have that the charge on the positive plate has increased from [tex]9.0 \mu C[/tex] to [tex]11.25 \mu C[/tex]. Therefore, the additional charge that moved to the positive plate is
[tex]\Delta Q = 11.25 \mu C-9.0 \mu C=2.25 \mu C[/tex]
In a uncharged capacitor when connected to 5.0 V battery, the additional charge flows to the positive plate, is [tex]2.25\rm \mu C[/tex] .
What is capacitance of capacitor?The capacitance of capacitor is the ratio of the electric charge stored inside the capacitance to the potential difference. The capacitance of capacitor can be given as,
[tex]C=\dfrac{Q}{V}[/tex]
Here, (Q) is the electric charge and (V) is the potential difference.
Given information-
The potential difference of the first battery is 4.0 V.
The charge of the first battery is 9.0 μC.
The potential difference of the second battery is 5.0 V.
The capacitance of the first battery is,
[tex]C=\dfrac{9.0\times10^{-6}}{4}\\C=2.25\times10^{-6} \rm F[/tex]
Let the charge of the second battery is (q). Thus The capacitance of the first battery is,
[tex]C=\dfrac{q}{5}\\[/tex]
As the capacitance of the capacitor remain same. Thus put the value of C in the above equation as,
[tex]2.25\times10^{-6}=\dfrac{q}{5}\\q=11.25\rm \mu C[/tex]
The additional charge flows to the positive plate is the difference of the charge flows to the positive plate and second battery to the first battery. Thus,
[tex]\Delta q=11.25-9\\\Delta q=2.25\rm \mu C[/tex]
Thus the additional charge flows to the positive plate is [tex]2.25\rm \mu C[/tex] .
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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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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.
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.
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!
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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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:
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?!!!!!
Answer:
33 m/s
Explanation:
By analyzing the vertical motion, we can find what is the time of flight of the projectile. The vertical position is
[tex]y(t) = h + v_{0y}t - \frac{1}{2}gt^2[/tex]
where
h = 20 m is the initial height
[tex]v_{0y} = 26 m/s[/tex] is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
By putting y(t)=0, we find the time t at which the projectile hits the ground:
[tex]0=20 + 26 t - 4.9t^2[/tex]
which has 2 solutions:
t = -0.7 s
t = 6.0 s
We discard the negative solution since it has no physical meaning. So, we know that the projectile hits the ground 6.0 s later after the launch.
The vertical velocity is given by
[tex]v_y (t)= v_{0y} -gt[/tex]
So we can find the vertical velocity when the projectile reaches point Q, by substituting t=6.0 s into this equation:
[tex]v_y = 26 m/s - (9.8 m/s^2)(6.0 s)=-32.8 m/s \sim -33 m/s[/tex]
and the negative sign means the direction is downward.
can someone help me?!!!!!
Answer:
Parabola
Explanation:
The motion of a projectile consists of two independent motions:
- A uniform motion along the horizontal (x) direction, with constant velocity [tex]v_x[/tex]. In fact, there are no forces acting along this direction (if we neglect air resistance), so the acceleration is zero and the velocity is constant
- An uniformly accelerated motion along the vertical (y) direction, with constant acceleration [tex]g=9.8 m/s^2[/tex] downward (acceleration due to gravity). This acceleration is due to the force of gravity that pulls the projectile downward.
The composition of these two motions gives a parabolic trajectory. In fact, the equations of the motion along the two directions are:
[tex]x(t) = v_x t[/tex] (1)
[tex]y(t) = h + v_y t - \frac{1}{2}gt^2[/tex] (2)
Solving for t in eq.(1),
[tex]t=\frac{x}{v_x}[/tex]
and substituting into (2)
[tex]y(t) = h + \frac{v_y}{v_x}x - \frac{1}{2}g(\frac{x}{v_x})^2[/tex]
which is the equation of a parabola.
How much heat is needed to raise the temperature of 8g of water by 15oC?
Explanation:
The heat (thermal energy) needed in to raise the temperature in a process can be found using the following equation:
[tex]Q=m.C.\Delta T[/tex] (1)
Where:
[tex]Q[/tex] is the heat
[tex]m=8 g[/tex] is the mass of the element (water in this case)
[tex]C[/tex] is the specific heat capacity of the material. In the case of water is [tex]C=4.184\frac{J}{g\°C}[/tex]
[tex]\Delta T=15\°C[/tex] is the variation in temperature (which is increased in this case)
Knowing this, let's rewrite (1) with these values:
[tex]Q=(8 g)(4.184\frac{J}{g\°C})(15\°C)[/tex] (2)
Finally:
[tex]Q=502.08 J[/tex]
A spring has a natural length of 14 cm. If a 23-N force is required to keep it stretched to a length of 20 cm, how much work W is required to stretch it from 14 cm to 17 cm? (Round your answer to two decimal places.)
Answer:
1.78 J
Explanation:
Find the spring coefficient using Hooke's law:
F = k Δx
23 N = k (0.20 m − 0.14 m)
k = 383.33 N/m
The work is the change in energy:
W = PE₂ − PE₁
W = ½ kx₂² − ½ kx₁²
W = ½ k (x₂² − x₁²)
W = ½ (383.33 N/m) ((0.17 m)² − (0.14 m)²)
W = 1.78 J
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
In which situation is the acceleration of the car negative?
A) The velocity of a car was 75 km/h over 4 hours.
B) The velocity of a car reduced from 50 km/h to 35 km/h over one minute.
C) The velocity of a car increased from 40 km/h to 75 km/h over 15 minutes.
Eliminate
D) The velocity of a car was 45 km/h at 2:00 pm and at 4:00 pm the velocity of the car was 85 km/h.
B) The velocity of the car reduced from 50k/m over one min.
Answer:
Answer b
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:
110 m
Explanation:
First of all, let's find the initial horizontal and vertical velocity of the projectile:
[tex]v_{x0}=v cos 30^{\circ}=(25 m/s)(cos 30^{\circ})=21.7 m/s[/tex]
[tex]v_{y0}=v sin 30^{\circ}=(25 m/s)(sin 30^{\circ})=12.5 m/s[/tex]
Now in order to find the time it takes for the projectile to reach the ground, we use the equation for the vertical position:
[tex]y(t)=h+v_{0y}t-\frac{1}{2}gt^2[/tex]
where
h = 65 m is the initial height
t is the time
g = 9.8 m/s^2 is the acceleration due to gravity
The time t at which the projectile reaches the ground is the time t at which y(t)=0, so we have:
[tex]0=65+12.5 t - 4.9t^2[/tex]
which has 2 solutions:
t = -2.58 s
t = 5.13 s
We discard the 1st solution since its negative: so the projectile reaches the ground after t=5.13 s.
Now we know that the projectile travels horizontally with constant speed
[tex]v_x = 21.7 m/s[/tex]
So, the horizontal distance covered (x) is
[tex]x=v_x t = (21.7 m/s)(5.13 s)=111.3 m[/tex]
So the closest option is
110 m
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]
The Sun has a lifetime of approximately 10 billion years. If you could determine the rate of nuclear fusion for a star with twice the mass of the Sun, which of the following would best describe how its fusion rate would compare to the Sun? Choose the best possible response to complete the sentence given below. A star with twice the mass of the Sun would have a rate of nuclear fusion that is ________ the rate of fusion in the Sun. a) less than b) a little more than c) twice d) more than twice Explain your reasoning for the choice you made.
Final answer:
A star with twice the mass of the Sun would have a nuclear fusion rate that is significantly more than twice the rate of the Sun. This occurs because the fusion rate is highly sensitive to core temperature and density, which increases more than linearly with mass. Additionally, the lifetime of such a star is inversely proportional to the square of its mass, resulting in a much shorter lifespan.
Explanation:
A star with twice the mass of the Sun would have a rate of nuclear fusion that is more than twice the rate of fusion in the Sun. This is because the rate of nuclear fusion increases with both the mass and the core temperature of the star. Since a star's core temperature and its density increase with greater mass, the rate of nuclear fusion increases exponentially, not linearly. The relationship between the temperature and the nuclear reaction rate is to the power of four. Therefore, a star with twice the mass of the Sun would have a considerably higher core temperature, leading to a significantly increased reaction rate, well over twice that of the Sun's.
To put this into perspective using the proportional relationship of mass (M), luminosity (L), and lifetime (T) of a star; for a star with twice the mass of the Sun, we use the equation T = 10¹⁰ y where M is the mass relative to the Sun, and L is the luminosity relative to the Sun. Considering that the luminosity increases dramatically with mass, the rate of fuel consumption would be much higher. Hence, while a massive star has more fuel, its higher fusion rate due to increased core temperature leads to a much shorter lifetime than less massive stars like our Sun.
Furthermore, we understand that the lifetime of a star is inversely proportional to its mass squared, as demonstrated by the worked example where the lifetime of a star with twice the Sun's mass is a quarter of the Sun's lifetime.
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.
Learn more about Oscillation:
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The highest frequency that a healthy ear can typically hear is 2.0 × 104 Hz. Assume that a sound wave with this frequency travels at 346 m/s and passes through a doorway that has a width of 0.94 m. (a) Determine the angle that locates the first minimum to either side of the central maximum in the diffraction pattern for the sound. (b) Suppose that yellow light (wavelength = 569 nm, in vacuum) passes through a doorway and that the first dark fringe in its diffraction pattern is located at the angle determined in part (a). How wide would this hypothetical doorway have to be?
The first part involves calculating the angle for the first minimum in the diffraction pattern of a sound wave and we get approximately 1.05°. For the second part, we calculate that the width of a hypothetical doorway for yellow light would need to be around 30.9 micrometers to match this diffraction angle.
To answer the student’s question, we need to determine the angle for the first minimum in the diffraction pattern for sound waves passing through a doorway. The formula for the angle of the first minimum is given by:
sin(θ) = λ / a
where λ is the wavelength and 'a' is the width of the doorway.
Part (a)
Calculate the wavelength (λ) of the sound using the formula: v = fλ:Part (b)
When yellow light (wavelength = 569 nm or 569 × 10-9 m) passes through a hypothetical doorway, and the first dark fringe is at the same angle:
Using the angle θ from part (a) which is approximately 1.05°, we calculate the width of the doorway 'a' for yellow light:sin(θ) = λ / aThus, the width of the hypothetical doorway for yellow light would need to be approximately 30.9 micrometers.
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!
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.
Learn more about combustion of methanol here:https://brainly.com/question/32198670
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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.
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.
Have a nice day!
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.
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.
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.