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3 years ago

Newton's second law of motion is also known as the law of.......

Physics

1 answer:

OverLord2011 [107]3 years ago

4 0

I think it is force of acceleration but I'm not sure.. hope this helps though!

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1000 kg car takes travels on a circular track having radius 100 m with speed 10 m/s. What is the magnitude of the acceleration o

Hoochie [10]

Answer: B. 1m/s^2

Explanation:

6 0

3 years ago

Part A What is the resistance of a 4.4 m length of copper wire 1.3 mm n diameter? The resistivity of copper is 1.68x 10-8 Ω-m Ex

Natalija [7]

Explanation:

It is given that,

Length of the copper wire, l = 4.4 m

Diameter of copper wire, d = 1.3 mm = 0.0013 m

Radius of copper wire, r = 0.00065 m

The resistivity of the copper wire, $\rho=1.68\times 10^{-8}\ \Omega-m$

We need to find the resistance of the copper wire. It is given by :

$R=\rho\dfrac{l}{A}$

$R=1.68\times 10^{-8}\ \times \dfrac{4.4\ m}{\pi (0.00065)^2}$

R =0.055 ohms

So, the resistance of the copper wire is 0.055 ohms. Hence, this is the required solution.

7 0

3 years ago

A bungee cord can stretch, but it is never compressed. When the distance between the two ends of the cord is less than its unstr

Ksju [112]

Answer:

Explanation:

Given that

g=9.8m/s²

The spring constant is

k=50N/m

The length of the bungee cord is

Lo=32m

Height of bridge which one end of the bungee is tied is 91m

A steel ball of mass 92kg is attached to the other end of the bungee.

The potential energy(Us) of the steel ball before dropped from the bridge is given as

P.E= mgh

P.E= 92×9.8×91

P.E= 82045.6 J

Us= 82045.6 J

Potential energy)(Uc) of the cord is given as

Uc= ½ke²

Where 'e' is the extension

Then the extension is final height extended by cord minus height of cord

e=hf - hi

e=hf - 32

Uc= ½×50×(hf-32)²

Uc=25(hf-32)²

Using conservation of energy,

Then,

The potential energy of free fall equals the potential energy in string

Uc=Us

25(hf-32)²=82045.6

(hf-32)² = 82045.6/25

(hf-32)²=3281.825

Take square root of both sides

√(hf-32)²=√(3281.825)

hf-32=57.29

hf=57.29+32

hf=89.29m

We neglect the negative sign of the root because the string cannot compressed

3 0

3 years ago

Why does a child in a wagon seem to fall backward when you give the wagon a sharp pull forward? 15. What force is needed to acc

slavikrds [6]

Answer:

Force, F = 77 N

Explanation:

A child in a wagon seem to fall backward when you give the wagon a sharp pull forward. It is due to Newton's third law of motion. The forward pull on wagon is called action force and the backward force is called reaction force. These two forces are equal in magnitude but they acts in opposite direction.

We need to calculate the force is needed to accelerate a sled. It can be calculated using the formula as :

F = m × a

Where

m = mass = 55 kg

a = acceleration = 1.4 m/s²

$F=55\ kg\times 1.4\ m/s^2$

F = 77 N

So, the force needed to accelerate a sled is 77 N. Hence, this is the required solution.

3 0

3 years ago

10. What is the acceleration of a 1000 kg car subject to a 500 N net force?

Simora [160]

Answer:

0.5m/s^2

Explanation:

We can use the formula [ F = ma ] but solve for "a" since that is what we are looking for.

F = ma

F/m = a

We know the net force and mass so substitute those values and simplify.

500/1000 = 0.5m/s^2

Best of Luck!

3 0

2 years ago