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2 years ago

A 50kg boy stands on rough horizontal ground. The coefficient

Physics

1 answer:

prisoha [69]2 years ago

8 0

Given :

A 50 kg boy stands on rough horizontal ground. The coefficient of static friction, us, is 0.68.

To Find :

The maximum static friction between the boy and the ground is _ N.

Solution :

We know maximum static friction is given by :

$F = \mu mg \\\\F= 0.68\times 50\times 9.8\\\\F = 333.2\ N$

Therefore, maximum static friction is 333.2 N.

Hence, this is the required solution.

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2 years ago

If a 3-kg rabbit's leg muscles act as imperfectly elastic springs, how much energy will they hold if the rabbit lands from a hei

DiKsa [7]

Answer;

- 15 J

Explanation;

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2 years ago

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masya89 [10]

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2 years ago

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snow_tiger [21]

Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

Speed of child in horizontal = 10 m/s

Length = 3.40 m

There is a rate of change of the perpendicular component of momentum.

Centripetal force acts always towards the center.

We need to calculate the magnitude of the rate of change of the child's momentum

Using formula of momentum

$\dfrac{dp}{dt}=F$

$\dfrac{dP}{dt}=\dfrac{mv^2}{r}$

Put the value into the formula

$\dfrac{dP}{dt}=\dfrac{27\times10^2}{3.40}$

$\dfrac{dP}{dt}=794.11\ N$

Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.

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2 years ago

While ice skating, you unintentionally crash into a person. Your mass is 60 kg, and you are traveling east at 8.0 m/s with respe

kaheart [24]

Answer:

6.18 m/s

Explanation:

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x-axis component form (+x east);

$P_{Miy}$ + $p_{Piy}$ + $j_{y}$ = $P_{Mfy}$ + $P_{pfy}$