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3 years ago

What net force would be required for a 50.5 kg wolf to accelerate at 5.0 m/s^2

Physics

1 answer:

Basile [38]3 years ago

4 0

Answer:

$\boxed {\boxed {\sf 252.5 \ Newtons }}$

Explanation:

Force is the push or pull on object that can cause different things, like acceleration. According to Newton's 2nd Law of Motion, it is the product of mass and acceleration.

$F=m*a$

The mass of the wolf is 50.5 kilograms and the acceleration is 5.0 meters per square second. Therefore:

$m= 50.5 \ kg$

$a= 5.0 \ m/s^2$

Substitute the values into the formula.

$F= 50.5 \ kg * 5.0 \ m/s^2$

Multiply.

$F=252.5 \ kg*m/s^2$

1 kilogram meter per square second is equal to 1 Newton.
Our answer of 252.5 kg*m/s² is equal to 252.5 Newtons.

$F= 252.5 \ N$

The force required is <u>252.5 Newtons</u>.

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The first part of the text is missing, you can find on google:

"A ball is thrown horizontally from the roof of a building 45 m. If it strikes the ground 56 m away, find the following values."

Let's now solve the different parts.

(a) 3.03 s

The time of flight can be found by analyzing the vertical motion only. The vertical displacement at time t is given by

$y(t) = h -\frac{1}{2}gt^2$

where

h = 45 m is the initial height

g = 9.8 m/s^2 is the acceleration of gravity

When y=0, the ball reaches the ground, so the time taken for this to happen can be found by substituting y=0 and solving for the time:

$0=h-\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(45)}{9.8}}=3.03 s$

(b) 18.5 m/s

For this part, we need to analyze the horizontal motion only, which is a uniform motion at constant speed.

The horizontal position is given by

$x=v_x t$

where

$v_x$ is the horizontal speed, which is constant

t is the time

At t = 3.03 s (time of flight), we know that the horizontal position is x = 56 m. By substituting these numbers and solving for vx, we find the horizontal speed:

$v_x = \frac{x}{t}=\frac{56}{3.03}=18.5 m/s$

The ball was thrown horizontally: this means that its initial vertical speed was zero, so 18.5 m/s was also its initial overall speed.

(c) 35.0 m/s at 58.1 degrees below the horizontal

At the impact, we know that the horizontal speed is still the same:

$v_x = 18.5 m/s$

we need to find the vertical velocity. This can be done by using the equation

$v_y = u_y -gt$

where

$u_y =0$ is the initial vertical velocity

g is the acceleration of gravity

t is the time

Substituting t = 3.03 s, we find the vertical velocity at the time of impact:

$v_y = -(9.8)(3.03)=-29.7 m/s$

So the magnitude of the velocity at the impact (so, the speed at the impact) is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.5^2+(-29.7)^2}=35.0 m/s$

The angle instead can be found as:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-29.7}{18.5})=-58.1^{\circ}$

so, 58.1 degrees below the horizontal.

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Answer:

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His potential energy to his point would be:

$E_{\phi} = mgh = 10ML(1 + sin(\phi))$

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