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bogdanovich [222]

3 years ago

15

Your friend just challenged you to a race through an obstacle course. You know in order to beat him, you must run 30 meters with

in 60 seconds in a southern direction. What does your average velocity need to be to win the race?
.5 meters per second, south
15 meters per second, south
2 meters per second, south
90 meters per second, south

1 answer:

seropon [69]3 years ago

3 0

Answer:

Velocity = 0.5 m/s South (A)

Explanation:

You need to determine the average rate of velocity.

The equation you will use is velocity = displacement/time

The displacement is 30m South.

The time is 60 seconds.

Plug into the equation Velocity = 30m South/60 s

Velocity = 0.5 m/s South

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The first law of thermodynamics can be given as ________.

MrRa [10]

The first law of thermodynamics can be written as
$\Delta U = Q-W$
where
$\Delta U$ is the variation of internal energy of the system
$Q$ is the amount of heat absorbed by the system
$W$ is the work done by the system on the surrounding.

Using this form, the sign convention for Q and W becomes:
Q > 0 --> heat absorbed by the system (because it increases the internal energy)
Q < 0 --> heat released by the system (because it decreases the internal energy)
W > 0 --> work done by the system (for instance, an expansion: when the system expands, it does work on the surrounding, and so the internal energy decreases, this is why there is a negative sign in the formula Q-W)
W < 0 --> work done by the surrounding on the system (for instance, a compression: when the system is compressed, the surrounding is doing work on the system, and so the internal energy of the system increases)

7 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 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago

A parallel-plate capacitor made of circular plates of radius 75 cm separated by 0.15 cm is charged to a potential difference of

ElenaW [278]

Answer:

See the attached pictures for detailed steps.

Explanation:

5 0

3 years ago

If objects at the same distance suddenly decreased in mass the gravitational force between them would

inn [45]

Answer:

The Gravitational force would decrease

Explanation:

To calculate FG the formula is:

Fg=mg

if mass decreases then the gravitational force would decrease

5 0

3 years ago