A 1000kg car uses a breaking force of 10,000N to stop in two second. What impulse acts on the car?

Two UFPD are patrolling the campus on foot. To cover more ground, they split up and begin walking in different directions. Offic

miss Akunina [59]

Answer:

0.256 hours

Explanation:

<u>Vectors in the plane </u>

We know Office A is walking at 5 mph directly south. Let $X_A$ be its distance. In t hours he has walked

$X_A=5t\ \text{miles}$

Office B is walking at 6 mph directly west. In t hours his distance is

$X_B=6t\ \text{miles}$

Since both directions are 90 degrees apart, the distance between them is the hypotenuse of a triangle which sides are the distances of each office

$D=\sqrt{X_A^2+X_B^2}$

$D=\sqrt{(5t)^2+(6t)^2}$

$D=\sqrt{61}t$

This distance is known to be 2 miles, so

$\sqrt{61}t=2$

$t =\frac{2}{\sqrt{61}}=0.256\ hours$

t is approximately 15 minutes

3 0

3 years ago

Assume a beam of light hits the boundary separating medium 1, with index of refraction n1 and medium 2, with index of refraction

Reptile [31]

Answer:

option C

Explanation:

Given,

Refractive index of medium 1 = n₁

Refractive index of medium 2 = n₂

For total internal reflection to take place light should move from denser medium to the rarer medium.

Here Total internal reflection take place at the boundary of medium 1 and medium 2 so, the refractive index of medium 1 is more than medium 2

n₁ > n₂

The correct answer is option C

3 0

3 years ago

Two basketballs of equal mass are rolling toward each other at constant velocities. The first basketball (B1) has a velocity of

slamgirl [31]

$v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)$

<u>Explanation:</u>

Velocity of B₁ = 4.3m/s

Velocity of B₂ = -4.3m/s

For perfectly elastic collision:, momentum is conserved

$m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2$

where,

m₁ = mass of Ball 1

m₂ = mass of Ball 2

v₁ = initial velocity of Ball 1

v₂ = initial velocity of ball 2

v'₁ = final velocity of ball 1

v'₂ = final velocity of ball 2

The final velocity of the balls after head on elastic collision would be

$v'_2 = \frac{2m_1}{m_1+m_2} v_1 - \frac{m_1-m_2}{m_1+m_2} v_2\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} v_1 + \frac{2m_2}{m_1+m_2} v_2$

Substituting the velocities in the equation

$v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)$

If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.

5 0

3 years ago

Energy that is transferred from a warmer object to a cooler object is called

alukav5142 [94]

Answer:

Heat

Explanation:

Energy that is transferred from a warmer object to a cooler object is called heat.

3 0

2 years ago

Read 2 more answers

6. Willingness to take turns is one way we can express our attitudes in

NNADVOKAT [17]

The answer is A ..........

5 0

3 years ago