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Marizza181 [45]
3 years ago
13

A shopper in a supermarket pushes a cart with a force of 35Ndirected at an angle of 25 degree below the horizontal. The forceis

just sufficient to overcome various frictional forces, so thecart moves at constant speed.
(A) Find the work done by the shopper as she moves down a 50.0mlength aisle.
(B) What is the net work done on the cart? why?
(C) The shopper goes down the next aisle, pushing horizontally andmaintaining the same speed as before. If the work don't byfrictional force doesn't chance, would the shopper's applied forcebe larger, smaller, or the same? What about the work done on thecart by the shopper?
Physics
1 answer:
Delicious77 [7]3 years ago
8 0
C the shipped hose down the next aisku pushing horizontally andmanitaining the sama speed as before if the work dont byfrictional Dorce dosent chance would the shopper appplied Dorce be larger smaller or the same what about the work dont on thecart BH the shopper
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(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

3 0
3 years ago
Read 2 more answers
A cannonball is fired at a 45.0° angle and an initial velocity of 670 m/s. Assume no air resistance. What is the vertical compon
elixir [45]

Answer:

<h3>473.8 m/s; 473.8 m/s</h3>

Explanation:

Given the initial velocity U = 670m/s

Horizontal velocity Ux = Ucos theta

Vertical component of the cannon velocity Uy = Usin theta

Given

U = 670m/s

theta = 45°

horizontal component of the cannonball’s velocity = 670 cos 45

horizontal component of the cannonball’s velocity = 670(0.7071)

horizontal component of the cannonball’s velocity = 473.757m/s

Vertical component of the cannonball’s velocity = 670 sin 45

Vertical component of the cannonball’s velocity  = 670 (0.7071)

Vertical component of the cannonball’s velocity  = 473.757m/s

Hence pair of answer is 473.8 m/s; 473.8 m/s

6 0
2 years ago
HELP!!!!!! Need Help !!!
denis23 [38]

Answer:

its y

Explanation:

3 0
2 years ago
This is a Physics practice question. How do i solve it?
Elena-2011 [213]

Answer:

P = 140000 [Pa]

Explanation:

To solve this problem we must remember that pressure is defined as the relationship between Force on the area of a body.

In this particular problem, we are given the force acting on the upper surface of the block, including the force exerted by the atmospheric pressure.

P = F/A

where:

P = pressure [Pa] (units of Pascals)

F = force = 3.5*10⁴ [N]

A = area = 0.25 [m²]

P = 3.5*10⁴/0.25

P = 140000 [Pa]

7 0
3 years ago
1. A buoyant force of 790 N lifts a 214 kg sinking boat.What is the boat's<br> net acceleration?
enot [183]

Answer:

The net acceleration of the boat is approximately 6.12 m/s² downwards

Explanation:

The buoyant or lifting force applied to the boat = 790 N

The mass of the boat lifted by the buoyant force = 214 kg

The force applied to a body is defined as the product of the mass and the acceleration of the body. Therefore, the buoyant force, F, acting on the boat can be presented as follows;

Fₐ = F - W

The weight of the boat = 214 × 9.81 = 2099.34 N

Therefore;

Fₐ = 790 - 2099.34  = -1309.34 N

Fₐ = Mass of the boat × The acceleration of the boat

Given that the buoyant force, Fₐ, is the net force acting on the boat, we have;

F = Mass of the boat × The net acceleration of the boat

F = -1309.34 N =  214 kg × The net acceleration of the boat

∴ The net acceleration of the boat = -1309.34 N/(214 kg) ≈ -6.12 m/s²

The net acceleration of the boat ≈ 6.12 m/s² downwards

7 0
3 years ago
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