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Mumz [18]
3 years ago
14

A rock is thrown horizontally off a cliff with an initial speed of 3 m/s. The initial position of the rock is 10 meters above th

e ground. How long is the rock in the air?
Physics
1 answer:
Valentin [98]3 years ago
3 0

Answer:

<em>1.43 s.</em>

Explanation:

Using one of the equations of motion,

S = ut + 1/2gt².......................... Equation 1

Where S = height of the cliff, u = initial velocity, t = time, g = acceleration due to gravity.

<em>Note: When the rock begins to fall from the maximum height, u = 0 m/s, g = positive</em>

<em>Given: S = 10 m, u = 0 m/s</em>

<em>Constant: g = 9.8 m/s²</em>

<em>Substituting these values into equation,</em>

<em>10 = 0(t) + 1/2(9.8)(t²)</em>

<em>10 = 0 + 4.9t²</em>

<em>t² = 10/4.9</em>

<em>t² = 100/49</em>

<em>t = √(100/49)</em>

<em>t = 10/7</em>

<em>t = 1.43 s.</em>

<em>Thus the rock spend 1.43 s in air</em>

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7 0
3 years ago
A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height
AleksandrR [38]

a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

E_i=\frac{1}{2}m_1v_1^2

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

v_3 = \frac{m_1 v_2}{m_1 + m_2}=\frac{(60 kg)(5.0 m/s)}{60 kg+15 kg}=4 m/s

c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J

While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:

E_f = (m_1 +m_2)gh_{max}

where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find

E_i = (m_1+m_2)gh_{max}\\h_{max}=\frac{E_i}{(m_1+m_2)g}=\frac{2070 J}{(75 kg)(9.8 m/s^2)}=2.8 m

4 0
3 years ago
This problem is based on the whole idea of pressure but I’m having trouble on when the area circle formula is included.
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Answer:

6.23x10^6Pa

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r (radius) = 0.005m

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First, we'll begin by calculating the area of the circlular heel. This is illustrated below:

Area of circle = πr^2

Area = 22/7 x (0.00)^2

Area = 7.86x10^-5m^2

Pressure is simply force per unit area. It represented mathematically as

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Pressure = 490/7.86x10^-5

Pressure = 6.23x10^6N/m2

Recall: 1N/m2 = 1Pa

Therefore, 6.23x10^6N/m2 = 6.23x10^6Pa

Therefore, the woman exert a pressure of 6.23x10^6Pa on the floor

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The volume of a cylinder is given by the formula v=pi r^2h, where r is the radius of the cylinder and h is the height.

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<h3>What is the cylinder's total surface area?</h3>

The sum of the curved surface areas makes up the cylinder's overall surface area.

To know more about  cylinder surface visit:-

brainly.com/question/22074027

#SPJ4

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