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2 years ago

A small car is pushing a large truck. They are speeding up.

Physics

1 answer:

Gnom [1K]2 years ago

3 0

The force of the truck on the car is equal to the force of the car on the truck.

<h3>What is Newton's third law of motion?</h3>

Newton's third law of motion states that action and reaction are equal and opposite.

Let the mass of the small car = m

Let the mass of the truck = M

The force exerted by the small car is calculated as;

F₁ = ma

The force exerted by the large truck is calculated as;

F₂ = Ma

According to Newton's third law, the magnitude of the two forces are equal but opposite in direction.

|F₁| = |F₂|

Thus, we can conclude that, the force of the truck on the car is equal to the force of the car on the truck.

Learn more about Newton's third law of motion here: brainly.com/question/25998091

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3 years ago

A dockworker applies a constant horizontal force of 90.0 N to a block of ice on a smooth horizontal floor. The frictional force

Oliga [24]

Answer:

The mass of the ice block is equal to 70.15 kg

Explanation:

The data for this exercise are as follows:

F=90 N

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x=13 m

t=4.5 s

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applying the equation of rectilinear motion we have:

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x = at^2/2

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4 years ago

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An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference

kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

$K=qV$ (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

$K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J$

The kinetic energy of the particle is also:

$K=\frac{1}{2}mv^2$ (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}$

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

$|F_B|=qvBsin(\theta)$

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

$|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N$

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

$r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm$

the radius of the trajectory of the electron is 10.1 cm

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3 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

3 years ago