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3 years ago

What is the force that gravity puts upon a 25kg person

Physics

1 answer:

yan [13]3 years ago

3 0

Answer:

245.25 Newtons

Explanation:

the gravitational force of the earth is 9.81 N/Kg

And so to work this out you would multiply 25 by 9.81 to give you 245.25

$25*9.81=245.25N$

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Both a 3 kg book and 5 kg bowling ball are sitting still on the floor. What is true about the kinetic energy of the objects? Exp

Charra [1.4K]

Answer:

The kinetic energy for both objects is the same.

Explanation:

While in other cases the kinetic energies of two objects that have different masses might be different depending on their velocities, in this case both the 3 kg book and 5 kg bowling ball have the same kinetic energy.

This is because kinetic energy is calculated using the formula: K = 1/2 * m * v^2, where m represents the mass and v represents the velocity of the object.

Since the book and the bowling ball are sitting still on the floor, their velocities are zero. Hence, when we plug in 0 for velocity into the equation for kinetic energy, we will get that the kinetic energy is 0 for the book and the bowling ball.

Hope this helps!

6 0

3 years ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

4 years ago

Gravity is the force that keeps us on the Earth. It pulls us towards the center of the Earth. If you were to move from the surfa

Nikitich [7]

<h2>Answer: B. Gravitational potential energy </h2>

Explanation:

<em>The gravitational potential energy is the energy that a body or object possesses, due to its position in a gravitational field. </em>

That is why this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.

In the case of the <u>Earth</u>, in which <u>the gravitational field is considered constant</u>, the value of the gravitational potential energy $U_{p}$ will be: