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Tom [10]
2 years ago
13

A force of 5.00 N to the left causes a 1.35 kg book to have a net acceleration of 0.76 m/s2 to the left. What is the frictional

force acting on the book?
Physics
1 answer:
Nuetrik [128]2 years ago
8 0

Answer:

4.0 N

Explanation:

Sum the forces in the x direction:

∑F = ma

F − Fr = ma

Fr = F − ma

Fr = 5.00 N − (1.35 kg) (0.76 m/s²)

Fr = 4.0 N

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An aircraft performs a maneuver called an "aileron roll." During this maneuver, the plane turns like a screw as it maintains a s
Dmitriy789 [7]

Answer:

0.17724 m/s²

Explanation:

D = Diameter of roll = Length of wing = 11 m

T = Time it takes to complete the circle = 35 s

Velocity

v=\frac{2\pi R}{T}\\\Rightarrow v=\frac{\pi D}{T}\\\Rightarrow v=\frac{\pi\times 11}{35}\\\Rightarrow v=0.98735\ m/s

Acceleration

a=\frac{v^2}{R}\\\Rightarrow a=\frac{0.98735^2}{\frac{11}{2}}\\\Rightarrow a=0.17724\ m/s^2

Acceleration of the tip of the plane is 0.17724 m/s²

3 0
2 years ago
Please help and explain if you can! Will give brainiest if I can! (Science)
serious [3.7K]
D ....Do you remember your teacher going over different physical and chemical reactions....It is a chemical reaction because it will never go back to the previous state...therefore your correct and it is a chemical reaction.
3 0
3 years ago
A 1.90-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. if you ignore the weight of th
gregori [183]

Answer: The center of gravity is 1.1338 m away from the left side of the barbell

Explanation:

Length of the barbell = 1.90 m

The distance center of gravity from left = x

Mass on the left side = 25 kg

The distance center of gravity from right = 1.90 - x

Mass on the right side = 37 kg

At the balance point: m_1x_1=m_2x_2

25 kg\times x=37 kg\times (1.90-x)

x=1.1338 m

The center of gravity is 1.1338 m away from the left side of the barbell

7 0
3 years ago
Read 2 more answers
What is the mass of a stone moving at a speed of
Serhud [2]
P = m*v 7.5 = m*15 m = 7.5/15 = 0.5 kg
6 0
3 years ago
Read 2 more answers
Please help with these questions as well! I need urgent help! I will give brainliest! God bless!
Radda [10]

6.  Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.  

F = ma

F = (15.3)(-9.8)

F = -150N

Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object.  So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.


7.  Same idea as question 2.  

First determine the weight of the object:

F = ma

F = (30)(-9.8)

F = -294N

The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.

-294 + 150N + x = 0

x = 144N  

So the person is exerting 144 N.


10.  First find the force of block B to the right due to its acceleration:

F = ma

F = (24)(0.5)

F = 12N

So block B is moving 12N to the right relative to block A due to block A's movement to the left.  However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A.  The force that is causing block B to experience the lower relative force to the right is because of the friction.  To find the friction:

The sum of the forces in the leftward and rightward direction for block B must equal 12N.

75 - x = 12

x = 63N

So the force of friction of block A on block B is 63N to the left.


5 0
2 years ago
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