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Anastasy [175]
3 years ago
11

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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2 years ago
3 bulbs are in series and the same 3 bulbs are in parallel with the same battery. Which bulbs will be dimmer?
Marina86 [1]

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3 0
1 year ago
A block of lead has dimensions of 4.50 cm by 5.20 cm by 6.00 cm. The block weighs 1587 g. From this information, calculate the d
mamaluj [8]

Answer:

11.3 g/cm^3

Explanation:

density = mass/volume

volume of rectangular prism = length * width * height

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2 years ago
Inez uses hairspray on her hair each morning before going to school. The spray spreads out before reaching her hair partly becau
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Answer:

7.0\cdot 10^{-13}C

Explanation:

The magnitude of the electrostatic force between two charged objects is

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

The force is attractive if the charges have opposite sign and repulsive if the charges have same sign.

In this problem, we have:

r=0.070 cm =7\cdot 10^{-4} m is the distance between the charges

q_1=q_2=q since the charges are identical

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Re-arranging the equation and solving for q, we find the charge on each drop:

F=\frac{kq^2}{r^2}\\q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(9.0\cdot 10^{-9})(7\cdot 10^{-4})^2}{8.99\cdot 10^9}}=7.0\cdot 10^{-13}C

8 0
2 years ago
Help me friends thank you
xenn [34]

Explanation:

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