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klasskru [66]
3 years ago
15

Two objects gravitationally attract with a force of 70N. If the masses of both of the objects are increased by 15x and the dista

nce between their centers change by a factor of 12x, then the new force of attraction is ___ N.
a. 1.56 N
b. 109.38 N
c. 7.29 N
d. 1312.5 N
Physics
1 answer:
erastova [34]3 years ago
7 0
(B is the answer!!
I took the test and I got a 100! Hope this helps
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inn [45]
Positive 2n to the right of the box
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3 years ago
Show that the acceleration of any object down an incline where friction behaves simply (that is, where fk=μkN ) is a=g(sinθ−μkco
11Alexandr11 [23.1K]

Answer:

a=g(sinθ-μkcosθ)

Explanation:

In an inclined plane the forces that interact with the object can be seen in the figure. The normal force, the weight w and the decomposition of the force vector of weight can be observed.

wx=m*g*sinθ

wy=m*g*cosθ

As the objects moves down an incline, acceleration in y axis is 0.

Then, by second Newton's Law:

Fy = m*ay

FN - m*g cos θ = 0,

FN=m*g cos θ

In x axis the forces that interacs are the x component of weight and friction force:

Fx = m*ax

mg sen u-FN*μk=m*a

Being friction force, Fr=FN*μk, we replace with its value in below formula:

m*g *sinθ-(m*g*cosθ*μk)=m*a

Then, isolating a:

a=(m*g sinθ-(m*g*cosθ*μk))/m

Solving, we have next equation:

a=g sinθ-(g*cosθ*μk)

Applying distributive property we have:

a=g*(sinθ-μk*cosθ)

5 0
3 years ago
The following graph shows the force exerted on and the displacement of object being pulled
Tomtit [17]

The work done to pull the object 7.0 m is the total area under the graph from 0.0 m to 7.0 m, determined as 245 J.

<h3>Work done by the applied force</h3>

The area under force versus displacement graph is work done.

The total work done by pulling the object 7 m, can be grouped into two areas;

  • First area, A1 = area of triangle from 0 m to 2.0 m
  • Second area, A2 = area of trapezium, from 2.0 m to 7.0 m

A1 = ¹/₂ bh

A1 = ¹/₂ x (2) x (20)

A1 = 20 J

A2 = ¹/₂(large base + small base) x height

A2  = ¹/₂[(7 - 2) + (7-3)] x 50

A2 = ¹/₂(5 + 4) x 50

A2 = 225 J

<h3>Total work done </h3>

W = A1 + A2

W = 20 J + 225 J

W = 245 J

Learn more about work done here: brainly.com/question/8119756

3 0
2 years ago
Corrosion is the tendency of a metal to revert back to its natural state.
algol13
Corrosion is the irreversible damage or destruction of living tissue or material due to a chemical or electrochemical reaction.
3 0
3 years ago
I NEED MAJOR HELP ON THIS AS WELL PLEASE SOMEONE HELP ME
yuradex [85]

Answer:

The total distance is  130.2 [m]

Explanation:

In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the cart starts from rest, i.e. its initial speed is zero.

v_{f} =v_{o} +(a*t)

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0

a = acceleration = 3 [m/s²]

t = time = 8[s]

Vf = 0 + (3*8)

Vf = 24 [m/s]

With this velocity we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

where

x = distance traveled [m]

24² = 0 + (2*3*x)

x = 576/(6)

x = 96 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.

We must calculate the final velocity.

v_{f}= v_{i} -(a*t)

Vf = final velocity [m/s]

Vi = initial velocity = 24 [m/s]

a = desacceleration = 1.6 [m/s²]

t = time = 15 [s]

Vf = 24 - (1.6*15)

Vf = 21.6 [m/s]

With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} -2*a*x

where

x = distance traveled [m]

21.6² = 24² - (2*1.6*x)

x = 109.44/(3.2)

x = 34.2 [m]

Note: The negative sign in the equations is because the car is desaccelerating, it means its velocity is decreasing.

Therefore the total distance is Xt = 34.2 + 96 = 130.2 [m].

5 0
3 years ago
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