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exis [7]
3 years ago
6

What is the acceleration of a 7 kg mass if the force of 70 N is used to move it toward the Earth?

Physics
1 answer:
Assoli18 [71]3 years ago
4 0

Answer:

<h2>10 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

m is the mass

f is the force

From the question we have

a =  \frac{70}{7}  = 10 \\

We have the final answer as

<h3>10 m/s²</h3>

Hope this helps you

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The table shows data for the planet Uranus. A 2 column table with 4 rows. The first column is labeled Quantity with entries, Esc
prohojiy [21]

Answer:

The answer is 218

Explanation:

Weight = mass * gravitational acceleration

weight is represented by F

F = 25kg (8.7)

(I'm pretty sure that you don't have to include the meters per second/per second thing)

4 0
3 years ago
A 2 L balloon filled with gas is warmed from 280 K to 700 K. What is the volume of the gas after it is heated?
irakobra [83]

Answer:

New volume, v2 = 0.8L

Explanation:

<u>Given the following data;</u>

Original Volume = 2L

Original Temperature = 280K

New Temperature = 700K

To find new volume V2, we would use Charles' law.

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles is given by;

VT = K

\frac{V1}{T1} = \frac{V2}{T2}

\frac{V1}{T1} = \frac{V2}{T2}

Making V2 as the subject formula, we have;

V_{2}= \frac{V1}{T1} * T_{2}

V_{2}= \frac{2}{700} * 280

V_{2}= 0.0029 * 280

V2 = 0.8L

Therefore, the volume of the gas after it is heated is 0.8L.

7 0
3 years ago
A ball is dropped from rest out of a high window in a tall building for 5 seconds. Assuming we ignore air resistance and assume
lara [203]

Answer:

what is the upwards force?

5 0
2 years ago
Read 2 more answers
Is there gravitational force between two students sitting in a classroom?
drek231 [11]

Answer:

Yes.

Explanation: the magnitude of the force is extremely small because the masses of the students are small relative to Earth's mass.

8 0
3 years ago
A horizontal force of magnitude 46.3 n pushes a block of mass 4.14 kg across a floor where the coefficient of kinetic friction i
IrinaVladis [17]
A) Calling F the intensity of the horizontal force and d the displacement of the block across the floor, the work done by the horizontal force is equal to
W=Fd = (46.3 N)(4.25 m)=196.8 J

b) The work done by the frictional force against the motion of the block is equal to:
W_f =  -F_f d =- (\mu mg) d =-(0.609)(4.14 kg)(9.81 m/s^2)(4.25 m)=
=-105.1 J
Part of these 105.1 Joules of work becomes increase of thermal energy of the block (\Delta E_B), and part of it becomes increase of thermal energy of the floor (\Delta E_F). We already know the increase in thermal energy of the block (38.2 J), so we can find the increase in thermal energy of the floor:
\Delta E_F = 105.1 J - \Delta E_B = 105.1 J-38.2 J=66.9 J

c) The net work done on the block is the work done by the horizontal force F minus the work done by the frictional force (the frictional force acts against the motion, so we must take it with a negative sign):
W_{net}=W-W_f=196.8 J-105.1 J=91.7 J
For the work-energy theorem, the work done on the block is equal to its increase of kinetic energy:
W_{net} = \Delta K
So, we have \Delta K=+91.7 J


5 0
3 years ago
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