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stiv31 [10]
3 years ago
12

A shopper at the Fries’ market place pushes a 13.0-kg shopping cart at a constant velocity for a distance of 58.7 m on a level s

urface. He pushes in a direction 24.50 below the horizontal. A 53.8 N frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts on the cart? (b) What is the work done by the pushing force? (c) What is the work done by the frictional force? (d) What is the work done by the gravitational force?
Physics
1 answer:
Gnom [1K]3 years ago
4 0

We do a force summatory, then,

a) \sum F=0

Fcos(24.5)-F_f= 0

Fcos(24.5)-53.8= 0

F= \frac{53.8}{cos(24.5)}

F=59.12 N

(b) For the Work we need the equiation with the variable in X.

W= F (d cos(24.5))

W= 59.12 (58.7) cos 24.5

W=3158.1 J

(c) However the Frictional Force Work,

W= F_f d

W= -53.8(58.7)

W=-3158.1 J

(d) Now the work by gravity

W_g= F d (cos(90))=0

W_g=0

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Help mee pleaseee :)))
Anettt [7]

Answer:

See the explanation below.

Explanation:

Solving the first image question:

C ) The resulting force is defined by Newton's second law which tells us that the sum of the forces on a body is equal to the product of mass by acceleration. That is, there must be a force that acts on a body to produce an acceleration. If there is no acceleration it is because there are no external forces or developed by the body. And if there is no acceleration the body moves at a constant speed, in a straight line, so the response is C.

For the second image, we must remember that weight is defined as the product of mass by gravitational acceleration.

W = m*g

where:

W = weight [N]

m = mass [kg]

g = gravity acceleration [m/s²]

Now we have

m = 50 [kg]

ge = Earth gravity acceleration = 10 [m/s²]

gp = Distant planet gravity acceleration = 4 [m/s²]

We = ge*m

We = 10*50 = 500 [N]

Wp =gp*m

Wp = 4*50 = 200 [N]

Therefore the answer is D

For the third image, The mass is always going to be preserved, regardless of where the body or object is in space, its weight is the only one that changes since the gravitational force is modified. That is, the mass on the moon and on Earth will always be the same.

m = 70 [kg]

First, we must calculate the acceleration, by means of the following equation of kinematics.

v_{f} =v_{o} +a*t

where:

Vf = final velocity = 20 [m/s]

Vo = initial velocity = 0 (because stars from the rest)

a = acceleration [m/s²]

t = time = 4 [s]

20 = 0 + a*4

20 = 4*a

a = 5 [m/s²]

Now using Newton's second law which tells us that the total force acting on a body is equal to the product of mass by acceleration.

F = m*a

where:

F = force [N] (units of Newtons)

m = mass = 2 [kg]

a = acceleration = 5 [m/s²]

F = 2*5

F = 10 [N]

The body of Figure D, since a total force of 25 [N] to the left acts on it, in the rest of cases the force is zero or much less than 25 [N]

50 + 40 - 35 - 30 = F

F = 25 [N]

8 0
3 years ago
Please help me ejnfmwkefwefm i dont wanna fail
Rasek [7]

Answer:B

Explanation:

6 0
3 years ago
Interactive LearningWare 10.1 reviews the concepts involved in this problem. A spring stretches by 0.0161 m when a 3.74-kg objec
yKpoI14uk [10]

Answer:

m = 3.91 kg

Explanation:

Given that,

Mass of the object, m = 3.74 kg

Stretching in the spring, x = 0.0161 m

The frequency of vibration, f = 3.84 Hz

When the object is suspended, the gravitational force is balanced by the spring force as :

mg=kx

k=\dfrac{mg}{x}

k=\dfrac{3.74\times 9.8}{0.0161}

k = 2276.52 N/m

The frequency of vibration is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

m=\dfrac{k}{4\pi^2f^2}

m=\dfrac{2276.52}{4\pi^2\times (3.84)^2}

m = 3.91 kg

So, the mass of the object is 3.91 kg. Hence, this is the required solution.

8 0
3 years ago
If 128 v is required to push 4.00 A of current though a resistor, what is the resistance
prisoha [69]

Answer:

32.0 Ω

Explanation:

Ohm's law:

V = IR

128 V = (4.00 A) R

R = 32.0 Ω

8 0
3 years ago
Scientists believe Earth is about how many billion years old
ehidna [41]
The earth is 4.54 billion years old.
7 0
3 years ago
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