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3 years ago

1. A 2500 kg car enters a curve with a radius of 45 m If the car is moving at a speed of 35 m's, what is the

Physics

1 answer:

Aleonysh [2.5K]3 years ago

3 0

Answer:

Banana Splits with tomatoes sauce on top (but no ketchup or Mustard) in my knee with a Balenciaga private jet.

Explanation:

There is no Explanation. Look at the answer.

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Two blocks of masses mA and mB are connected by a massless spring. The blocks are moved apart, stretching the spring, and subseq

WINSTONCH [101]

Answer:

Part a)

$\frac{v_A}{v_B} = -\frac{m_B}{m_A}$

Part b)

$\frac{K_A}{K_B} = \frac{m_B}{m_A}$

Explanation:

Part a)

As we know that initially the two blocks are connected by a spring and initially stretched by some amount

Since the two blocks are at rest initially so its initial momentum is zero

since there is no external force on this system so final momentum is also zero

$m_Av_{1i} + m_Bv_{2i} = m_Av_A + m_Bv_B$

now for initial position the speed is zero

$0 = m_Av_A + m_Bv_B$

now we have

$\frac{v_A}{v_B} = -\frac{m_B}{m_A}$

Part b)

now for ratio of kinetic energy we know that the relation between kinetic energy and momentum is given as

$K = \frac{P^2}{2m}$

now for the ratio of energy we have

$\frac{K_A}{K_B} = \frac{P^2/2m_A}{P^2/2m_B}$

since we know that momentum of two blocks are equal in magnitude so we have

now we have

$\frac{K_A}{K_B} = \frac{m_B}{m_A}$

7 0

3 years ago

a projectile is launched straight up at 141 m/s . How fast is it moving at the top of its trajectory? suppose it is launched upw

BARSIC [14]

The velocity of projectile has 2 components, horizontal component vcosθ and vertical component vsinθ, where v is the velocity of projection and θ is the angle between +ve X-axis and projectile motion.

In case 1, θ = 90⁰

So horizontal component is vcos90 = 0

Vertical component at maximum height = 0

So velocity at maximum height = 0 m/s

In case 2, θ = 45⁰

So horizontal component is 141cos45 = 100m/s

Vertical component at maximum height = 0

So velocity at maximum height = 100 m/s

7 0

3 years ago

A 0.25 kg book falls off a 2 m shelf on to a 0.5 m chair. What was the change in GPE?

Snowcat [4.5K]

The equation of GPE is mgH, where m is mass, g is gravitational acceleration, and H is the height.

If we're solving for the change in GPE, then:

∆ $U_{g}$ = mg∆H

Input our given values for m and g:

∆ $U_{g}$ = 0.25 * 9.80 * ∆H

The book falls from 2 meters high to 0.5 meters high, so:

∆ $U_{g}$ = 0.25 * 9.80 * (2.0 - 0.5)

∆ $U_{g}$ = 0.25 * 9.80 * 1.5

∆ $U_{g}$ = 3.675 (J)

Adjust for significant figures:

∆ $U_{g}$ = 3.7 (J)

The change in gravitational potential energy was 3.7 (J)

If you have any questions on anything I did to get to the answer, just ask!

- breezyツ

6 0

2 years ago

THIS IS ME BEWARE UGLINESS

Nadusha1986 [10]

Answer:

DONT POST PICS OF YOURSELF ON THIS!!!!!!!!!!!!! WHAT IF I WAS A CREEPY OLD MAN? STAY SAFE ONLINE!!!!!!!!!!!!!

Explanation:

3 0

2 years ago

Read 2 more answers

What is the chemical formula for ammonium hydroxide

hoa [83]

NH4OH is the answer. Hope this helps you.

8 0

3 years ago