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Sauron [17]
3 years ago
9

Two small balls, A and B, attract each other gravitationally with a force of magnitude F. If we now double both masses and the s

eparation of the balls, what will now be the magnitude of the attractive force on each one?
a) 16F
b) 8F
c) 4F
d) F
e)F/4
Physics
2 answers:
PSYCHO15rus [73]3 years ago
6 0

Answer:

After studying the law of gravitational attraction, students constructed a model to illustrate the relationship between gravitational attraction (F) and distance. If the distance between two objects of equal mass is increased by 2, then the gravitational attraction (F) is 1/4F or F/4. How would this model, situation A, change if the mass of the spheres is doubled?

A)  A

B)  B

C)  C

D)  D

If you came here from usa test prep it is:

Actually A

But for the question given right now is D.

Explanation:

stich3 [128]3 years ago
3 0

Answer:

New force, F' = F

Explanation:

Given that, two small balls, A and B, attract each other gravitationally with a force of magnitude F. It is given by :

F=G\dfrac{m_Am_B}{r^2}

If we now double both masses and the separation of the balls, the new force is given by :

F'=G\dfrac{2m_A\times 2m_B}{(2r)^2}

F' = F

So, the new force remains the same as previous one. Hence, the correct option is (d) "F"

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An athlete in a gym applies a constant force of 50 N to the pedals of a bicycle to keep the rotation rate of the wheel at 10 rev
marishachu [46]

Answer:

Explanation:

Given that,

Force applied to pedal F = 50N

Angular velocity ω = 10rev/s

We know that, 1rev = 2πrad

Then, ω = 10rev/s = 10×2π rad/s

ω = 20π rad/s

Length of pedal r = 30cm = 0.3m

Power?

Power is given as

P = τ×ω

We need to find the torque τ

τ = r × F

Since r is perpendicular to F

Then, τ = 0.3 × 50

τ = 15 Nm

Then,

P = τ×ω

P = 15 × 20π

P = 942.48 Watts

power delivered to the bicycle by the athlete is 942.48 W

6 0
3 years ago
The amount of energy required to raise the temperature of one gram of water by one celsius degree is called a
Elena L [17]

The amount of energy required to raise the temperature of one gram of water by one celsius degree is called a calorie.

<h3>How can you define calorie?</h3>

In physics, we can define it as a unit of measurement that defines the amount of energy to raise the temperature of 1 gram  of water by 1°C. Each material has its unique calorie, however by definition we use water as it is the lowest of all the elements.

Normally, calorimetry is directly linked with the thermal capacity that an element has to contain heat.

See more about calorie at brainly.com/question/22374134

#SPJ1

5 0
2 years ago
how might countries go about changing their CO2 emissions? What, if anything should be done about these countries C02 emissions?
svetlana [45]
1. Use renewable energy sources (eg. wind, solar etc.)
2. Encourage citizens to use public transport
3. Better waste management systems (instead of burning waste, recycle etc.)
that’s all i could think of :)
3 0
3 years ago
Two blocks connected by a light string are being pulled across a frictionless horizontal tabletop by a hanging 16.2-N weight (bl
Artemon [7]

Newton's second law allows us to find the results for the string tensions are:

  • T₁ = 6.7 N
  • T₂ = 16.54 N

Newton's second law gives a relationship between force, mass and acceleration of bodies

            ∑ F = ma

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration.

Free-body diagrams are representations of the forces applied to bodies without the details of them.

The reference system is a coordinate system with respect to which the forces decompose, in this case the x-axis is parallel to the plane and the positive direction in the direction of movement, the y-axis is perpendicular to the plane.

In the attachment we see a free-body diagram of the three-block system.

Let's apply Newton's second law to each body.

Block C

Y-axis

       W_c -T_2 = m_c a

Block A

X axis  

       T_2 - T_1 - W_a_x = m_a a  

Y axis  

       N_a - W_a_y = 0  

Block B

X axis

      T_1 - W_b_x = m_b a  

Y axis

      N_b - W_b_y =0

Let's  use trigonometry to find the components of the weight.

Block A

         cos θ = \frac{W_a_y}{W_a}  

         sin θ = \frac{W_a_x}{W_a}

         W_a_y = W_a cos \theta

         W_a_x= W_a sin \theta

Block B

        cos θ = \frac{W_b_y}{W_b}

        sin θ =  \frac{W_b_x}{W_b}

        W_b_y = W_b cos \theta \\W_b_x = W_b sin \theta

Let's write our system of equations.

     W_c - T_2 = m_c a \\           T_2 - T_1 - W_a_x = m_a a \\T_1 - W_b_x = m_b a

 

Let's find the acceleration of the bodies, adding the equations.

     W_c - W_a_x - W_b_x = ( m_a+m_b+m_c) a\\  

         

The weight is

    W = mg

Let's  substitute

         (m_c - m_a -m_b ) g \ sin \theta = ( m_c+m_a+m_b)  \ a  \\a= \frac{ m_c-m_a-m_b  }{ m_a+m_b+m_c} \ g sin \theta

Indicate ma mass of the block a ma = 1.00 kg, the mass of the block b mb = 2.2 kg and the weight of the block c Wc = 16.2 N, let's find the mass of block c.

          m_c = Wc / g

          m_c = 16.2 / 9.8

          m_c = 1.65 kg

we substitute the values

          a= \frac{1.65 -2.20 -1.00}{1.65+2.20+1.00} \ 9.8 \ sin \theta  \\a= -0.3096 sin \theta

The negative sign indicates that the system is descending, to be able to give a specified value an angle is needed, they assume that the angle of the ramp is 45º

          a = - 0.3196 sin 45

          a = -0.226 m / s

Taking the acceleration we are going to look for the tensions.

From the equation of block C

           W_c - T_2 = m_c a \\T_2 = m_c ( g-a)\\T_2 = 1.65 ( 9.8 + 0.226)

            T₂ = 16.54 N

From the equation of block B

          T_1 - W_b_x = m_b a\\T_1 = m_b (a + g sin \theta)\\T_1 = 1.00 (-0.226 + 9.8 \ sin 45)

           T₁ = 6.7 N

In conclusion using Newton's second law we can find the results for the string tensions are:

  •  T₁ = 6.7 N
  •  T₂ = 16.54 N

Learn more here:  brainly.com/question/20575355

7 0
3 years ago
Anna applies a force of 19.5 newtons to push a book placed on a table. If the normal force of the book is 51.7 newtons, what is
GarryVolchara [31]

that would be given by

[email protected]

@ representing coefficient of kinetic friction.

thus 19.5/51.7 = 0.377

6 0
3 years ago
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