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3 years ago

Adam and Natalie push on a cart using 150 N of force.The cart does not move.What is static friction on the cart?

Physics

2 answers:

jeyben [28]3 years ago

5 0

Answer:

150N

Explanation:

Static friction is the friction that exists between two surfaces in contact. For a static body, the applied force (Fm) used to move the body is always equal to the frictional force(Ff) since frictional force is a force of opposition that tends to oppose the moving force acting on a static body.

According to the question, if Adam and Natalie push on a cart using 150N force, then the force of friction on the cart will be equal and opposite to the applied force and will also be equal to 150N

Roman55 [17]3 years ago

4 0

When force is applied on an object the object will not slide

So here object will remain in equilibrium after applying the force

So in this case the condition of equilibrium is maintained due to friction force that is applied on the object.

So here for equilibrium be can say

$F_{net} = 0$

also we know that

$F_{net} = F - F_f$

we are given that

$F = 150 N$ ( applied force on the box)

$F_{net} = 150 - F_f = 0$

$F_f = 150 N$

So here this friction force on the cart is known as static friction

$F_s = 150 N$

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Read 2 more answers

A 8.2-V battery is connected in series with a 38-mH inductor, a 150-Î© resistor, and an open switch.A 8.2-V battery is connected

tigry1 [53]

Answer:

(A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.

Explanation:

Given that,

Voltage = 8.2 V

Inductor = 38 mH

Resistance = 150 Ω

Time t = 0.110 ms

The battery has negligible internal resistance, so that the total resistance in the circuit is 150 ohms. Then use this equation for current at time t in terms of inductance

We need to calculate the current

Using formula of current

$I(t)=\dfrac{V}{R}\times(1-e^{-t\times\dfrac{R}{L}})$