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Mazyrski [523]
3 years ago
15

I drop a penny from the top of the tower at the front of billy jones high school and it takes 30 seconds to hit the ground calcu

late the velocity in m/s after 25 seconds of free fall
Physics
1 answer:
Sedaia [141]3 years ago
3 0

Answer:

The velocity of the penny at 25th second is, v = 245 m/s

Explanation:

Given,

The time taken by the penny to hit the ground is, t = 30 s

The initial velocity of the penny, u = 0 m/s

The velocity of the penny at 25th second, v = ?

Using the II equation of motion, the displacement is given by the formula

                              <em> S = ut + ½ gt²</em>

                                     = 0 + ½ x 9.8 ( 25²)

                                      = 3062.5 m

Using the III equations of motion

                                <em>v² = u² + 2gs</em>

Substituting the given values,

                                v² = 0 + 2 x 9.8 x 3062.5

                                     = 60025

                                v = 245 m/s

Hence, the velocity of the penny at 25th second is, v = 245 m/s

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The quantities that have only magnitude are called vector quantities​.(T or F)
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Read 2 more answers
A ball is thrown with an initial velocity of 20 m/s at an angle of 60° above the horizontal. If we can neglect air resistance, (
Neporo4naja [7]

Answer:

10 m/s

1.87914 s

18.7914 m

Explanation:

v = Initial velocity = 20 m/s

\theta = Angle = 60°

Horizontal component is given by

v_x=20cos60\\\Rightarrow v_x=10\ m/s

The horizontal component is 10 m/s

y direction final displacement is zero

s=vsin \theta t+\frac{1}{2}at^2\\\Rightarrow 0=20sin60+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2\times 20sin 60}{9.81}}\\\Rightarrow t=1.87914\ s

The time the ball is in the air is 1.87914 s

Range is given is by

R=vcos\theta t\\\Rightarrow R=10\times 1.87914\\\Rightarrow R=18.7914\ m

The range is 18.7914 m

6 0
4 years ago
Jose pushes a cart with a force of 12 N to the right. Lucy also pushes the cart with a force of 5 N to the right. John pushes th
cricket20 [7]

So, <u>their net force on the cart is 10 N to the right</u>.

<h3>Introduction</h3>

Hi ! Here, I will help you with the net forces (results of forces) acting on a two-dimensional area and in opposite directions. Steps that can be taken are as follows :

  1. Determine where the force will go, the important thing is that you are consistent until the end.
  2. Count the values of the force acting, the force against the direction of your mind in number 1 is given a negative sign.
  3. Look at the results, if it's marked (-), then choose the opposite direction from your thoughts at number 1.

The equation for calculating the net force from this two-dimensional straight line is as follows:

\boxed{\sf{\bold{\sum F = F_1 + F_2 + ... + F_n}}}

With the following condition :

  • \sf{\sum F} = net force (N)
  • \sf{F_1} = first force and its direction (N)
  • \sf{F_2} = second force and its direction (N)
  • \sf{... + F_n} = You can add up the force values as many times as the question (N).

<h3>Problem Solving</h3>

We know that :

In my mind, I determined that the force will go to the right. So :

  • \sf{F_1} = Jose's force = 12 N >> Because he already walk to the right.
  • \sf{F_2} = Lucy's force = 5 N >> Because she already walk to the right.
  • \sf{F_3} = John's force = -7 N >> Because he is walked to the left.

What was asked :

  • \sf{\sum F} = net force = ... N

Step by step :

\sf{\sum F = F_1 + F_2 + F_3}

\sf{\sum F = 12 + 5 + (-7)}

\sf{\sum F = 17 - 7}

\boxed{\sf{\sum F = 10 \: N \: to \: the \: right}}

<h3>Conclusion</h3>

The movement of the cart is to the right because the net force value that I calculated is not opposite (with negative sign) to the right direction. So, their net force on the cart is 10 N to the right.

5 0
2 years ago
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