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3 years ago

A force of 50 N stretches a string by 4 cm,calculate the elastic constant.

Physics

1 answer:

murzikaleks [220]3 years ago

8 0

Answer:

50/0.04= answer

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Hari drag a load of 60 kg along a distance of 12 metre what amount of work does he do also mention the type of work

Vlada [557]

Answer:

W = 7056 J

Explanation:

Given that,

Mass of a load, m = 60 kg

Distance moved, d = 12 m

We need to find the amount of work. The work done by an object is given by :

W = Fd

So,

W = mg×d

W = 60 kg × 9.8 m/s²×12 m

W = 7056 J

So, the amount of work done is 7056 J.

6 0

3 years ago

A ball of mass 0.15 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.960 m. b)

Art [367]

Answer:

Explanation:

If v be the velocity just after the rebound

Kinetic energy will be converted into potential energy

1/2 m v² = mgh

v² = 2gh

v = √ 2gh

= √ 2 x 9.8 x .96

= 4.33 m / s

4 0

3 years ago

A person is standing on a level floor. His head, upper

BabaBlast [244]

Answer:

$y_{cg} = 1.03 m$

Explanation:

Given data:

weigh (head+arms + head) w_1 = 438 N

centre of gravity y_1= 1.28 m

weigh (upper leg) w_2 = 144 N

Center of gravity y_2 = 0.760 m

weigh ( lower leg + feet) = 87 N

centre of gravity = y_3 = 0.250 m

location of center of gravity $= \frac{w_1 y_1 + w_2 y_2 + w_3 y_3}{w_1 +W_2 +w_3}$

$y_{cg} = \frac{438 \times 1.28 + 144\times 0.760 + 87 \times 0.250}{438+144+87}$

$y_{cg} = 1.03 m$

8 0

3 years ago

1. In an experiment, a positively charged oil droplet weighing 6.5 x 10-15 N is held stationary

blagie [28]

The charge on the droplet must be $1.2\cdot 10^{-18} C$

Explanation:

In order for the oil droplet to be in equilibrium, the force of gravity acting on it (the weight, acting downward) must be equal to the electrical force acting upward.

The gravitational force is:

$F_g = 6.5 \cdot 10^{-15} N$

which is equivalent to the weight of the droplet.

The electric force is given by:

$F_E = qE$

where

q is the charge on the oil droplet

$E=5.3\cdot 10^3 N/C$ is the magnitude of the electric field

Since the two forces must be equal,

$F_G = F_E\\F_G = qE$

And solving for q,

$q=\frac{F_G}{E}=\frac{6.5\cdot 10^{-15} N}{5.3\cdot 10^3}=1.2\cdot 10^{-18} C$

Learn more about gravitational force and electric force:

brainly.com/question/1724648

brainly.com/question/12785992

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

3 0

3 years ago

a _ is an explanation that unifies the findings of many experiments, and it can be changed when new experimental results need to

MakcuM [25]

That's a pretty good intuitive definition of a <em>theory </em>.

3 0

3 years ago

Read 2 more answers