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
blagie [28]
2 years ago
7

Holding onto a tow rope moving parallel to a frictionless ski slope, a 61.8 kg skier is pulled up the slope, which is at an angl

e of 6.8° with the horizontal. What is the magnitude Frope of the force on the skier from the rope when (a) the magnitude v of the skier's velocity is constant at 2.47 m/s and (b) v = 2.47 m/s as v increases at a rate of 0.109 m/s2?

Physics
1 answer:
nata0808 [166]2 years ago
8 0

Answer:

a) Frope= 71.7 N

b) Frope=6.7 N

Explanation:

In the figure the skier is simulated as an object, "a box".

a) At constant velocity we can say that the object is in equilibrium, so we apply the Newton's first law:

∑F=0

Frope=w*sen6.8°

Frope=71.71N

Take into account that w is the weight that is calculated as mass per gravitiy constant:

w=m*g

w=61.8Kg*9.8\frac{m}{s^{2} }

w=605.64N

b) In this case the system has an acceleration of 0.109m/s2.  Then, we apply Newton's second law of motion:

F=m*a

F=61.8Kg*0.109m/s2

Frope=6.73N

You might be interested in
Two objects were lifted by a machine. One object had a mass of 5000 kg, and was lifted at a speed of 2 m/sec. The other had a ma
Leokris [45]
Equation for ke = 1/2mv^2
1) ke = 1/2 x 5000 x (2x2)
= 10,000J
2) ke = 1/2 x 4000 x (3x3)
= 18,000J
So Object 2 has more Kinetic Energy
3 0
3 years ago
2. A 75 kg runner accelerates from 0.00 m/s to 10.0 m/s in 1.5 seconds.
raketka [301]
A :-) A.a.) Given - u = 0.00 m/s
v = 10.0 m/s
t = 1.5 sec
m = 75 kg
Solution -
a = v - u by t
a = 10 - 0 by 1.5
a = 10 by 1.5
a = 6.6 m/s^2

A.b.) sorry ! I don’t no how to do this question
3 0
3 years ago
You are walking from your math class to your science class. You are carrying books
kolezko [41]

Answer:

1800J

Explanation:

Given parameters:

Weight of the book  = 20N

Total distance covered  = 45m + 15m + 30m  = 90m

Unknown:

Total work performed on the books  = ?

Solution:

To solve this problem we must understand that work done is the force applied to move a body through a certain distance.

So;

    Work done  = Force x distance

  Work done  = 20 x 90  = 1800J

8 0
3 years ago
A 2kg block has 70J of KE. It then travels 1.5 meters up a hill. As it travels up the hill friction does -12J of work on the blo
Dima020 [189]

Answer:

v = 5.34[m/s]

Explanation:

In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.

Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.

E₁ = mechanical energy at initial state [J]

E_{1}=E_{pot}+E_{kin}+E_{elas}\\

In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.

In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.

E₂ = mechanical energy at final state [J]

E_{2}=E_{kin}+E_{pot}

Now we can use the first statement to get the first equation:

E_{1}+W_{1-2}=E_{2}

where:

W₁₋₂ = work from the state 1 to 2.

E_{k}=\frac{1}{2} *m*v^{2} \\

E_{pot}=m*g*h

where:

h = elevation = 1.5 [m]

g = gravity acceleration = 9.81 [m/s²]

70 - 12 = \frac{1}{2}*2*v^{2}+2*9.81*1.5

58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]

4 0
3 years ago
A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.
MakcuM [25]

Answer:

b. 9.5°C

Explanation:

m_i = Mass of ice = 50 g

T_i = Initial temperature of water and Aluminum = 30°C

L_f = Latent heat of fusion = 3.33\times 10^5\ J/kg^{\circ}C

m_w = Mass of water = 200 g

c_w = Specific heat of water = 4186 J/kg⋅°C

m_{Al} = Mass of Aluminum = 80 g

c_{Al} = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C

The final equilibrium temperature is 9.50022°C

4 0
3 years ago
Other questions:
  • Please Please Please Can Help Me On This Question!!!!! I Give Thanks!!!!
    5·1 answer
  • A man stands on a platform that is rotating (without friction) with an angular speed of 2.4 rev/s; his arms are outstretched and
    14·1 answer
  • Does gravity always include the Earth’s pull or can it be between other objects?
    13·2 answers
  • A uniform electric field is pointing in x direction. The magnitude of the electric field is 10 N/C. The filed makes an angle of
    8·1 answer
  • At each corner of a square of side there are point charges of magnitude Q, 2Q, 3Q, and 4Q
    13·1 answer
  • What is machine. what is mechanica advantage​
    5·2 answers
  • A vaulter holds a 26.90-N pole in equilibrium by exerting an upward force U with her leading hand and a downward force D with he
    7·2 answers
  • Moving current has electrical energy.
    6·1 answer
  • Please Help with thia question​
    6·1 answer
  • The most massive stars will eventually collapse entirely giving us what<br> unique structure?
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!