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NeX [460]
3 years ago
5

Below are a pair of wires holding a hanging light that weighs 50N. Is the tension in each of the wires less than 50N, 50N, or mo

re than 50N?
Physics
1 answer:
Nookie1986 [14]3 years ago
7 0

Answer:

Less than 50N

Explanation:

Given that a pair of wires holding a hanging light that weighs 50N.

Since the hanging weight is equal to 50 N, the sum of the tension in the wires will also experience the same magnitude of force acting on the weight.

Considering each of the wire, the tension on each of the wire will surely be less than 50N

Therefore, the tension in each wire is less than 50N

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Describe how the gravitational force between 2 objects depends on the mass of the objects and the distance between them. your an
ollegr [7]
The strength of the gravitational force between two objects depends<span> on </span>two<span>factors, </span>mass<span> and </span>distance<span>. the </span>force<span> of gravity the </span>masses<span> exert on each other. If one of the </span>masses<span> is doubled, the </span>force<span> of gravity </span>between<span> the </span>objects<span> is doubled. increases, the </span>force<span> of gravity decreases</span>
3 0
3 years ago
A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac
RoseWind [281]

Answer:

a) \Delta U_{g} = 12.945\,J, b) \Delta U_{k} = 12.945\,J, c) k = 2930.059\,\frac{N}{m}

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)

\Delta U_{g} = 12.945\,J

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

\Delta U_{k} = 12.945\,J

c) The spring constant of the gun is:

\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}

k = \frac{2\cdot \Delta U_{k}}{x^{2}}

k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}

k = 2930.059\,\frac{N}{m}

4 0
2 years ago
Need help with stu*pid science thing that I keep getting wrong, augh!!
Kryger [21]

Answer:

i think it is iron

Explanation:

its the only one that makes sense to me

4 0
2 years ago
Read 2 more answers
Why to astronauts appear weightless while they are filmed performing activities inside the orbiting space shuttle?
vagabundo [1.1K]
<h2>Answer: The astronauts are falling at the same rate as the space shuttle as it orbits around earth</h2>

The astronauts seem to float because they are in free fall just like the spacecraft.

However, although they are constantly falling on the Earth, they do not fall because the ship orbits at a sufficient speed (in the same direction of rotation of the Earth) so that the centrifugal force is balanced with the Earth's gravitational pull.

In other words:

The spaccraft and the astronauts are in free fall but the Earth's surface will never be reached as long as they does not decrease the speed.

Then, as they accelerate toward Earth (regardless of their mass), it curves beneath them and never comes close.

That's why astronauts, having the same acceleration as the spacecraft, feel weightless and see themselves floating.

8 0
2 years ago
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A bicycle rider pushes a 13kg bicycle up a steep hill. the incline is 24 degree and the road is 275m long. the rider pushes the
Digiron [165]

Answer:

A. W = 6875.0 J.

B. W = -14264.6 J.

Explanation:

A. The work done by the rider can be calculated by using the following equation:

W_{r} = |F_{r}|*|d|*cos(\theta_{1})

Where:                

F_{r}: is the force done by the rider = 25 N

d: is the distance = 275 m

θ: is the angle between the applied force and the distance

Since the applied force is in the same direction of the motion, the angle is zero.

W_{r} = |F_{r}|*|d|*cos(0) = 25 N*275 m = 6875.0 J

Hence, the rider does a work of 6875.0 J on the bike.

B. The work done by the force of gravity on the bike is the following:

W_{g} = |F_{g}|*|d|*cos(\theta_{2})  

The force of gravity is given by the weight of the bike.

F_{g} = -mgsin(24)     

And the angle between the force of gravity and the direction of motion is 180°.

W_{g} = |mgsin(24)|*|d|*cos(\theta_{2})  

W_{g} = 13 kg*9.81 m/s^{2}*sin(24)*275 m*cos(180) = -14264.6 J  

The minus sign is because the force of gravity is in the opposite direction to the motion direction.

Therefore, the magnitude of the work done by the force of gravity on the bike is 14264.6 J.  

I hope it helps you!                                                                                          

3 0
2 years ago
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