What is the speed of a bobsled whose distance-time graph indicates that it traveled 122m in 27s?

What type of material is good at transferring heat?

MatroZZZ [7]

The answer is A

Materials that are good conductors of thermal energy are called thermal conductors. Metals are very good thermal conductors. Materials that are poor conductors of thermal energy are called thermal insulators. Gases such as air and materials such as plastic and wood are thermal insulators

5 0

3 years ago

A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:

madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take $g = 10\; \rm m \cdot s^{-2}$ .

a. The momentum of the ball would be approximately $60\;\rm kg \cdot m \cdot s^{-1}$ two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately $\rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum $p$ of an object is equal its mass $m$ times its velocity $v$ . That is: $\vec{p} = m \cdot \vec{v}$ .

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . In other words, its velocity would become approximately $10\; \rm m \cdot s^{-1}$ more negative every second.

The initial velocity of the ball is $60\; \rm m \cdot s^{-1}$ . After two seconds, its velocity would have become $60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}$ . The momentum of the ball at that time would be around $p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}$ .

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be $0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}$ . Accordingly, the ball's momentum at that moment would be $p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ .

3 0

3 years ago

I need help someone help me

GaryK [48]

what do you need , i mean your help , let me see if i can help

5 0

3 years ago

If a tank filled with water contains a block and the height of the water above point A within the block is 0.6 meter what’s the

LenaWriter [7]

The answer is C I believe

4 0

3 years ago

A firecracker in a coconut blows the coconut into three pieces. Twopieces of equal mass fly off south and west, perpendicular to

loris [4]

Answer:

V = 13m/s

North-East (i.e. 45 degree from North)

Explanation:

This question deals with the idea of momentum.

Since the directions of a compass are fixed, we can take south as horizontal and west as vertical. Since both pieces of coconut are of equal mass, then the resultant of the two pieces would be in between South and West and the third piece would be opposite this direction and be in the North-East (i.e. 45 degree from North) direction

To find the speed of the third piece, we first find the speed on the two pieces of the coconut in terms of $V_{x}$ (horizontal component of velocity) and $V_{y}$ (vertical component of velocity)

We have

$V_{y}$ = Velocity in South direction = 18m/s

$m_{y}$ = Mass of piece in South direction = m

$V_{x}$ = Velocity in West direction = 18m/s

$m_{x}$ = Mass of piece in West direction = m

$V_{ry}$ = Velocity of Third Piece in North direction = unknown

$V_{rx}$ = Velocity of Third Piece in East direction = unknown

$m$ = Mass of third piece = 2m

$mV_{ry} = m_{y} V_{y} \\ 2mV_{ry} = m18\\ V_{ry} = 9$

$mV_{rx} = m_{x} V_{x} \\ 2mV_{rx} = m18\\ V_{rx} = 9$

Since we now know the horizontal and vertical component of the velocity of the third piece of coconut we find the resultant velocity. Since we know the direction is North-East, we can imagine a right angled triangle with base of 9 and height 9 and the hypotenuse equal to the resultant velocity .

We can simply apply the Pythagoras theorem and find the hypotenuse

$Hypotenuse^{2} = Height^{2} + Base^{2} \\ V_{3}^2 = 9^{2} + 9^{2} \\ V = 12.728 m/s$

V = 13m/s

6 0

3 years ago