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4 years ago

14

If a person weighs 600N on earth, how much does he weigh on the moon

1 answer:

Snezhnost [94]4 years ago

5 0

Answer:

His weight would be 100 N

Explanation:

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A negative charge is at rest at the origin of an axis system. Location x is at coordinate point (2m,3m) while location y is at (

Sergeu [11.5K]

The magnitude of the E-field decreases as the square of the distance from the charge, just like gravity.

Location ' x ' is √(2² + 3²) = √13 m from the charge.

Location ' y ' is √ [ (-3)² + (-2)² ] = √13 m from the charge.

The magnitude of the E-field is the same at both locations.

The direction is also the same at both locations ... it points toward the origin.

5 0

3 years ago

A cable is 100-m long and has a cross-sectional area of 1.0 mm2. A 1000-N force is applied to stretch the cable. Young's modulus

Blizzard [7]

Answer:

1 m

Explanation:

L = 100 m

A = 1 mm^2 = 1 x 10^-6 m^2

Y = 1 x 10^11 N/m^2

F = 1000 N

Let the cable stretch be ΔL.

By the formula of Young's modulus

$Y=\frac{F\times L}{A\times\Delta L}$

$\Delta L=\frac{F\times L}{A\times\Y}$

$\Delta L=\frac{1000\times 100}{10^{-6}\times10^{11}}$

ΔL = 1 m

Thus, the cable stretches by 1 m.

4 0

3 years ago

A single loop of wire with an area of 0.0900 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendic

erica [24]

Answer:

(a) 0.0171 V

Explanation:

A = 0.09 m^2, dB/dt = 0.190 T/s

(a) According to the law of electromagntic induction

e = dФ / dt

e = A dB / dt

e = 0.09 x 0.190 = 0.0171 V

(b)

as we know

i = e / R

we can find induced current by dividing induced emf by resistance

5 0

3 years ago

An iron ball is dropped at a height of 10 m from the surface of the moon.

galben [10]

Answer:

3.51s

Explanation:

There are many students who can not get answers step by step and on time

So there are a wats up group where you can get help step by step and well explained by the trusted experts.

3 0

3 years ago

If a ball that is 10 meters above the ground is thrown horizontally at 5.51 meters per second. a. how long will it take for the

GalinKa [24]

Answer:

a. t = 1.43 s

b. d = 7.88 m

Explanation:

a. The time of flight can be found using the following equation:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height = -10 m

$y_{0}$ : is the initial height = 0

$v_{0_{y}}$ : is the initial speed in the vertical direction = 0

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

By solving the above equation for "t" we have:

$t = \sqrt{\frac{2y_{f}}{g}} = \sqrt{\frac{2*10 m}{9.81 m/s^{2}}} = 1.43 s$

Hence, the ball will hit the ground in 1.43 s.

b. The distance in the horizontal direction can be found as follows:

$x_{f} = x_{0} + v_{0}t + \frac{1}{2}at^{2}$

Where:

x₀: is the initial position in the horizontal direction = 0

a: is the acceleration in the horizontal direction = 0 (it is moving at constant speed)

$x_{f} = 5.51 m/s*1.43 s = 7.88 m$

Therefore, the ball will travel 7.88 m before it hits the ground.

I hope it helps you!

4 0

3 years ago