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3 years ago

What upward gravitational force does a 5600kg elephant exert on the earth?

2 answers:

8 0

Force is determined by multiplying mass and gravity (F= mg). To determine the answer, the mass of the elephant (5600 kg) is multiplied with the gravity (9.8 m/s²). The answer is 5800 N. This is the upward gravitational force that the elephant exerts on the earth.

Inessa [10]3 years ago

6 0

<u>Answer</u>

54,880 N

<u>Explanation</u>

The Gravitational force is the force exited by an object towards the center of the earth. It is called the weight of an object. The Newton's 2nd law of motion says that for every action there is equal and opposite reaction. So, if the elephant exerts a force on the earth, the earth exerts an equal force to the elephant.

Weight = mass × gravity

= 5600 × 9.8

= 54,880 N

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A person jumps out a fourth-story window 14 m above a firefighter safety net. The survivor stretches the net 1.8 m before coming

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Answer:

The deceleration is $a = - 76.27 m/s^2$

Explanation:

From the question we are told that

The height above firefighter safety net is $H = 14 \ m$

The length by which the net is stretched is $s = 1.8 \ m$

From the law of energy conservation

$KE_T + PE_T = KE_B + PE_B$

Where $KE_T$ is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

and $PE_T$ is the potential energy of the before jumping which is mathematically represented at

$PE_T = mg H$

and $KE_B$ is the kinetic energy of the person just before landing on the safety net which is mathematically represented at

$KE_B = \frac{1}{2} m v^2$

and $PE_B$ is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

So the above equation becomes

$mgH = \frac{1}{2} m v^2$

=> $v = \sqrt{2 gH }$

substituting values

$v = 16.57 m/s$

Applying the equation o motion

$v_f = v + 2 a s$

Now the final velocity is zero because the person comes to rest

$0 = 16.57 + 2 * a * 1.8$

$a = - \frac{16.57^2 }{2 * 1.8}$

$a = - 76.27 m/s^2$

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3 years ago

You are viewing a read out of an oscilloscope that shows you when different voltages in a circuit maximize. If the period of the

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Answer:

The time is $t = 0.02 \ s$

Explanation:

From the question we are told that

The period of the circuit is $T = 0.04$

Generally voltage maximization of the capacitor occurs during the voltage minimization of the inductor and vise versa

So the time between the voltage maximization of the capacitor and that of the inductor is mathematically represented as

$t = \frac{T}{2}$

=> $t = \frac{0.04}{2}$

=> $t = 0.02 \ s$

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3 years ago

A ball thrown vertically upward is caught by the thrower after 2.93 s. Find the initial velocity of the ball. The acceleration o

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Answer:

The initial velocity of the ball is 28.714 m/s

Explanation:

Given;

time of flight of the ball, t = 2.93 s

acceleration due to gravity, g = 9.8 m/s²

initial velocity of the ball, u = ?

The initial velocity of the ball is given by;

v = u + (-g)t

where;

v is the final speed of the ball at the given time, = 0

g is negative because of upward motion

0 = u -gt

u = gt

u = (9.8 x 2.93)

u = 28.714 m/s

Therefore, the initial velocity of the ball is 28.714 m/s

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3 years ago

You have a 3.00-liter container filled with N₂ at 25°C and 4.45 atm pressure connected to a 2.00-liter container filled with Ar

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