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Thepotemich [5.8K]
2 years ago
8

What force acts on all objects, all the time on Earth?

Physics
1 answer:
TiliK225 [7]2 years ago
7 0

The force that acts on all objects, all the time on Earth is gravitational force.

The force that surface exert on an object perpendicularly is normal reaction.

<h3>What force acts on all objects, all the time on Earth?</h3>
  • Force due to gravity is gravitational pull on objects due to its position on earth's surface.

The force due to gravity on object's is calculated by applying Newton's second law of motion as follows;

F = mg

where;

  • m is the mass of the object
  • g is acceleration due to gravity

The force that surface exert on an object perpendicularly is normal reaction.

Thus, the force that acts on all objects, all the time on Earth is gravitational force.

Learn more about force of gravity here: brainly.com/question/2537310

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You connect a 100-resistor, a 800-mH inductor, and a 10.0-uF capacitor in series across a 60.0-Hz, 120-V (peak) source. The impe
steposvetlana [31]

Answer:

Impedance, Z = 107 ohms

Explanation:

It is given that,

Resistance, R = 100 ohms

Inductance, L=800\ mH=800\times 10^{-3}\ H=0.8\ H

Capacitance, C=10\ \mu F=10\times 10^{-6}\ F=10^{-5}\ F

Frequency, f = 60 Hz

Voltage, V = 120 V

The impedance of the circuit is given by :

Z=\sqrt{R^2+(X_C-X_L)^2}...........(1)

Where

X_C is the capacitive reactance, X_C=\dfrac{1}{2\pi fC}

X_C=\dfrac{1}{2\pi \times 60\times 10^{-5}}=265.65\ \Omega

X_L is the inductive reactance, X_L={2\pi fL}

X_L={2\pi \times 60\times 0.8}=301.59\ \Omega

So, equation (1) becomes :

Z=\sqrt{(100)^2+(265.65-301.59)^2}

Z = 106.26 ohms

or

Z = 107 ohms

So, the impedance of the circuit is 107 ohms. Hence, this is the required solution.

8 0
3 years ago
NEED HELP NOW DUE TODAY Which of these statements is most likely correct about Newton's law on gravity? (2 points)
yKpoI14uk [10]
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3 years ago
A ball is projected into the air with 100 j of kinetic energy which is transformed to gravitational potential energy at the top
raketka [301]
<span>when it returns to its original level after encountering air resistance, its kinetic energy is decreased. 
In fact, part of the energy has been dissipated due to the air resistance.

The mechanical energy of the ball as it starts the motion is:
</span>E=K = 100 J
<span>where K is the kinetic energy, and where there is no potential energy since we use the initial height of the ball as reference level.
If there is no air resistance, this total energy is conserved, therefore when the ball returns to its original height, the kinetic energy will still be 100 J. However, because of the presence of the air resistance, the total mechanical energy is not conserved, and part of the total energy of the ball has been dissipated through the air. Therefore, when the ball returns to its original level, the kinetic energy will be less than 100 J.</span>
3 0
3 years ago
Which feature of a chemical equation represents the law of conservation of mass?
N76 [4]

Answer:

it shows the products of a chemical reaction to the right of the reaction arrow

4 0
2 years ago
A bicycle racer inflates their tires to 7.1 atm on a warm autumn afternoon when temperatures reached 27 °C. By morning the tempe
natulia [17]

Answer:

The required pressure is 6.4866 atm.

Explanation:

The given data : -

In the afternoon.

Initial pressure of tire ( p₁ ) = 7 atm = 7 * 101.325 Kpa =  709.275 Kpa

Initial temperature ( T₁ ) = 27°C = (27 + 273) K = 300 K

In the morning .

Final temperature ( T₂ ) = 5°C = ( 5 + 273 ) K = 278 K

Given that volume remains constant.

To find final pressure ( p₂ ).

Applying the ideal gas equation.

p * v = m * R * T

\frac{p}{T}  = constant

\frac{p_{1} }{T_{1} }  = \frac{p_{2} }{T_{2} }

p_{2}  = \frac{T_{2} }{T_{1} } *p_{1}  

p_{2}  = \frac{278}{300}  * 709.275  = 657.2615 Kpa = 6.486 atm

8 0
2 years ago
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