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Luba_88 [7]
3 years ago
15

If f3=0 and f1=12n, what does the magnitude of f? 2 have to be for there to be rotational equilibrium? answer numerically in new

tons to two significant figures.
Physics
1 answer:
lisov135 [29]3 years ago
4 0
Not to good with math im a language and history kinda guy sorry
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The mass of an object is 60kg on the surface of the earth what will be its weight on the surface of the moon
iris [78.8K]

Answer:

Wm = 97.2 [N]

Explanation:

We must make it clear that mass and weight are two different terms, the mass is always preserved that is to say this will never vary regardless of the location of the object. While weight is defined as the product of mass by gravitational acceleration.

W = m*g

where:

m = mass = 60 [kg]

g = gravity acceleration = 10 [m/s²]

But in order to calculate the weight of the body on the moon, we must know the gravitational acceleration of the moon. Performing a search of this value on the internet, we find that the moon's gravity is.

gm = 1.62 [m/s²]

Wm = 60*1.62

Wm = 97.2 [N]

8 0
3 years ago
Which pulls harder the earth or the student
Nana76 [90]

Answer:

wha

Explanation:

5 0
3 years ago
An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. How high does it rise (v = 0 cm
Anit [1.1K]

Answer:

The value of d is 20.4 m.

(C) is correct option.

Explanation:

Given that,

Initial velocity = 20 m/s

Final velocity = 0

We need to calculate the time

Using equation of motion

v = u+gt

Where, u = Initial velocity

v = Final velocity

Put the value into the formula

t = \dfrac{20}{9.8}

t= 2.04\ sec

We need to calculate the distance

Using equation of motion

s = ut+\dfrac{1}{2}gt^2

s=0+\dfrac{1}{2}\times9.8\times(2.04)^2

s=20.4\ m

Hence, The value of d is 20.4 m.

4 0
4 years ago
How much heat is required to convert 5.53 g of ice at -12.0 ∘C to water at 24.0 ∘C? (The heat capacity of ice is 2.09 J/(g⋅∘C),
fredd [130]

Answer:

2.55 × 10³ J =2.55 kJ

Explanation:

Specific heat capacity of ice =  37.8 J / mol °C

Specific heat capacity of water = 76.0 J/ mol °C

Ice at -12 °C is converted to ice at 0 °C by absorbing heat Q₁

Ice at 0°C melts to water at 0 °C. Let Heat  absorbed during this phase change be Q₂ .

Let heat  absorbed to raise the temperature of water from 0 C to 24°C be Q₃ .

Total heat = Q = Q₁ + Q₂ + Q₃

Q₁ = (37.8 j/mol C )(5.53 g /18.01532 g/ mol )( 0-(-12)) = 139.23749 j

Q₂ =(5.53 g/18.01532 g H₂O / mol ) (6.02 x10³ j) = 1847.905 j

Q₃ = (76 j/mol C) ( (5.53 g/18.01532 g H₂O / mol )(24-0) = 559.8968 j

Total Heat required = Q = 139.23749 j + 1847.905 j + 559.8968 j

= 2547.039 j = 2.55 × 10³ J =2.55 kJ

5 0
4 years ago
Which of the following shows resistors in a parallel circuit
Alexeev081 [22]

The bottom drawing do

7 0
3 years ago
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