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Degger [83]
3 years ago
10

Calculate the gravitational potential energy of the interacting pair of the Earth and a 11 kg block sitting on the surface of th

e Earth. You would need to supply the absolute value of this result to move the block to a location very far from the Earth (actually, you would need to use even more energy than this due to the gravitational potential energy associated with the Sun-block interacting pair).
Physics
1 answer:
trasher [3.6K]3 years ago
5 0

Answer:

Gravitational potential energy (GPE) = 107.8J

Explanation:

Gravitational potential energy (GPE) = mgh

Where mass(m) = 11kg

Acceleration due to gravity(g) = 9.8m²/s

height = assumed to be 1m

Force(F) = mg

Force(F) = 11×9.8 = 107.8N

Gravitational potential energy (GPE) = 107.8×1

= 107.8J

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Bob and John are pulling in different directions. If Bob is pulling to the right with a force of 10N, and John is pulling to the
Murljashka [212]

Answer:

The net force on the box is 2 N to the left.

The box will move to the left.

The acceleration on the box is 0.5 m/s^2 to the left.

Explanation:

Let's say movement to the right is positive and left is negative.

Bob: +10 N

John: -12 N

Add those together and you get a net force of -2 N, and the negative sign means that the box is moving to the left.

For the acceleration:

Fnet = ma

-2 = (4 kg)a

a = -0.5 m/s^2

Again, the negative sign in this answer means the box is being accelerated to the left.

4 0
4 years ago
Using the
xxMikexx [17]

Answer:

The principle of momentum conservation states that if there no external force the total momentum of the system before and after the collision is conserved.

Since momentum is a vector, we should investigate the directions and magnitudes of initial and final momentum.

\vec{P}_{initial} = \vec{P}_{final}\\\vec{P}_{initial} = m_1\vec{v}_1 + 0\\\vec{P}_{final} = m_1\vec{v}_1' + m_2 \vec{v}_2'

If the first ball hits the second ball with an angle, we should separate the x- and y-components of the momentum (or velocity), and apply conservation of momentum separately on x- and y-directions.

6 0
3 years ago
A 2300 Kg car accelerates from rest to 6.00 m/s in 12.00 seconds. What is the net force acting in the car?
pav-90 [236]
F=ma
a=(v2-v1)/(t2-t1)
a=(6-0)/(12-0)
a=6/12
a= .5 m/s^2
f=2300kg*.5m/s^2
f=1150N
f=1200N if using correct sig figs
4 0
3 years ago
Read 2 more answers
En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a
Maru [420]

Answer:

c=0.14J/gC

Explanation:

A.

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

B.

We can use the expression for heat transmission

Q=mc(T_2-T_1)

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

Q_1=-Q_2

for water we have to

c = 4.18J / g ° C

replacing we have

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

I hope this is useful for you

A.

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

B.

Podemos usar la expresión para la transmisión de calor

Q=mc(T_2-T_1)

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

Q_1=-Q_2

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

7 0
3 years ago
What is the formula for conservation of momentum
olga55 [171]

Answer:

The general equation for conservation of momentum during a collision between n number of objects is given as: [m i ×v i a ] = [m i ×v i b ] Where m i is the mass of object i , v i a is the velocity of object i before the collision, and v i b is the velocity of object i after the collision.

Explanation:

6 0
3 years ago
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