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10 days ago

6

A physics student stands on the rim of the canyon and drops a rock. The student measures the time for it to reach the bottom to

be 3.2 seconds. How deep is the canyon?

1 answer:

inna [2.2K]10 days ago

5 0

Answer:

The canyon measures 50.176 meters deep.

Explanation:

The student drops a rock from the rim of the canyon, requiring us to ascertain the depth of the canyon—name how far the ground is below the cliff.

The data we have:

Time = t = 3.2 s

Initial velocity = $v_{i}$ = 0 m/s

Gravitational acceleration = g = 9.8 m/s²

Height = h =?

According to the second equation of motion

$h = v_{i}t + \frac{1}{2}gt^{2}$

Given the initial velocity is zero, the right-hand side of the equation simplifies to zero

$h = \frac{1}{2}gt^{2}$

h = (0.5)(9.8)(3.2)²

h = 50.176 m

This calculation indicates that the rock dropped a distance of 50.176 meters to reach the canyon's base. Thus, the canyon depth is 50.176 meters.

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3.113 A heat pump is under consideration for heating a research station located on Antarctica ice shelf. The interior of the sta

ValentinkaMS [2425]

Answer:

a. β = 8.23 K

b. β = 28.815 K

Explanation:

The performance of the heat pump can be calculated using the formula

β = TH / (TH - TC)

a.

TH = 15 ° C + 273.15 K = 288.15 K

TC = - 20 ° C + 273.15 K = 253.15 K

β = 288.15 K / (288.15 K - 253.15 K)

β = 8.23 K

b.

TH = 15 ° C + 273.15 K = 288.15 K

TC = 5 ° C + 273.15 K = 278.15 K

β = 288.15 K / (288.15 K - 278.15 K)

β = 28.815 K

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14 days ago

9. A 2 liter bottle of Coke weighs about 2 kilograms. If that bottle were combined with an equal amount of anti-Coke, how many J

Maru [2360]

Answer:

The energy expected to be released is calculated to be 4182 Joules.

Explanation:

The total mass of coke is 2 kg, which is equivalent to 2000 g

1 calorie per gram corresponds to 4.184 Joules of energy

4.184 J/gC * 2000g results in 8368 J

1 food calorie approximates to 4186 J

By subtracting, we find 8368 - 4186

Hence, the total energy that will be released amounts to 4182 Joules.

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A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at

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Answer

Given:

Wavelength = λ = 18.7 cm

= 0.187 m

Amplitude, A = 2.34 cm

Velocity, v = 0.38 m/s

A) Calculate the angular frequency.

$f = \dfrac{v}{\lambda}$

$f = \dfrac{0.38}{0.187}$

$f =2.03\ Hz$

Angular frequency,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) Calculate the wave number:

$K = \dfrac{2\pi}{\lambda}$

$K= \dfrac{2\pi}{0.187}$

$K =33.59\ m^{-1}$

C)

Since the wave is traveling in the -x direction, the sign is positive between x and t

y (x, t) = A sin(k x - ω t)

y (x, t) = 2.34 sin(33.59 x - 12.75 t)

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29 days ago

A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of 25.60 rad/s2. During a 4.20-s time inte

Yuliya22 [2438]

We will use the equations of rotational kinematics,

$\theta =\theta _{0} + \omega_{0} t+ \frac{1}{2}\alpha t^2$ (A)

$\omega^2= \omega^2_{0} +2\alpha\theta$ (B)

Here, $\theta$ and $\theta _{0}$ denote the final and initial angular displacements, respectively, whereas $\omega$ and $\omega_{0}$ represent final and initial angular velocities, and $\alpha$ is the angular acceleration.

We are provided with $\alpha = - 25.60 \ rad/s^2, \theta = 62.4 \ rad$ and $t = 4.20 \ s$ .

By substituting these values into equation (A), we have

$62.4 \ rad = 0 + \omega_{0} 4.20 \ s + \frac{1}{2} (- 25.60 \ rad/s^2) ( 4.20)^2 \\\\ \omega_{0} = \frac{220.5+ 62.4 }{4.20} =67.4 \ rad/s$

Now, using equation (B),

$\omega^2=(67.4 \ rad/s)^2 + 2 (- 25.60 \ rad/s^2)62.4 \ rad \\\\\ \omega = 36.7 \ rad/s$

This indicates that the wheel's angular speed at the 4.20-second mark is 36.7 rad/s.

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29 days ago

Mosses don't spread by dispersing seeds; they disperse tiny spores. The spores are so small that they will stay aloft and move w

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Solution:

$Em_{f}$ / Em₀ = 0.30

Explanation:

In this problem, we apply the connection between mechanical energy, kinetic energy, and gravitational potential energy.

K = ½ m v²

U = mgh

We assess the mechanical energy at two positions:

Initial. Lower

Em₀ = K = ½ m v²

At its highest point

$Em_{f}$ = U = mg and

Now let's compute

Em₀ = ½ m 3.6²

Em₀ = m 6.48

$Em_{f}$ = m 9.8 × 0.2

$Em_{f}$ = m 1.96

Thus the energy lost is given by:

$Em_{f}$ / Em₀ = m 1.96 / m 6.48

$Em_{f}$ / Em₀ = 0.30

This means that 30% of the sun's energy is transformed into potential energy.

There are various conversion possibilities.

This energy changes into thermal energy affecting the spores and air, since it cannot be regained.

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