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natali 33 [55]
3 years ago
14

2. A student drives 7.8-km trip to school and averages a speed of

Physics
1 answer:
Alekssandra [29.7K]3 years ago
6 0

Answer:

<em>The total time is: t=451.22 sec</em>

<em>The average speed is: V=34.57 m/s</em>

Explanation:

<u>Average speed</u>

The average speed is calculated by dividing the total distance traveled by an object (x) by the total time it took it to travel that distance (t).

\displaystyle V=\frac{x}{t}

Since the student makes the trip in two parts, we have to calculate the total distance and the total time.

We know the distance to school is 7.8 Km = 7,800 m. The student makes his way home over the same distance, thus the total distance is

x=2*7,800 m=15,600 m

The first trip to school was done at an average speed of v1=32.6 m/s. Knowing the distance and speed, we can calculate the time:

\displaystyle t1=\frac{x1}{v1}=\frac{7,800}{32.6}=239.26\ sec

The second trip back home was done at an average speed of v2=36.8 m/s. Let's calculate the second time:

\displaystyle t2=\frac{x2}{v2}=\frac{7,800}{36.8}=211.96\ sec

The total time is:

t=239.26\ sec+211.96\ sec=451.22\ sec

\boxed{t=451.22\ sec}

The average speed is:

\displaystyle V=\frac{15,600}{451.22}=34.57\ m/s

\boxed{\displaystyle V=34.57\ m/s}

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Answer:

B. 450 feet

Explanation:

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3 years ago
If the torque required to loosen the nut that is holding a flat tire in a place on a car has a magnitude of 35 N•m what minimum
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Answer: F = 130 N

Explanation: Solution:

Convert first 27 cm to m.

27 cm  x 0.01 m / 1 cm = 0.27 m

Calculate the torque using T = Fd

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4 0
3 years ago
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Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude
DENIUS [597]

Answer:

The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

Explanation:

Given;

initial velocity of proton, v_p_i = 3 x 10⁵ m/s

distance moved by the proton, d = 3.5 m

electric field strength, E = 120 N/C

The kinetic energy of the proton at the end of the motion is calculated as follows.

Consider work-energy theorem;

W = ΔK.E

W =K.E_f - K.E_i

where;

K.Ef is the final kinetic energy

W is work done in moving the proton = F x d  = (EQ) x d = EQd

K.E_f =EQd + \frac{1}{2}m_pv_p_i^2

m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C

K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5   \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\

K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J

Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

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3 years ago
What is the distance time how can we find the speed of an object from its distance time graph​
CaHeK987 [17]

Answer:

speed is the gradient of the graph

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3 years ago
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Gravitational force is affected by: a. mass c. distance b. weight d. both a and c
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<span>Gravitational force is affected by: a. mass c. distance b. weight d. both a and c

Mass.

</span>When an object is above the Earth's surface it hasgravitational potential<span> energy (GPE). The amount of GPE an object has depends on its mass and its height above the Earth's surface</span><span>


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