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BaLLatris [955]

3 years ago

13

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg

e toy rocket to the back of a sled and take the modified sled to a large, flat, snowy field. You ignite the rocket and observe that the sled accelerates from rest in the forward direction at a rate of 14.5 m/s2 for a time period of 3.10 s. After this time period, the rocket engine abruptly shuts off, and the sled subsequently undergoes a constant backward acceleration due to friction of 5.65 m/s2.
After the rocket turns off, how much time does it take for the sled to come to a stop?time: ____sBy the time the sled finally comes to a rest, how far has it traveled from its starting point?distance traveled: ____m

1 answer:

IRINA_888 [86]3 years ago

7 0

Answer:

Time : <u>7.96 s</u>

Distance Traveled : <u>357.8 m</u>

Explanation:

In order to solve this problem, we first consider the accelerated motion of rocket. We will be using the subscript 1 for accelerated motion.

So, for accelerated motion, we have:

Acceleration = a₁ = 14.5 m/s²

Time Period = t₁ = 3.1 s

Initial Velocity = Vi₁ = 0 m/s (Since, it starts from rest)

Final Velocity = Vf₁

Distance covered by sled during acceleration motion = s₁

Now, using 1st equation of motion:

Vf₁ = Vi₁ + (a₁)(t₁)

Vf₁ = 0 m/s + (14.5 m/s²)(3.1 s)

Vf₁ = 44.95 m/s

Now, using 2nd equation of motion:

s₁ = (Vi₁)(t) + (0.5)(a₁)(t₁)

s₁ = (0 m/s)(3.1 s) + (0.5)(14.5 m/s²)(3.1 s)

s₁ = 22.5 m

Now, we first consider the decelerated motion of rocket. We will be using the subscript 2 for decelerated motion.

So, for accelerated motion, we have:

Deceleration = a₂ = - 5.65 m/s²

Time Period = t₂ = ?

Initial Velocity = Vi₂ = Vf₁ = 44.95 m/s (Since, decelerate motion starts, where accelerated motion ends)

Final Velocity = Vf₂ = 0 m/s (Since, rocket will eventually stop)

Distance covered by sled during deceleration motion = s₂

Now, using 1st equation of motion:

Vf₂ = Vi₂ + (a₂)(t₂)

0 m/s = 44.95 m/s + (- 5.65 m/s²)(t₂)

t₂ = (44.95 m/s)/(5.65 m/s²)

<u>t₂ = 7.96 s</u>

Now, using 2nd equation of motion:

s₂ = (Vi₂)(t₂) + (0.5)(a₂)(t₂)

s₂ = (44.95 m/s)(7.96 s) + (0.5)(- 5.65 m/s²)(7.96 s)

s₂ = 357.8 m - 22.5 m

s₂ = 335.3 m

Thus, the total distance covered by sled will be:

Total Dustance = S = s₁ + s₂

S = 22.5 m + 335.3 m

<u>S = 357.8 m</u>

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3 years ago

A 1.0-kg block of aluminum is at a temperature of 50 Celsius. How much thermal energy will it lose when it’s temperature is redu

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Answer:

The lose of thermal energy is, Q = 22500 J

Explanation:

Given data,

The mass of aluminium block, m = 1.0 kg

The initial temperature of block, T = 50° C

The final temperature of the block, T' = 25° C

The change in temperature, ΔT = 50° C - 25° C

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7 0

3 years ago

Coherent light of frequency 6.37×1014 Hz passes through two thin slits and falls on a screen 88.0 cm away. You observe that the

IgorC [24]

Answer:

The distance between the two slits is 40.11 μm.

Explanation:

Given that,

Frequency $f= 6.37\times10^{14}\ Hz$

Distance of the screen l = 88.0 cm

Position of the third order y =3.10 cm

We need to calculate the wavelength

Using formula of wavelength

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

where, c = speed of light

f = frequency

Put the value into the formula

$\lambda=\dfrac{3\times10^{8}}{6.37\times10^{14}}$

$\lambda=471\ nm$

We need to calculate the distance between the two slits

$m\times \lambda=d\sin\theta$

$d =\dfrac{m\times\lambda}{\sin\theta}$

Where, m = number of fringe

d = distance between the two slits

Here, $\sin\theta =\dfrac{y}{l}$

Put the value into the formula

$d=\dfrac{3\times471\times10^{-9}\times88.0\times10^{-2}}{3.10\times10^{-2}}$

$d=40.11\times10^{-6}\ m$

$d = 40.11\ \mu m$

Hence, The distance between the two slits is 40.11 μm.

7 0

3 years ago

A model plane has a mass of 0.75 kg and is flying 12 m above the ground

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Answer:

Option C. 210 J.

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 0.75 Kg

Height (h) = 12 m

Velocity (v) = 18 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Total Mechanical energy (ME) =?

Next, we shall determine the potential energy of the plane. This can be obtained as follow:

Mass (m) = 0.75 Kg

Height (h) = 12 m

Acceleration due to gravity (g) = 9.8 m/s²

Potential energy (PE) =?

PE = mgh

PE = 0.75 × 9.8 × 12

PE = 88.2 J

Next, we shall determine the kinetic energy of the plane. This can be obtained as follow:

Mass (m) = 0.75 Kg

Velocity (v) = 18 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.75 × 18²

KE = ½ × 0.75 × 324

KE = 121.5 J

Finally, we shall determine the total mechanical energy of the plane. This can be obtained as follow:

Potential energy (PE) = 88.2 J

Kinetic energy (KE) = 121.5 J

Total Mechanical energy (ME) =?

ME = PE + KE

ME = 88.2 + 121.5

ME = 209.7 J

ME ≈ 210 J

Therefore, the total mechanical energy of the plane is 210 J.

8 0

2 years ago

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S_A_V [24]

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3 years ago

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