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EleoNora [17]
3 years ago
13

Which requires more energy: lifting a 50 kg sack vertically 2 meters or lifting a 25 kg sack vertically 4 meters?

Physics
2 answers:
gtnhenbr [62]3 years ago
5 0

Answer: C) both require the same amount of work

Explanation:

If you think about 25 is half of 50 and 2 is half of 4 and 50 times 2 =100 and

25 times 4 =100

elena-s [515]3 years ago
3 0

Answer:

c. Both require the same amount of work

Explanation:

GPE = mgh

50 * 10 * 2 = 1000 J

25 * 10 * 4 = 1000 J

Therefore, both are same and require the same amount of work.

Thenk and mark me brainliest :)

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The velocity of an object includes its speed and
a_sh-v [17]
Velocity is a vector meaning it includes an objects speed and direction.
3 0
3 years ago
A 27.0-m steel wire and a 48.0-m copper wire are attached end to end and stretched to a tension of 145 N. Both wires have a radi
algol13

Answer:

The time taken by the wave to travel  along the combination of two wires is 458 ms.

Explanation:

Given that,

Length of steel wire= 27.0 m

Length of copper wire = 48.0 m

Tension = 145 N

Radius of both wires = 0.450 mm

Density of steel wire \rho_{s}= 7.86\times10^{3}\ kg/m^{3}

Density of copper wire \rho_{c}=8.92\times10^{3}\ kg/m^3

We need to calculate the linear density of steel wire

Using formula of linear density

\mu_{s}=\rho_{s}A

\mu_{s}=\rho_{s}\times\pi r^2

Put the value into the formula

\mu_{s}=7.86\times10^{3}\times\pi\times(0.450\times10^{-3})^2

\mu_{s}=5.00\times10^{-3}\ kg/m

We need to calculate the linear density of copper wire

Using formula of linear density

\mu_{c}=\rho_{s}A

\mu_{c}=\rho_{s}\times\pi r^2

Put the value into the formula

\mu_{c}=8.92\times10^{3}\times\pi\times(0.450\times10^{-3})^2

\mu_{c}=5.67\times10^{-3}\ kg/m

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

v_{s}=\sqrt{\dfrac{T}{\mu_{s}}}

v_{s}=\sqrt{\dfrac{145}{5.00\times10^{-3}}}

v_{s}=170.3\ m/s

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

v_{c}=\sqrt{\dfrac{T}{\mu_{c}}}

v_{c}=\sqrt{\dfrac{145}{5.67\times10^{-3}}}

v_{c}=159.9\ m/s

We need to calculate the time taken by the wave to travel  along the combination of two wires

t=t_{s}+t_{c}

t=\dfrac{l_{s}}{v_{s}}+\dfrac{l_{c}}{v_{c}}

Put the value into the formula

t=\dfrac{27.0}{170.3}+\dfrac{48.0}{159.9}

t=0.458\ sec

t=458\ ms

Hence, The time taken by the wave to travel  along the combination of two wires is 458 ms.

4 0
4 years ago
A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is at rest at a stopligt. The car comes to a
Burka [1]

Answer:

5 m/s, moving to the South.

Explanation:

Parameters given:

Mass of car, m = 1500 kg

Initial velocity of car, u = 15 m/s

Mass of truck, M = 4500 kg

Initial velocity of truck, v = 0 m/s (Truck is at rest)

Final velocity of car, U = 0 m/s (Car comes to a stop)

Final velocity of truck = V

Because the collision is elastic, we can apply the principle of conservation of momentum, we have that:

Total initial momentum = Total final momentum

m*u + M*U = m*v + M*V

(1500 * 15) + (4500 * 0) = (1500 * 0) + (4500 * V)

22500 + 0 = 0 + 00V

=> V = 22500/4500

V = 5 m/s

The velocity carries a positive sign, hence, it's moving in the same direction as the car was moving initially.

That is, it's moving to the South.

8 0
3 years ago
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7 0
4 years ago
1) A pendulum is configured to have a period of 2 seconds.
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3 years ago
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