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4 years ago

An object weighs 237 N on Earth's surface. What is its mass?

1 answer:

5 0

To get the object’s mass on Earth, use the formula mass = force/acceleration. On Earth, the acceleration is due to gravity, wherein the value is 9.81m/s². An N (Newton) is the measurement of force, which equal to kg m/s². Hence, 237N = 237 kg m/s².

Solution:

m = F/a
m = 237 kg m/s² ÷ 9.81 m/s²
m = 24.16kg

The object’s mass on the Earth’s surface is 24.16 kg.

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Calculate the drop voltage for a battery of internal resistance 2 ohm and emf of 24 volt

Ivenika [448]

Answer:

The drop voltage is 0.3 V

Explanation:

Electromotive Force EMF

When connecting a battery of internal resistance Ri and EMF ε to an external resistance Re, the current through the circuit is:

$\displaystyle i=\frac{\varepsilon }{R_e+R_i}$

The battery has an internal resistance of Ro=2 Ω, ε=24 V and is connected to an external resistance of Re=158 Ω. Thus, the current is:

$\displaystyle i=\frac{24 }{158+2}$

$\displaystyle i=\frac{24 }{160}$

i = 0.15 A

The drop voltage is the voltage of the internal resistance:

$V_i = i.R_i$

$V_i = 0.15*2$

$\boxed{V_i = 0.3\ V}$

The drop voltage is 0.3 V

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3 years ago

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3 years ago

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

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

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.

I hope it helps you!

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3 years ago

The mass of an object on earth is 20 kg what will be the mass of that object on moon? And why?

Alla [95]

Answer:

20kg

Explanation:

Mass is a measure of the amount of matter in an object. The mass of an object, the amount of matter inside it does not change based on location. E.g. Objects do not lose matter when they travel to the moon.

Weight, on the other hand is the downward force you exert on the ground. Weight is calculated by multiplying the mass by the gravitational field strength and changes in different places with different gravitational strength. E.g. The moon's gravitational strength is 1/5 of Earth's so the mass of the object would stay the same but the weight would be only 20% of the weight is had on earth.

Hope this helped!

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Sonbull [250]

Answer:

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Explanation:

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3 years ago

Read 2 more answers