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nevsk [136]
3 years ago
5

Help me with 5 daily life situations to find the resultant force.

Physics
1 answer:
ser-zykov [4K]3 years ago
3 0

Answer:

Ok! :)

Explanation:

"A resultant force is the single force and associated torque obtained by combining a system of forces and torques acting on a rigid body. The defining feature of a resultant force, or resultant force-torque, is that it has the same effect on the rigid body as the original system of forces."

Also known as net force.

Ball pushed one was with 5 N right, other way 1 N left.

Net force = 4 N to right.

Box thrown 10 up, pushed 20 down. Net force = 10 down.

Can't think of 3 more...   :)

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5 0
2 years ago
Charge is flowing through a conductor at the rate of 420 C/min. If 742 J. of electrical energy are converted to heat in 30 s., w
Y_Kistochka [10]

Answer:

3.53 V

Explanation:

Electric charge: The is the rate of flow of electric charge along a conductor.

The S.I unit of electric charge is C.

Mathematically it is expressed as,

Q = It ............................ Equation 1

Where Q = electric charge, I = current, t = time.

I = Q/t.......................... Equation 2

From the question, charge flows through the conductor at the rate of 420 C/mim

Which means in 1 min, 420 C of charge flows through the conductor.

Hence,

Q = 420 C, t = 1 min = 60 seconds

Substitute into equation 2

I = 420/60

I =7 A

Also

P = VI......................... Equation 3

Where P = power, V = potential drop, I = current.

V = P/I................... Equation 4

Note: Power = Energy/time

From the question, P = 742/30 = 24.733 W. and I = 7 A.

Substitute these values into equation 4

V = 24.733/7

V = 3.53 V

Hence the potential drop across the conductor =  3.53 V

4 0
3 years ago
When a voltage difference is applied to a piece of metal wire, a current flows through it. if this metal wire is now replaced wi
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The resistance of the cylindrical wire is R=\frac{\rho l}{A}.

Here R is the resistance, l is the length of the wire and A is the area of cross section. Since the wire is cylindrical A=\frac{\pi d^2}{4} .

Comparing two wires,

R_1=\frac{\rho_1 l}{A_1} \\ R_2=\frac{\rho_2 l}{A_2}

Dividing the above 2 equations,

\frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2}  \frac{A_2 }{A_1}  \\ \frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2}  \frac{d_2^2 }{d_1^2}  \\

Since d_2=2d_1

The above ratio is

\frac{R_1}{R_2}=\frac{1.68(10^{-8})  }{1.59(10^{-8}) } (4)\\ \frac{R_1}{R_2}=4.2264

We also have,

\frac{E/R_1}{E/R_2} =\frac{I_1}{I_2} \\ I_2=\frac{R_1}{R_2}I_1 \\ I_2=4.23I_1

The current through the Silver wire will be 4.23 times the current through the original wire.

8 0
3 years ago
Case 1: A 0.780-kg silver pellet with a temperature of 85 oC is added to 0.150 kg of water in a copper cup of unknown mass. The
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Answer: The silver pellet will release heat

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Extremely large main sequence stars consume their fuel quickly and burn hot and bright. As they consume all their hydrogen, they
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Answer:  Their temperature decreases dramatically, but their luminosity increases only slightly.

Explanation: Edmentum answer

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3 years ago
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