1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
mars1129 [50]
3 years ago
13

A moving rope (parallel to the slope) is used to pull skiers up the mountain. If the slope of the hill is 37" and friction is ne

gligible, what force must the rope provide to pull a 500 N person up the hill at a constant rate?
Physics
1 answer:
Charra [1.4K]3 years ago
5 0

Since rope is parallel to the inclined plane so here we can say that net force parallel to the person which is pulling upwards must counterbalance the component of weight of the person.

Now here we will do the components of the weight of the person

given that weight of the person = 500 N

now its components are

W_x = 500 cos37

W_y = 500 sin37

now here as we can say that one of the component is balanced here by the normal force perpendicular to plane

while the other component of the weight is balanced by the force applied on the rope

So here the force applied on the rope will be given as

F = W_y = 500 sin37

F = 300 N

so it apply 300 N force along the inclined plane

You might be interested in
____ charges are attract each other
timama [110]

Answer:

One positive and one negative

3 0
3 years ago
Read 2 more answers
A baseball is batted. It's a long fly ball. 4 seconds later the ball reaches the outfield 100 meters away and returns to the hei
Vitek1552 [10]

Answer:

25 m/s

Explanation:

First we should define the variables

T=4

Dx = 100

ay=-9.8

ax=0

We can use formula 1 from the BIG 5

x=(v+v0)t/2

By plugging in our variables we can get 100=4(v+v0)/2

Which is 50=v+v0

v=v0 since horizontal acceleration always equals zero

so 2v0 = 50

v0 = 25

8 0
3 years ago
In what subject could we see cross cutting concepts
meriva
A Framework for K–12 Science Education: Practices, Crosscutting Concepts, and Core Ideas (Framework) recommends science education in grades K–12 be built around three major dimensions: science and engineering practices, crosscutting concepts that unify the study of science and engineering through their common application across fields, and core ideas in the major disciplines of natural science.
4 0
3 years ago
What planet are people working on so they can move there
Nata [24]

Answer:

mars

Explanation:

just search it up

8 0
3 years ago
Read 2 more answers
Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.
SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

8 0
3 years ago
Other questions:
  • Carlos is baking a cake. The last step in the directions is to put the cake batter in the oven. Why does Carlos need to put the
    5·2 answers
  • What is the theory of punctuated equilibrium answers?
    15·2 answers
  • Compare the inertia of a car to the inertia of a bicycle
    9·2 answers
  • Which of the following pairs of terms directly relates to the actual brightness of a star?
    9·2 answers
  • 1 2 3 4 5 6 7 8 9 10
    7·2 answers
  • A woman walks into a carpet store wearing high-heeled shoes with a circular heel of diameter 0.987 cm. To the dismay of the stor
    5·1 answer
  • Which condition describes an object having terminal velocity?
    8·2 answers
  • Describe the arguments supporting the idea that quasars are at the distances indicated by their redshifts.
    8·1 answer
  • In the solar panel system presented in the video which of the following was necessary to generate usable electrical current for
    11·1 answer
  • Quantitative data is _____________ Lesson 1.11
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!