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2 years ago

What was world war 2 about?

Physics

1 answer:

GrogVix [38]2 years ago

8 0

Answer:

it was about a group of people that thought that Jews should be elimanited but the Jews fought back

Explanation:

You might be interested in

What does training for hypertrophy do to your body

VikaD [51]

Answer:

The answer is A.

Explanation:

Hypertrophy is an increase and growth of muscle cells. Hypertrophy refers to an increase in muscular size achieved through exercise. When you work out, if you want to tone or improve muscle definition, lifting weights is the most common way to increase hypertrophy.

Hope that helped.

5 0

2 years ago

Read 2 more answers

A motorcycle is traveling east at a rate of 56mph.it take 6.7 s for it to decrease its speed to 35mph. How far does the motorcyc

Phantasy [73]

Initial speed = 56mph
Final speed = 35mph

Time taken = 6.7seconds...

Converting the time to hour.. Divide by 3600..
= 6.7/3600

=0.00186hour..

Acceleration = v-u/t

a = 35-56/0.00186
a = -11283.6mph²

The negative sign shows that it decelerated...

V² = u²+2as

(35)² = (56)² + 2×-11283.6×s
Where s is the distance covered within that time...

1225 = 3136 - 22567.2s

22567.2s = 3136-1225

22567.2s = 1911

S = 1911/22567.2

S = 0.08468miles...

But at the end of the question we were made to understand that 1miles = 5280ft

Therefore 0.08468miles = (0.08468×5280)ft

= 447. 11feets...

Which is approximately 447ft.....

Hope this helped.... ?

8 0

3 years ago

A book is sitting on a desk. What best describes the normal force acting on the book?

maxonik [38]

Gravitational force is the result of the earth pulling down on the book, so the normal force is best described as the earth or more accurately the table pushing up on the book

8 0

3 years ago

Read 2 more answers

The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were obser

ELEN [110]

Answer:

a) v² = G M R³, b) T = 2π / $\sqrt{GMR}$ , c) $m \sqrt{GMR^5 }$

Explanation:

a) The kinetic energy is

K = ½ m v²

to find the velocity let's use Newton's second law

F = m a

acceleration is centripetal

a = v² / R

force is the universal force of attraction

F = G m M / r²

we substitute

G m M R² = m v² R

v² = G M R³

the kinetic energy is

K = ½ m G M R³

b) angular and linear velocity are related

v = w R

w = v / R

w = $\frac{\sqrt{GMR^3 }}{R}$

w = $\sqrt{GMR}$

the angular velocity is related to the period

w = 2π / T

T = 2π / w

we substitute

T = 2π / $\sqrt{GMR}$

c) the angular moeomto is

L = m v r

L = m RA G M R³ R

L = $m \sqrt{GMR^5 }$

4 0

3 years ago

Two carts, one twice as heavy as the other, are at rest on a horizontal track. A person pushes each cart for 8 s. Ignoring frict

GalinKa [24]

Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = $m$

Mass of the heavier cart = $2m$

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ $KE = \frac{mv^2}{2}$

⇒ $KE = \frac{mv^2}{2}\times \frac{m}{m}$

⇒ $KE = \frac{m^2v^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(mv)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}$

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

KE (K) of mass m.

⇒ $K=\frac{(F\times t)^2}{2m}$ ...equation (i)

KE (K1) of mass 2m.

⇒ $K_1=\frac{(F\times t)^2}{2\times 2m}$

⇒ $K_1=\frac{(F\times t)^2}{4m}$ ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ $\frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}$

⇒ $\frac{K_1}{K} =\frac{2m}{4m}$

⇒ $\frac{K_1}{K} =\frac{2}{4}$

⇒ $\frac{K_1}{K} =\frac{1}{2}$

⇒ $K_1=\frac{K}{2}$

⇒ $K_1=\frac{1}{2}K$

The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

7 0

2 years ago