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3 years ago

12

Questions to consider:

1 answer:

Margaret [11]3 years ago

7 0

Answer: 2352 J

Explanation:

A body's<u> gravitational potential energy</u> $U$ depends on its position and is mathematically expressed as follows:

$U=mgh$

Where:

$m=60 kg$ is the mass of the skater

$g=9.8 m/s^{2}$ is the acceleration due gravity

$h=4 m$ is the skater's current height

Solving:

$U=(60 kg)(9.8 m/s^{2})(4 m)$

$U=2352 J$

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Given: F = k· m. g<br> Solve for "k"

gulaghasi [49]

Answer:

$F = kmg \\ k = \frac{F}{mg}$

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3 years ago

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A 41.0 g marble moving at 2.30 m/s strikes a 25.0 g marble at rest. What is the speed of each marble immediately after the colli

Gre4nikov [31]

Answer:

speed of each marble after collision will be 1.728 m/sec

Explanation:

We have given mass of the marble $m_1=41gram=0.041kg$

Velocity of marble $v_1=2.30m/sec$

Its collides with other marble of mass 25 gram

So mass of other marble $m_2=25gram=0.025kg$

Second marble is at so $v_2=0m/sec$

We have to find the velocity of second marble

From momentum conservation we know that

$m_1v_1+m_2v_2=(m_!+m_2)v$ , here v is common velocity of both marble after collision

So $0.041\times 2.30+0.025\times 0=(0.041+0.025)v$

v = 1.428 m /sec

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3 years ago

Please give a explanation if possible Tysm!

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2 years ago

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A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h

Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

$v_{1}=\dfrac{3.0\times10^{3}}{3600}$

$v_{1}=0.833\ m/s$

The force F₁is constant acceleration is also a constant.

$F_{1}=ma_{1}$

We need to calculate the acceleration

Using formula of acceleration

$a_{1}=\dfrac{v}{t}$

$a_{1}=\dfrac{0.833}{10}$

$a_{1}=0.083\ m/s^2$

Similarly,

$F_{2}=ma_{2}$

For total force,

$F_{3}=F_{2}+F_{1}$

$ma_{3}=ma_{2}+ma_{1}$

The speed of second tugboat is

$v=\dfrac{11\times10^{3}}{3600}$

$v=3.05\ m/s$

We need to calculate total acceleration

$a_{3}=\dfrac{v}{t}$

$a_{3}=\dfrac{3.05}{10}$

$a_{3}=0.305\ m/s^2$

We need to calculate the acceleration a₂

$0.305=a_{2}+0.083$

$a_{2}=0.305-0.083$

$a_{2}=0.222\ m/s^2$

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

$\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}$

$\dfrac{F_{1}}{F_{2}}=3.7$

$F_{1}=3.7F_{2}$

Hence, The magnitude of F₁ is 3.7 times of F₂

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