All categories

3 years ago

I don’t know how to answer this question? Can anyone help?

Physics

1 answer:

VMariaS [17]3 years ago

7 0

Answer:

F=ma

here F is force, m is mass and a is accelaration,

According to the question,

F=3*F= 3F

m= 1/3 of m= m/3

a= ?

so the equation becomes,

3F= m/3*a

3F*3= ma

9F=ma

F= ma/9

Therefore accelaration reduces by 1/9.

I am not very sure.

You might be interested in

A uniform marble rolls down a symmetric bowl, starting from rest atthe top of the left side. The top of each side is a distanceh

Debora [2.8K]

Answer:

A. 5/7h

B. Same height

C. See attachment below

Explanation:

Find the attachment below for better understanding.

7 0

3 years ago

What is the kinetic energy of a bicycle with a mass of 14 kg traveling at a<br> velocity of 3.0 m/s?

Sav [38]

46.6666 that is the mass number

Explanation:

14 divided 3.0

5 0

3 years ago

A block with mass m is pulled horizontally with a force F_pull leading to an acceleration a along a rough, flat surface.

Simora [160]

Answer:

$\mu_k=\frac{a}{g}$

Explanation:

The force of kinetic friction on the block is defined as:

$F_k=\mu_kN$

Where $\mu_k$ is the coefficient of kinetic friction between the block and the surface and N is the normal force, which is always perpendicular to the surface that the object contacts. So, according to the free body diagram of the block, we have:

$N=mg\\F_k=F=ma$

Replacing this in the first equation and solving for $\mu_k$ :

$ma=\mu_k(mg)\\\mu_k=\frac{a}{g}$

6 0

3 years ago

The students in the diagram wanted to calculate the speed of sound. They

topjm [15]

Isolate the variable by dividing each side by factors that don't contain the variable.

32.021

8 0

3 years ago

A 0.70-kg basketball dropped on a hardwood floor rises back up to 65% of its original height. (a) If the basketball is dropped f

dem82 [27]

Answer:

Part a)

$Loss = 3.6 J$

Part b)

$Loss = 0.99 J$

Part C)

This is loss in terms of thermal energy due to collision with the floor

Explanation:

Part a)

Since we know that the ball rises up by 65% of initial height

so after first bounce it will lose 35% of its initial energy

so we will have

$U = mgH$

Energy Loss = 0.35 mgH[/tex]

$Loss = 0.35(0.70)(9.81)(1.5)$

$Loss = 3.6 J$

Part b)

Energy of the ball after first bounce

$U_1 = 0.65 mgH$

energy of ball after 2nd Bounce

$U_2 = 0.65(0.65 mgH)$

energy of the ball after 3rd bounce

$U_3 = (0.65)(0.65^2)mgH$

$U_3 = 0.65^3(0.70)(9.81)(1.5)$

$U_3 = 2.83 J$

Now we will have energy loss in fourth bounce given as

$Loss = 0.35 U_3$

$Loss = 0.35(2.83)$

$Loss = 0.99 J$

Part C)

This is loss in terms of thermal energy due to collision with the floor

7 0

3 years ago