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m_a_m_a [10]
3 years ago
6

Which of the balls will exert the smallest force on object A?Why

Physics
1 answer:
olganol [36]3 years ago
6 0
The 1kg ball would exert the smallest force.

As force = mass x gravity, this means that the smaller the mass (kg), the lesser the force.

When the mass is lighter (1kg):

Force = mass x gravity
Force = 1 x 9.8
Force = 9.8N

Compared to when the mass is heavier (10kg)

Force = mass x gravity
Force = 10 x 9.8
Force = 98N

Where this proves that the lighter the mass, the smaller the force exerted.
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Plz help with this paper.
lord [1]

Answer:

1. a) Draw a line towards the right side from the engine

b) This force pushes the boat forward and helps it accelerate further

2. a) Fixed volume for both solid and liquid

Compressible for only solid

Fixed shape is also for only Solid

b) The answer is 'c'

c) Solids, because they have their particles closely packed therefore they can be compressed (not so sure bout this answer)

4 0
3 years ago
A system does 506 kj of work, and loses 266 kj of heat to the surroundings. what is the change in internal energy of the system?
babunello [35]

Answer:

The change in internal energy of the system = -772kJ

Explanation:

Given :

Heat lost by the system , a = -266KJ

Workdone by the system, W = -506KJ

The first law of thermodynamics states that:

Change in internal energy = q + w

Substituting values into the equation

Change in internal energy = (-266KJ) + (-506KJ)

Change in internal energy = -722KJ

7 0
3 years ago
Read 2 more answers
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

4 0
4 years ago
A wheel rotates 3 times. What is its angular displacement?
tia_tia [17]

Answer:

The angular displacement would be 6 Pi radians or 1080 degrees.

Explanation:

Every individual rotation is 2 Pi (this applies to anything regardless of radius).

Since there are three rotations you are going to multiply 2 Pi by 3.

This will give you 6 Pi which you can convert to equal 1080 degrees (1 Pi = 180 degrees).

5 0
4 years ago
Read 2 more answers
A watt is a unit of<br><br>light<br>power<br>force<br>motion<br>energy
Crazy boy [7]

The answer would be Power

3 0
3 years ago
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