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sertanlavr [38]

2 years ago

8

if a person can jump 2m in earth surface how high can he jump in the moon (g of moon = 1.66m/s, g of earth = 9.8 m/s) [hint: use

w=mgh since work done and mass of a person is same everywhere

1 answer:

julia-pushkina [17]2 years ago

5 0

Answer:

$h_{moon} = 11.8\ m$

Explanation:

Since the work done is same everywhere in the universe. Hence, the work done in jumping will be same for the person on moon and earth:

$W_{moon} = W_{earth}\\\\P.E_{moon} = P.E_{earth}\\\\mg_{moon}h_{moon} = mg_{earth}h_{earth}\\\\g_{moon}h_{moon} = g_{earth}h_{earth}\\\\(1.66\ m/s^2)h_{moon} = (9.8\ m/s^2)(2\ m)\\\\h_{moon} = \frac{(9.8\ m/s^2)(2\ m)}{1.66\ m/s^2}\\\\h_{moon} = 11.8\ m$

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$514.288$

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V=IR

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2 years ago

Dust particles in air have a typical mass of 5.0 x 10-16 kg. They undergo irregular motion due to collisions with air molecules.

Mars2501 [29]

Answer:

Rms speed of the particle will be $38.68\times 10^8m/sec$

Explanation:

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Temperature is given T = $27^{\circ}C=273+27=300K$

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3 years ago

What accounts for the two precipitation peaks in mbandaka?

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2 years ago

A horizontal force of 40N is needed to pull a 60kg box across the horizontal floor at which coefficient of friction between floo

Mazyrski [523]

Answer:

Coefficient of friction is $0.068$ .

Work done is $320~J$ .

Explanation:

Given:

Mass of the box ( $m$ ): $60$ kg

Force needed ( $F$ ): $40$ N

The formula to calculate the coefficient of friction between the floor and the box is given by

$F=\mu mg...................(1)$

Here, $\mu$ is the coefficient of friction and $g$ is the acceleration due to gravity.

Substitute $40$ N for $F$ , $60$ kg for $m$ and $9.80$ m/s² for $g$ into equation (1) and solve to calculate the value of the coefficient of friction.

$40 N=\mu\times60 kg\times9.80 m/s^{2} \\~~~~~\mu=\frac{40 N}{60 kg\times9.80 m/s^{2}}\\~~~~~~~=0.068$

The formula to calculate the work done in overcoming the friction is given by

$W=Fd..........................(2)$

Here, $W$ is the work done and $d$ is the distance travelled.

Substitute $40$ N for $F$ and $8 m$ for $d$ into equation (2) to calculate the work done.

$W=40~N\times8~m\\~~~~= 320~J$

8 0

2 years ago

A 6.00-kg block starts from rest and slides down a frictionless incline. When the block has slid a distance 2.00 m, its speed is

Sauron [17]

Answer:

The angle above horizontal is 13.3°.

Explanation:

The angle can be calculated with Newton's third law:

$\Sigma F = ma$

Where:

ΣF: is the forces acting on the object

m: is the mass of the object = 6.00 kg

a: is the acceleration of the object

The only force acting on the object is the weight since there is no friction, so:

$mgsin(\theta) = ma$

$gsin(\theta) = a$ (1)

Where:

θ: is the angle

g: is the acceleration due to gravity = 9.81 m/s²

We can find the acceleration from the following kinematic equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

Where:

$v_{f}$ : is the final speed = 3.00 m/s

$v_{0}$ : is the initial speed = 0 (the block starts from rest)

d: is the distance traveled = 2.00 m

The acceleration is:

$a = \frac{v_{f}^{2}}{2d} = \frac{(3.00 m/s)^{2}}{2*2.00 m} = 2.25 m/s^{2}$

Finally, the angle is (equation 1):

$\theta = sin^{-1}(\frac{a}{g}) = sin^{-1}(\frac{2.25 m/s^{2}}{9.81 m/s^{2}}) = 13.3$

Therefore, the angle above horizontal is 13.3°.

I hope it helps you!

4 0

2 years ago