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
Vika [28.1K]
3 years ago
13

A boy pulls his 9.0 kg sled, applying a horizontal force of 14.0 N (rightward). The coefficient of friction between the snow and

the sled is 0.12. Determine the net force and the acceleration of the sled. Draw a free-body diagram for the sled; use appropriate force symbols to label the type of each force acting on the sled.
Physics
1 answer:
Mama L [17]3 years ago
5 0

Here is the answer, dud

You might be interested in
A general contractor drives 5 miles to the hardware store in 10 minutes. He shops for needed tools for 30 minutes, then drives t
Alex17521 [72]

Given parameters:

Distance to hardware shop = 5 miles

Time to reach hardware shop = 10 minutes

Time spent at the shop = 30 minutes

Average speed to customer home = 45 mph

Time taken for the travel = 20 minutes

Unknown:

Average speed of the contractor = ?

Solution:

 Average speed is the total distance covered divided by the total time taken.

   Average speed = \frac{total distance}{total time taken}  

     total distance = distance to hardware shop + distance to customer's home

We do not know the distance to customer's home but we have been given the speed and time, so we can find the distance.

  Distance  = speed x time

 Convert the time to hrs and solve;  

                       60 minutes  = 1 hr

                       20 minutes  = \frac{20}{60} hr  = \frac{1}{3} hr

So, Distance  = 45mph x \frac{1}{3} hr   = 15miles

Now;

   Total distance  = 5 + 15 = 20miles

Total time = time to reach hardware shop + time to reach customer's house

                  = 10 + 20

                  = 30 minutes

Convert the time from minutes to hrs;

                 60 minutes  = 1hr

                 30 minutes  = \frac{30}{60}   = 0.5hr

So;

    Average speed  = \frac{20}{0.5} = 40mph

The average speed is 40mph

3 0
3 years ago
Consider massive gliders that slide friction-free along a horizontal air track. Glider A has a mass of 1 kg, a speed of 1 m/s, a
mamaluj [8]

Answer:

0.167m/s

Explanation:

According to law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision. The bodies move with a common velocity after collision.

Given momentum = Maas × velocity.

Momentum of glider A = 1kg×1m/s

Momentum of glider = 1kgm/s

Momentum of glider B = 5kg × 0m/s

The initial velocity of glider B is zero since it is at rest.

Momentum of glider B = 0kgm/s

Momentum of the bodies after collision = (mA+mB)v where;

mA and mB are the masses of the gliders

v is their common velocity after collision.

Momentum = (1+5)v

Momentum after collision = 6v

According to the law of conservation of momentum;

1kgm/s + 0kgm/s = 6v

1 =6v

V =1/6m/s

Their speed after collision will be 0.167m/s

6 0
3 years ago
What are three common forms of work in science
Sonja [21]
Biology, Chemistry and Physics
5 0
2 years ago
Read 2 more answers
100 points!!! Please help!!!
Greeley [361]

Answer:

41°

Explanation:

Kinetic energy at bottom = potential energy at top

½ mv² = mgh

½ v² = gh

h = v²/(2g)

h = (2.4 m/s)² / (2 × 9.8 m/s²)

h = 0.294 m

The pendulum rises to a height of  above the bottom.  To determine the angle, we need to use trigonometry (see attached diagram).

L − h = L cos θ

cos θ = (L − h) / L

cos θ = (1.2 − 0.294) / 1.2

θ = 41.0°

Rounded to two significant figures, the pendulum makes a maximum angle of 41° with the vertical.

7 0
3 years ago
At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown strai
zhannawk [14.2K]

Answer:

The center of mass of the two-ball system is 7.05 m above ground.

Explanation:

<u>Motion of 0.50 kg ball:</u>

Initial speed, u = 0 m/s

Time = 2 s

Acceleration = 9.81 m/s²

Initial height = 25 m

Substituting in equation s = ut + 0.5 at²

                 s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m

Height above ground = 25 - 19.62 = 5.38 m

<u>Motion of 0.25 kg ball:</u>

Initial speed, u = 15 m/s

Time = 2 s

Acceleration = -9.81 m/s²

Substituting in equation s = ut + 0.5 at²

                 s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m

Height above ground = 10.38 m

We have equation for center of gravity

          \bar{x}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

          m₁ = 0.50 kg

          x₁ = 5.38 m        

          m₂ = 0.25 kg

          x₂ = 10.38 m    

Substituting

         \bar{x}=\frac{0.50\times 5.38+0.25\times 10.38}{0.50+0.25}=7.05m

The center of mass of the two-ball system is 7.05 m above ground.  

8 0
3 years ago
Other questions:
  • An object that’s charged has more electrons than protons. An object that’s charged has fewer electrons than protons. An object t
    7·1 answer
  • 8. A student does 1,000 J of work when she moves to her dormitory. Her internal energy is decreased by 3,000 J. Determine the he
    12·1 answer
  • Rutherford's gold foil experiment revealed the atom has what subatomic particle
    6·1 answer
  • A scientist is investigating protons, neutrons, and electrons. Which topic is she studying?
    7·2 answers
  • How many regions does the deltoid muscle have
    5·2 answers
  • Which of these period three elements from the periodic table would you predict to be the MOST metallic and why? A) Al B) Si C) P
    10·1 answer
  • What is the kinetic energy of the ball just before it hits the ground?
    10·1 answer
  • A city is going to build a small power plant to supply extra electricity. Which of the following is a cost associated with using
    13·2 answers
  • Need help ASAP please and thank you
    13·1 answer
  • 3) Скорость автомобиля увеличилась от 10 м/с до 20 м/с. Во сколько раз увеличилась его
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!