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andrezito [222]

3 years ago

12

Clive's car has a top speed of 150 mph. He attaches a roof box to his car. How will this affect its top speed? Explain your answ

er.

1 answer:

vovikov84 [41]3 years ago

5 0

It will affect the top speed by the velocity

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A 50-g chunk of a common metal at 80°C is placed into a calorimeter with 50 g of water at 50°C. At thermal equilibrium, which ef

Blizzard [7]

Answer:

The metal will have changed temperature more than the water.

Explanation:

As we know that when two objects are mixed at different temperature then due to heat exchange between the two they both reach to same final temperature

This is known as thermal equilibrium

So here we have

Heat given by metal = heat absorbed by water

so we have

$m s_{metal} \Delta T_1 = m s_{water}\Delta T_2$

since mass of metal is same as that of water

so we have

$s_{metal} \Delta T_1 = s_{water}\Delta T_2$

so here we know that

$s_{water} > s_{metal}$

so temperature change in water must be smaller than that of metal

8 0

3 years ago

Why do different atoms or molecules preferentially absorb different wavelengths of light?

Paladinen [302]

Answer:

The electrons in an atom can only occupy certain allowed energy levels. ... Only certain energy levels are allowed, so only certain transitions are possible and hence specific wavelengths are emitted when an electron drops to a lower energy level.

6 0

3 years ago

A ball is rolled uphill a distance of 13 meters before it slows, stops, and begins to roll back. The ball rolls downhill 27 mete

yarga [219]

Magnitude of displacement means the total displacement. Basically taking your final distance and subtracting it from your initial distance so it would be 14m

8 0

3 years ago

A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.

Alexxandr [17]

Answer:

Kinetic energy of the projectile at the vertex of the trajectory: $900\; {\rm J}$ .

Work done when firing this projectile: $2500\; {\rm J}$ .

Explanation:

Since the drag on this projectile is negligible, the horizontal velocity $v_{x}$ of this projectile would stay the same (at $30\; {\rm m\cdot s^{-1}}$ ) throughout the flight.

The vertical velocity $v_{y}$ of this projectile would be $0\; {\rm m\cdot s^{-1}}$ at the vertex (highest point) of its trajectory. (Otherwise, if $v_{y} > 0$ , this projectile would continue moving up and reach an even higher point. If $v_{y} < 0$ , the projectile would be moving downwards, meaning that its previous location was higher than the current one.)

Overall, the velocity of this projectile would be $v = 30\; {\rm m\cdot s^{-1}}\!$ when it is at the top of the trajectory. The kinetic energy $\text{KE}$ of this projectile (mass $m = 2.0\; {\rm kg}$ ) at the vertex of its trajectory would be:

$\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (30\; {\rm m\cdot s^{-1}})^{2} \\ &= 900\; {\rm J} \end{aligned}$ .

Apply the Pythagorean Theorem to find the initial speed of this projectile:

$\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \left(\sqrt{900 + 1600}\right)\; {\rm m\cdot s^{-1}} \\ &= 50\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Hence, the initial kinetic energy $\text{KE}$ of this projectile would be:

$\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (50\; {\rm m\cdot s^{-1}})^{2} \\ &=2500\; {\rm J} \end{aligned}$ .

All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be $2500\; {\rm J}$ .

7 0

2 years ago

Two horses on opposite sides of a narrow stream are pulling a barge up the stream. Each hors pulls with a force of 720N. The rop

Sonja [21]

For the answer to the question above, each horse's force forms a right angle triangle with the barge and subtends an angle of 60/2 = 30°. The resultant in the direction of the barge's motion is:
Fx = Fcos(∅)
We can multiply this by 2 to find the resultant of both horses.
Fx = 2Fcos(∅)
Fx = 2 x 720cos(30)
Fx = 1247 N

8 0

3 years ago