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3 years ago

5

Five metals that are more useful for electroplating

1 answer:

Radda [10]3 years ago

6 0

5 meteals that are more useful for electroplating carbon,seth,

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An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.

SOVA2 [1]

Answer:

Same direction: t=234s; d=6.175Km

Opposite direction: t=27.53s; d=0.73Km

Explanation:

If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be $v_{AT}=v_A-v_T$ (<em>the train must see the car advancing at a lower speed</em>), where $v_A$ is the speed of the automobile and $v_T$ the speed of the train.

So we have $v_{AT}=(95km/h)-(75Km/h)=20Km/h$ .

So the train (<em>anyone in fact</em>) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:

$t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s$

And in that time the car would have traveled (<em>relative to the ground</em>):

$d=v_At=(95Km/h)(0.065h)=6.175Km$

If they are traveling in opposite directions, <u>we have to do all the same</u> but using $v_{AT}=v_A+v_T$ (<em>the train must see the car advancing at a faster speed</em>), so repeating the process:

$v_{AT}=(95km/h)+(75Km/h)=170Km/h$

$t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s$

$d=v_At=(95Km/h)(0.00765h)=0.73Km$

5 0

3 years ago

A block is pulled along a horizontal surface at a constant speed by a force (14.1i 0 j 5.1k). The direction k is perpendicular t

Pepsi [2]

Answer:

$W =84.6\ Nm$

Explanation:

given,

F = 14.1 i + 0 j + 5.1 k

displacement = 6 m

Assuming block is moving in x- direction

we know,

dW = F dx

$\int dW = F\int dx$

$W = F\int_0^6 dx$

$W = F[x]_0^6$

$W = 14.1 \times 6$

$W =84.6\ Nm$

hence, work done by the force is equal to $W =84.6\ Nm$

7 0

3 years ago

a snooker ball travels 4 meters, diagonally across the table it travels at a speed of 5m/s. calculate how long the ball is trave

Korvikt [17]

Answer:

4 secs

Explanation:

I am right trust me

6 0

3 years ago

The sun emits electromagnetic waves with a power of 4.0 × 10²⁶ W. Determine the intensity of electromagnetic waves from the sun

Sphinxa [80]

Answer:

I_v = 2,700 W / m^2

I_m = 610 W / m^2

I_s = 16 W / m^2

Explanation:

Given:

- The Power of EM waves emitted by Sun P_s = 4.0*10^26 W

- Radius of Venus r_v = 1.08 * 10^11 m

- Radius of Mars r_m = 2.28 * 10^11 m

- Radius of Saturn r_s = 1.43 * 10^12 m

Find:

Determine the intensity of electromagnetic waves from the sun just outside the atmospheres of (a) Venus, (b) Mars, and (c) Saturn.

Solution:

- We know that Power is related to intensity and surface area of an object follows:

I = P / 4*pi*r^2

Where, A is the surface area of a sphere models the atmosphere around the planets.

a)

- The intensity at the surface of Venus is calculated as:

I_v = P_s / 4*pi*r^2_v

I_v = 4.0*10^26 / 4*pi*(1.08*10^11)^2

I_v = 2,700 W / m^2

b)

- The intensity at the surface of Mars is calculated as:

I_m = P_s / 4*pi*r^2_m

I_m = 4.0*10^26 / 4*pi*(2.28*10^11)^2

I_m = 610 W / m^2

c)

- The intensity at the surface of Saturn is calculated as:

I_s = P_s / 4*pi*r^2_s

I_s = 4.0*10^26 / 4*pi*(1.43*10^12)^2

I_s = 16 W / m^2

7 0

4 years ago

The hydrogen and helium under the clouds of Jupiter are in liquid form due to _____.

asambeis [7]

<span>high pressure produced by the clouds because its the most likely!!!!!!!!!!</span>

8 0

3 years ago

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