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julia-pushkina [17]

3 years ago

8

A railroad car of mass 3.45 ✕ 104 kg moving at 2.60 m/s collides and couples with two coupled railroad cars, each of the same ma

ss as the single car and moving in the same direction at 1.20 m/s. (a) What is the speed of the three coupled cars after the collision? 1.27 Incorrect: Your answer is incorrect. m/s (b) How much kinetic energy is lost in the collision?

1 answer:

Alex3 years ago

8 0

Answer:

a) 1.67 m/s

b) 23kJ

Explanation:

We need to apply the linear momentum conservation formula, that states:

$m1*v_{o1}+m2*v_{o2}=m1*v_{f1}+m2*v_{f2}$

in this case:

$3.45*10^4kg*2.60m/s+2*3.45*10^4kg*1.20m/s=3*m1*v_{f}\\v_f=1.67m/s$

the initital kinetic energy is:

$K_i=\frac{1}{2}*3.45*10^4kg*(2.60m/s)^2+2(\frac{1}{2}*3.45*10^4kg*(1.20m/s)^2\\K_i=167kJ$

and the final:

$K_f=3*\frac{1}{2}*3.45*10^4kg*(1.67m/s)^2\\K_f=144kJ$

The energy lost is given by:

$E_l=|K_f-K_i|\\E_l=23kJ$

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Which part of the brain controls body temperature

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3 years ago

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A baseball thrown by a pitcher is hit by a batter. At the moment when the ball hits the bat, the force exerted on the bat by the

Slav-nsk [51]

Newton's Laws of motion describe the motion of an object based on the applied force

The correct option that gives the relationships between the forces is the the option;

$\underline{F_{ball}}$ <u> on bat</u> = $\underline{F_{bat}}$ <u> on ball</u>

<em>(Either option A or B without the minus symbol before </em> $F_{bat}$ <em>, likely typographical error)</em>

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Reason:

Force exerted on the bat by the ball = $F_{ball}$

Force exerted on the ball by the bat = $F_{bat}$

Given that the batter hits the ball with the bat, the force exerted by the bat on the ball, $F_{bat}$ , is reacted to by the force of the ball acting on the bat, $F_{ball}$

According to Newton's third law of motion, action and reaction are equal and opposite. Therefore, at the moment when the ball hits the bat, we have;

$\underline{F_{ball}}$ <u> on bat</u> = $\underline{F_{bat}}$ <u> on ball</u>

<em>Where the force of the bat is high, the ball is accelerated to travel at high speed</em>

Learn more Newton's Laws of motion here:

brainly.com/question/24522313

8 0

3 years ago

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

julsineya [31]

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$ .

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .

$v^2 - u^2 = 2\, g \cdot h$ .

In this case, $v$ would be the velocity of the ball just before it hits the ground. Solve for

$v^2 = 2\, a\, x + u^2$ .

$\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}$ .

3 0

3 years ago

A helicopter flies with an air speed of 175 km/h, heading south. The wind is blowing at 85 km/h to the east relative to the grou

spayn [35]

Answer:

154° at 195 km/h

Explanation:

The helicopter is moving south at 175 km/h, relative to the wind.

But the wind is moving east at 85 km/h, relative to the ground.

This means that the helicopter is moving south east relative to the ground.

Every hour, the helicopter will move 175 km to the south and 85 km to the east, relative to the ground.

This means that we can determine the speed and direction of the helicopter using a right triangle and simple trigonometry.

Refer to the triangle b1.

The distance traveled by the helicopter in 1 hour is denoted by d.

d is the hypotenuse of the right triangle.

Using the Pythagorean Theorem, we can calculate d to be 195 km (rounded to 3 s. f.)

Hence the helicopter is traveling at 195 km/h relative to the ground.

To calculate the direction we use,

tan (x) = opposite/adjacent = 85/175

So the angle x is,

$x = arctan (\frac{85}{175} )$ = 25.9°

Relative to the North, the helicopter is moving at 180° - 25.9° = 154° (rounded to 3 s. f.)

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2 years ago

A 43.9-g piece of copper (CCu= 0.385 J/g°C) at 135.0°C is plunged into 254 g of water at 39.0°C. Assuming that no heat is lost t

Semmy [17]

Answer:

$T = 40.501\,^{\textdegree}C$

Explanation:

The interaction of the piece of copper and water means that the first one need to transfer heat in order to reach a thermal equilibrium with water. Then:

$-Q_{out,Cu} = Q_{in,H_{2}O}$

After a quick substitution, the expanded expression is:

$-(43.9\,g)\cdot (0.385\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-135^{\textdegree}C) = (254\,g)\cdot (4.187\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-39\,^{\textdegree}C)$

$-16.902\,\frac{J}{^{\textdegree}C}\cdot (T-135^{\textdegree}C) = 1063.498\,\frac{J}{^{\textdegree}C} \cdot (T-39^{\textdegree}C)$

$43758,192\,J = 1080.4\,\frac{J}{^{\textdegree}C}\cdot T$

The final temperature of the system is:

$T = 40.501\,^{\textdegree}C$

8 0

3 years ago

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