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3 years ago

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

Note: It has to be done with the Newton's 2nd Law of Motion.

1 answer:

ale4655 [162]3 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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Number of vibrations (waves) in an amount of time

m_a_m_a [10]

Answer:

I think frequency not sure though

How frequently a wave or vibration occurs during a span of time, determines the waves frequency. Frequency is the number of waves per unit time. The unit for frequency if a Hertz ( 1/second). The speed a wave travels is the wavelength multiplied by this frequency. The amplitude of a wave is the maximum distance the wave is displaced.

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2 years ago

In the ground state of hydrogen, according to the Bohr model, an electron orbits 5.3 x 10-11 m from the nucleus. It undergoes a

Readme [11.4K]

Answer:

Explanation:

Given

radius of electron $(r)=5.3\times 10^{-11} m$

centripetal acceleration $(a_c)=9\times 10^22 m/s^2$

we know

$a_c=\frac{v^2}{r}$

$v=\sqrt{r\times a_c}$

$v=\sqrt{5.3\times 10^{-11}\times 9\times 10^{22}}$

$v=\sqrt{47.7\times 10^11}$

$v=21.84\times 10^5 m/s$

(b)For n=10

$r=100\times 5.3\times 10^{-11} m\approx 5.3\times 10^{-9} m$

$a_c=10^4\times 9\times 10^{22} m/s^2$

$a_c=9\times 10^{26} m/s^2$

$v=\sqrt{r\times a_c}$

$v=\sqrt{9\times 10^{26}\times 5.3\times 10^{-9}}$

$v=21.84\times 10^8 m/s$

8 0

3 years ago

A batter swings at a baseball. The action force is the bat hitting the ball with a force of 5N. What is the reaction force?

omeli [17]

The ball hitting the bat

6 0

3 years ago

Read 2 more answers

Lucia kicks a ball on a level playing field with an initial velocity of 11.3 m/s at an angle of 35° above the horizontal. Find:

grandymaker [24]

Explanation:

Given that,

Initial velocity, u = 11.3 m/s

Angle above the horizontal, $\theta=35^{\circ}$

Time of flight :

$t=\dfrac{2u\sin\theta}{g}\\\\t=\dfrac{2\times 11.3\times \sin(35)}{9.8}\\\\t=1.32\ s$

Horizontal distance traveled is given by :

x = ut

x = 11.3 m/s × 1.32 s

x = 14.916 m

Maximum height is given by :

$H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(11.3)^2\times \sin^2(35)}{2\times 9.8}\\\\H=2.14\ m$

Hence, time of flight is 1.32 s, horizontal distance is 14.916 m and maximum height is 2.14 m.

6 0

2 years ago

The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.

irina [24]

Part a)

At t = 0 the position of the object is given as

$x = 0$

At t = 2

$x = 2 sin(\pi/2) = 2cm$

so displacement of the object is given as

$d = 2 - 0 = 2cm$

so average speed is given as

$v_{avg} = \frac{2}{2} = 1 cm/s$

Part b)

instantaneous speed is given by

$v = \frac{dy}{dt}$

$v = 2cos(\pi t/4 ) * \frac{\pi}{4}$

now at t= 0

$v = \frac{\pi}{2} cm/s$

at t = 1

$v = 2 cos(\pi/4) * \frac{\pi}{4}$

$v = \frac{\pi}{2\sqrt2}$

at t = 2

$v = 0$

Part c)

Average acceleration is given as

$a_{avg} = \frac{v_f - v_i}{t}$

$a_{avg} = \frac{0 - \frac{\pi}{2}}{2}$

$a = -\frac{\pi}{4} cm/s^2$

Part d)

Now for instantaneous acceleration

As we know that

$a =- \omega^2 y$

at t = 0

$a = -\frac{\pi^2}{16} * 0 = 0 cm/s^2$

at t = 1

$y = \sqrt2 cm$

now we have

$a = -\frac{\pi^2}{16}*\sqrt2$

At t = 2 we have

$y = 2 cm$

$a = -\frac{\pi^2}{16}*2$

$a = -\frac{\pi^2}{8}$

<em>so above is the instantaneous accelerations</em>

7 0

3 years ago