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kozerog [31]
3 years ago
8

In Alamogordo, New Mexico, falling rocks are a hazard along US Highway 82. If a rock falls from a cliff with 157 N of force,what

was the mass of the rock?
Physics
1 answer:
jeka943 years ago
5 0

Answer:

Mass, m = 16.02kg

Explanation:

Given the following data;

Force, F = 157N

Since it's a free fall, acceleration, a = g = 9.8m/s² (acceleration due to gravity).

To find the mass;

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

F = ma

Where;

  • F represents force measured in Newton.
  • m represents the mass of an object measured in kilograms.
  • a represents acceleration measured in meter per seconds square.

<em>Making mass (m) the subject of formula, we have;</em>

Mass (m) = \frac{F}{a}

<em>Substituting into the equation;</em>

Mass (m) = \frac{157}{9.8}

Mass, m = 16.02kg

<em>Therefore, the mass of the rock is 16.02kg.</em>

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Elina [12.6K]

Answer:

C. 4.0 J

Explanation:

Work is equal to the change in kinetic energy or kinetic energy final - kinetic energy initial. 1/2 mv^2 is the equation for kinetic energy. SInce the mass is the same, you don't have to include it when solving. 1/2 times 1^2 = 1/2. 1/2 times 3^2 = 4.5. 4.5 - 0.5 = 4 J

8 0
3 years ago
Richardson pulls a toy 3.0 m across the floor by a string, applying a force of0.50 N. During the first meter, the string is para
Anastasy [175]

Answer:

Total Work done =0.65 joule

Explanation:

Work done is given Mathematically as

W=F *d

Where w=work done in joules

F=applied force

d= distance moved

The work done to move the toy accros the first meter is

W1=0.5*1

W1=0.5joule

The work done to move the toy across the next 2m at an angle of 30° is

.W2=0.5*2cos30

W2=0.5*2*0.154

W2=0.154joule

Hence total work done is

W1+W2=0.5+0.154

Total Work done =0.65 joule

7 0
3 years ago
Nvm I dont need help anymore
pychu [463]

Answer:

no problem it's okay if you need any problem then ask

5 0
3 years ago
A ball is thrown 20.0 m/s at an angle of 40.0° with the horizontal. Assume the ball is thrown at ground level.
TEA [102]
The ball's horizontal component of velocity (ie it's horizontal speed) is 20 cos 40degrees. Without knowing the distance of the ball to the wall it's difficult to go further ...
8 0
3 years ago
Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with an
Alona [7]

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

F=\dfrac{kQ^2}{r^2} ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}.....(2)

Dividing equation (1) and (2), we get :

\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}

Hence, the correct option is (d) i.e. " 4F/9"

7 0
3 years ago
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