A rock is tossed straight up from the ground with a speed of 21 m/s . When it returns, it falls into a hole 10 m deep.a.) What is the rock's velocity as it hits the bottom of the hole?b.) How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

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.

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.

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

(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]

Answer:

Explanation:

6.7 amps

The total current of the circuit is 6.7 amps.

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.

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.

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

Final Answer:

The magnitude of the block's velocity just after impact is 0.008 times the initial velocity of the bullet. The initial speed of the bullet is 1.25 m/s.

Explanation:

To solve this problem, we can use the principle of conservation of momentum. The mass of the bullet is 8.00 g, or 0.008 kg, and the mass of the block is 0.992 kg. Let's denote the velocity of the block just after impact as V and the initial velocity of the bullet as u. The equation for conservation of momentum is:

m_bullet * u = (m_bullet + m_block) * V

Substituting the given values, we have:

0.008 kg * u = (0.008 kg + 0.992 kg) * V

0.008 u = 1 * V

Now, let's find the magnitude of the block's velocity just after impact:

V = 0.008 u

Given that the impact compresses the spring 15.0 cm, which requires a force of 0.750 N, we can use Hooke's Law to find the spring constant:

Force = spring constant * displacement

0.750 N = k * 0.150 m

k = 5 N/m

Now, to find the initial speed of the bullet, we can use the principle of conservation of mechanical energy. The energy stored in the spring when compressed 0.250 cm is given as 5.0 J. Let's denote the initial speed of the bullet as v_i. The equation for conservation of mechanical energy is:

0.5 * k * (0.250 m)^2 = 0.5 * m_bullet * (v_i)^2

Substituting the given values, we have:

0.5 * 5 N/m * (0.0025 m^2) = 0.5 * 0.008 kg * (v_i)^2

0.00625 J = 0.004 kg * (v_i)^2

(v_i)^2 = 0.00625 J / 0.004 kg

(v_i)^2 = 1.5625 m^2/s^2

v_i = 1.25 m/s

a) The final velocity of the block is approximately 0.0000808 * v_bullet.

b) The initial speed of the bullet is approximately 3.68 m/s.

a) Final Velocity of the Block (v_final):

Given data:

Mass of the bullet (m_bullet): 8.00 g (convert to kg: 0.008 kg)

Mass of the block (m_block): 0.992 kg

Compression of the spring (x): 15.0 cm (convert to meters: 0.15 m)

Initial velocity of the bullet (v_bullet): to be determined

Using the conservation of linear momentum:

m_bullet * v_bullet = (m_bullet + m_block) * v_final

0.008 kg * v_bullet = (0.008 kg + 0.992 kg) * v_final

Solving for v_final:

v_final = (0.008 kg * v_bullet) / (1 kg + 0.992 kg)

v_final ≈ 0.0000808 * v_bullet

b) Initial Speed of the Bullet (v_bullet):

Given data:

Force required to compress the spring (F): 0.750 N

Compression of the spring (x): 0.250 cm (convert to meters: 0.0025 m)

Spring constant (k): to be determined

The work done in compressing the spring is given by W = (1/2) * k * x^2, and this work is equal to the initial kinetic energy (KE_initial) of the bullet:

KE_initial = W = (1/2) * k * x^2

The kinetic energy is also related to the initial speed of the bullet:

KE_initial = (1/2) * m_bullet * v_bullet^2

Setting the two expressions for kinetic energy equal to each other:

(1/2) * k * x^2 = (1/2) * m_bullet * v_bullet^2

Solving for v_bullet:

v_bullet = sqrt((k * x^2) / m_bullet)

Now, using the given information that F = kx, where F is the force required to compress the spring:

k = F / x

Substitute this value of k back into the equation for v_bullet:

v_bullet = sqrt((F * x) / m_bullet)

Substitute the known values:

v_bullet = sqrt((0.750 N * 0.0025 m) / 0.008 kg)

v_bullet ≈ 3.68 m/s

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).

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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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.

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,

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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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).

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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Answer:

they did it

Explanation:

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.

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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(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.

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 °

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.

Answer:

3. Madison, WI

4. Quito, Equador

6. Main Sequence stars are becoming red giants

Explanation:

Answer:

1) The angle at which spider-man points his web shooter

2) The distance spider-man’s web shoots

3) Madison, WI

4) Quito, Equador

5) Upper left

6) Main sequence stars are becoming red giants.

Explanation:

1) The experimenters are choosing the angle at which to point the web shooter. This means the angle at which they shoot is independent of the other variables.

2) The distance the web shoots is not chosen by the experimenters and depends on the angle at which it is shot.

3) July is the 7th month and the graph for Madison, WI shows temperatures exceeding all other locations at this time.

4) Fluctuations refer to a changing of temperatures over seasons. Since Quito, Equador has a flat graph, this indicates no fluctuations.

5) This graph plots the hottest (blue) stars on the left and the highest luminosity (bright) stars at the top. Thus, bright blue stars are plotted in the top left.

6) At 5 billion years there are 0 giants and 100 main sequence stars. At 10 billions years there are 94 giants and 6 main sequence stars. This shows that the main sequence stars are becoming giants.

Answer: Last option

2.27 m/s2

Explanation:

As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.

If we call a_c to the centripetal acceleration then, by definition

[tex]a_c =w^2r = \frac{v^2}{r}[/tex]

in this case we know the speed of the runner

[tex]v =8.00\ m/s[/tex]

The radius "r" will be the distance from the runner to the center of the track

[tex]r = 28.2\ m[/tex]

[tex]a_c = \frac{8^2}{28.2}\ m/s^2[/tex]

[tex]a_c = 2.27\ m/s^2[/tex]

The answer is the last option

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 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.

The answer is:

The first option, the force tending to lift Rover is equal to 14.5 N.

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!

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]

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.

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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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.

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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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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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.

Answer:

56 m/s

Explanation:

The time we are considering is

t = 15 s

The vertical velocity of the projectile is given by

[tex]v_y(t) = v_{0y}-gt[/tex]

where

[tex]v_{0y}=100 m/s[/tex] is the initial vertical velocity

[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity

Substituting t=15 s, we find the vertical velocity of the projectile at that time:

[tex]v_y = 100 m/s - (9.8 m/s^2)(15 s)=-47 m/s[/tex]

where the negative sign means the direction is now downward.

The horizontal velocity does not change since there are no forces acting along that direction, so it remains constant:

[tex]v_x = 30 m/s[/tex]

So, the magnitude of the velocity at the moment of impact is

[tex]v=\sqrt{v_x^2 +v_y^2}=\sqrt{(30 m/s)^2+(-47 m/s)^2}=55.8 m/s \sim 56 m/s[/tex]

The answer is:

The third option, the approximate magnitude of the given vector is 9.9 units.

[tex]|A|=9.9units[/tex]

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!

Answer:

C

Explanation:

Objects in free-fall (also known as projectiles) follow a parabolic curve.  So the answer is C.

The answer is:

The time took for the rock to reach its maximum height is 0.110 seconds.

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!

(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]

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 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;

Learn more about the law of universal gravitation here: brainly.com/question/2347945