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3 years ago

Difference between rest and motion?

Physics

1 answer:

Yuki888 [10]3 years ago

6 0

Answer:

Rest and motion are relative terms. In simple terms, an object that changes its position is said to be in motion while the opposite action causes an object to be at rest.

Explanation:

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Adam and Natalie push on a cart using 150 N of force.The cart does not move.What is static friction on the cart?

Roman55 [17]

When force is applied on an object the object will not slide

So here object will remain in equilibrium after applying the force

So in this case the condition of equilibrium is maintained due to friction force that is applied on the object.

So here for equilibrium be can say

$F_{net} = 0$

also we know that

$F_{net} = F - F_f$

we are given that

$F = 150 N$ ( applied force on the box)

$F_{net} = 150 - F_f = 0$

$F_f = 150 N$

So here this friction force on the cart is known as static friction

$F_s = 150 N$

4 0

4 years ago

Read 2 more answers

Which cell structure is found in plant cells, but not in animal cells?

andre [41]

Answer:

chloroplasts

Explanation:

5 0

3 years ago

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A star is known to be moving at 8.46km/s toward the earth. If you observe the spectral line to be at 5.02nm, at what wavelength

anastassius [24]

Answer:

$\lambda_x=5.019858nm$

Explanation:

From the question we are told that

Speed of star $S=8.46km/s$

Distance of spectral line $\lambda_0= 5.02nm$

Generally the equation for wavelength with respect to spectral lines is mathematically given by

$\lambda=\lambda _0 *\frac{v}{c}$

where

$\lambda_0= length\ of\ spectral\ line$

$c=The\ speed\ of\ light$

$v= speed\ of\ moving\ object$

therefore

$\lambda=5.02*10^{-9} *\frac{8.46*10^{12}}{299 792 458*10^9}$

$\lambda=1.42*10^{-4} nm$

Generally the equation for new wavelength is mathematically given as

$\lambda_x=\lambda _0-\lambda$

$\lambda_x=5.02 nm-1.42*10^{-4} nm$

$\lambda_x=5.02-1.42*10^{-4}$

Therefore

$\lambda_x=5.019858nm$

5 0

4 years ago

If the water in a lake is everywhere at rest, what is the pressure as a function of distance from the surface? The air above the

Rudiy27

Answer:

101325 + 10055.25h

h = 10.1 m

Explanation:

the pressure at sea level = 1 atm = 101325 Pa

density of sea water = 1025 kg/ m^(3)

pressure due to fluid height = pgh

Absolute pressure = 101325 + 1025*9.81*h

= 101325 + 10055.25h

where h= 0 at sea level at increases downwards

101325 = 1025* 9.81* h

h = 10.1 m

Explanation:

7 0

3 years ago

A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?

dexar [7]

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

3 0

3 years ago