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3 years ago

What if multiple forces act on an object in equilibrium??

Physics

1 answer:

tatuchka [14]3 years ago

4 0

Unless if all forces cancel each other out , the object will no longer be in equilibrium

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Three identical resistors are connected in parallel to a battery. If the current of 12. A flows from the battery, how much curre

Doss [256]

Answer:

4 A

Explanation:

We are given that

$R_1=R_2=R_3=4\Omega$

I=12 A

We have to find the current flowing through each resistor.

We know that in parallel combination current flowing through different resistors are different and potential difference across each resistor is same.

Formula :

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

Using the formula

$\frac{1}{R}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}$

$R=\frac{4}{3}\Omega$

$V=IR$

Substitute the values

$V=12\times \frac{4}{3}=16 V$

$I_1=\frac{V}{R_1}=\frac{16}{4}=4 A$

$I_1=I_2=I_3=4 A$

Hence, current flows through any one of the resistors is 4 A.

7 0

2 years ago

A rectangular conducting loop of wire is approximately half-way into a magnetic field B (out of the page) and is free to move. S

Black_prince [1.1K]

Answer:

. The loop is pushed to the right, away from the magnetic ﬁeld

Explanation

This decrease in magnetic strength causes an opposing force that pushes the loop away from the field

8 0

2 years ago

A spring has a equilibrium length of 10.0 cm. When a force of 40.0 N is applied to the spring, the spring has a length of 14.0 c

mote1985 [20]

Answer:

The value of the spring constant of this spring is 1000 N/m

Explanation:

Given;

equilibrium length of the spring, L = 10.0 cm

new length of the spring, L₀ = 14 cm

applied force on the spring, F = 40 N

extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm

From Hook's law

Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.

F ∝ e

F = ke

where;

k is the spring constant

k = F / e

k = 40 / 0.04

k = 1000 N/m

Therefore, the value of the spring constant of this spring is 1000 N/m

7 0

3 years ago

A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24

alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

$I=0.8*m*r^{2}$

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

$N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}$

$m=40kg$

moment of inertia

$M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}$

Kinetic energy of the rotation motion

$K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J$

Kinetic energy translational

$K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J$

Total kinetic energy

$K=3317.79J+4147.2J\\K=7464.99J$

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

$K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m$

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3 years ago

Does mars has a bulge near its equator ?

Keith_Richards [23]

Yessir it sure does

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3 years ago