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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​
Physics
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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A battery of voltage 9.0 V produces a current of 0.0175 A across a circuit. What is the resistance of the circuit?
Natalka [10]

Answer:

514.288

Explanation:

V=IR

R=V/I

R=9/0.0175

8 0
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:

We have given mass of the air particle m=5\times 10^{-16}kg

Gas constant R = 8.314 J/mol-K

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

We have to find the root mean square speed of the particle

Which is given by v_{rms}=\sqrt{\frac{3RT}{m}}=\sqrt{\frac{3\times 8.314\times 300}{5\times 10^{-16}}}=38.68\times 10^8m/sec

So rms speed of the particle will be 38.68\times 10^8m/sec

5 0
3 years ago
What accounts for the two precipitation peaks in mbandaka?
slava [35]

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<span>Since Mbandaka is located at the cented of Tumba-Ngiri-Maindombe area, which is named as a Wetland of International importance, there is really a bigger chance that this place experience above 60mm precipitation in a year, temperatures averaging from 23 – 26 degrees Celsius.</span>

7 0
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
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