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2 years ago

A student pushes on a crate with a force of 100 N directed to the right. What force does the crate exert on the student?

Physics

1 answer:

sattari [20]2 years ago

6 0

Answer:

The Force exerted on the student by crate will be 100 N.

Explanation:

As per the given Question student is trying to push the crate in right direction with the force of 100N.

And we know that, Newton's third law states that every action has equal and opposite reaction.

So from the Newton's 3rd law of motion it is very clear the student must be experiencing the same amount of force which he is applying.

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During the first 14 minutes of a 1.0 hour trip, a car has an average speed 36 km/h. What must the average speed in km/h of the c

oksian1 [2.3K]

Answer:

85.5 km/h

Explanation:

$t_{1}$ = time interval for first phase = 14 min = $\frac{14}{60}$ h = 0.233 h

$t_{2}$ = time interval for second phase = 46 min = $\frac{46}{60}$ h = 0.767 h

$v$ = average speed for the entire trip = 74 km/h

$v_{1}$ = average speed in first phase = 36 km/h

$v_{2}$ = average speed in second phase

$d_{1}$ = distance traveled in first phase

$d_{2}$ = distance traveled in first phase

average speed is given as

$v = \frac{d_{1} + d_{2}}{t_{1} + t_{2}}$

$v = \frac{v_{1} t_{1} + v_{2} t_{2}}{t_{1} + t_{2}}$

$74 = \frac{(36) (0.233) + v_{2} (0.767)}{0.233 + 0.767}$

$74 = (36) (0.233) + v_{2} (0.767)$

$v_{2} = 85.5$ km/h

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3 years ago

What upward gravitational force does a 5600kg elephant exert on the earth?

Anastaziya [24]

Force is determined by multiplying mass and gravity (F= mg). To determine the answer, the mass of the elephant (5600 kg) is multiplied with the gravity (9.8 m/s²). The answer is 5800 N. This is the upward gravitational force that the elephant exerts on the earth.

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3 years ago

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Select all that apply.

Crank

Answer:B

Explanation:explanation

4 0

1 year ago

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A bird flies 3.6 km due west and then 1.8 km due north. Another bird flies 1.8 km due west and 3.6 km due north. What is the ang

kondor19780726 [428]

Define unit vectors as follows:
$\hat{i}$ is in the eastern direction.
$\hat{j}$ is in the northern direction.

The position of the first bird is
$\vec{a} = -3.6 \, \hat{i} + 1.8 \, \hat{j}$

The position of the second bird is
$\vec{b} = - 1.8 \, \hat{i} + 3.6 \, \hat{j}$

Let θ = the angle between the net displacement vector for the two birds.
By definition,
$\vec{a} . \vec{b} = |a| |b| cos\theta \\\\ \theta = cos^{-1} ( \frac{\vec{a}.\vec{b}}{|a||b|} )$

$\vec{a}.\vec{b} = (-3.6)(-1.8)+(3.6)(1.8) = 12.96$
$|a| = \sqrt{3.24+12.96} =4.025$
Similarly,
|b| = 4.025

Therefore
$\theta = cos^{-1} \frac{12.96}{4.025^{2}} =36.9^{o}$

Answer: 36.9°

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3 years ago

Be sure to answer all parts. Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 k

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Explanation:

Given that,

(a) Speed, $v=6.66\times 10^6\ m/s$