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mihalych1998 [28]
3 years ago
10

11). Determine the acceleration due to gravity if a 6kg book was on top of

Physics
1 answer:
DerKrebs [107]3 years ago
6 0

Answer:

g = 9.8 [m/s²]

Explanation:

To solve this problem we must remember that the potential energy is defined as the product of mass by gravitational acceleration by the height or elevation with respect to a reference level.

Ep = m*g*h

where:

Ep = potential energy = 235.2 [J]

m = mass of the book = 6 [kg]

g = gravity acceleration [m/s²]

h = elevation = 4 [m]

Now replacing these values:

235.2 = 6*g*4

g = 235.2/(24)

g = 9.8 [m/s²]

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Inertia: tendency of an object to resist changes in its velocity. An object at rest has zero velocity - and (in the absence of an unbalanced force) will remain with a zero velocity. Such an object will not change its state of motion (i.e., velocity) unless acted upon by an unbalanced force.

~done by cessly420

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8 0
4 years ago
An egg drops from a nest at a height of 2.3 m. Which equation can you use to calculate the impact speed of the egg?
igomit [66]

Answer:

V2^2=V1^2+2a/\d

Explanation:

3 0
4 years ago
You throw a basketball and a tennis ball across the classroom so that each ball has the exact same momentum. Explain how this is
Helga [31]

Answer:

It is possible by increasing the speed of the tennis ball by a factor of (Mass of the tennis ball)/(Mass of the basketball)

Explanation:

The momentum of a body = The bod's mass × The body's velocity

Therefore, the momentum of a given mass of an object, such as a tennis ball can be increased by increasing the velocity or speed of the object. Whereby the speed of the ball, v₁, is increased such that the momentum of the basketball and the tennis ball will be the same, is given by the following equation

Mass of the basketball × v₂ = Mass of the tennis ball × v₁

Therefore, v₁/v₂ = (Mass of the tennis ball)/(Mass of the basketball)

7 0
3 years ago
A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s
Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8^{-3}

Explanation:

m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg

v_{bullet}=341\frac{m}{s}

Momentum of the motion the first part of the motion have a momentum that is:

P_{1}=m_{bullet}*v_{bullet}

P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529

The final momentum is the motion before the action so:

a).

P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}

P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}

P_{1}=P_{2}

2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}

b).

kinetic energy

K=\frac{1}{2}*m*(v)^{2}

Kinetic energy after

Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J

Kinetic energy before

Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J

Ratio =\frac{Ka}{Kb}

R=\frac{1.14}{406.4}\\R=2.8x10^{-3}

3 0
4 years ago
. Two people are pushing a car of mass 2000 kg.
Viktor [21]

Answer:

two people who are not going to be able to make it to class today because of the day and then I will be there at the house and then we can go

5 0
4 years ago
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