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
patriot [66]
2 years ago
7

A 1400 kg car traveling in the positive direction takes 11.0 seconds to slow from 25.0 meters per second to 12.0 meters per seco

nd. What is the average force on the car during this time?
Physics
1 answer:
Alex777 [14]2 years ago
6 0

Answer:

–1652 N. The negative sign indicate that the force is in opposite direction to car.

Explanation:

The following data were obtained from the question:

Mass (m) of car = 1400 kg

Time (t) = 11 s

Initial velocity (u) = 25 m/s

Final velocity (v) = 12 m/s

Force (F) applied on the car =?

Next, we shall determine the acceleration of the car. This can be obtained as follow:

Time (t) = 11 s

Initial velocity (u) = 25 m/s

Final velocity (v) = 12 m/s

Acceleration (a) =?

a = (v – u) /t

a = (12 – 25) /11

a = –13/11

a = –1.18 m/s²

Finally, we shall determine the force on the car as follow:

Acceleration (a) = –1.18 m/s²

Mass (m) of car = 1400 kg

Force (F) applied on the car =?

F = ma

F = 1400 × –1.18

F = –1652 N

Thus, the force on the car is –1652 N. The negative sign indicate that the force is in opposite direction to car.

You might be interested in
A 6 kg block initially at rest is pulled to East along a horizontal surface with coefficient of kinetic friction μk=0.15 by a co
ozzi

Answer:

1.8 m/s

Explanation:

Draw a free body diagram of the block.  There are four forces:

Normal force Fn up.

Weight force mg down.

Applied force F to the east.

Friction force Fn μ to the west.

Sum the forces in the y direction:

∑F = ma

Fn − mg = 0

Fn = mg

Sum the forces in the x direction:

F − Fn μ = ma

F − mg μ = ma

a = (F − mg μ) / m

a = (12 N − 6 kg × 9.8 m/s² × 0.15) / 6 kg

a = 0.53 m/s²

Given:

Δx = 3 m

v₀ = 0 m/s

a = 0.53 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (0.53 m/s²) (3 m)

v = 1.8 m/s

8 0
3 years ago
A stretched rubber band is an example of which type potential energy
Alenkinab [10]

Answer: elastic potential energy

Explanation:

6 0
2 years ago
Read 2 more answers
What type of speed looks at a particular point in time?
bearhunter [10]
The answer is D because instantaneous means at a particular point in time
4 0
3 years ago
you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k
alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}

→ v_{p}=150 km/h ,  v_{w}=50 km/h

→ v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11 km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is tan^{-1}\frac{50}{150}=18.4°

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

8 0
3 years ago
A car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh
Temka [501]

Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

speed of the car as it approaches the stationary observer, Vs = 20 m/s

Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

F_s = F_o [\frac{v}{v_s + v} ] \\\\

where;

F₀ is the observed frequency

v is the speed of sound in air = 340 m/s

F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\

F₀ ≅ 424 Hz.

Therefore, the observed frequency by the pedestrian is 424 Hz.

8 0
3 years ago
Other questions:
  • An object has a kinetic energy of 268 j and a momentum of magnitude 22.4 kg·m/s. find the speed and the mass of the object.
    14·1 answer
  • Make a general statement concerning how large bodies of water affect the climate of nearby coastal communities.
    5·1 answer
  • If the astronaut throws the tool with a force of 16.0 n , what is the magnitude of the acceleration a of the astronaut during th
    5·2 answers
  • Even though forces are acting on this box, it remains at rest on the table. Which force is represented by vector A.
    5·2 answers
  • _____ ions are those ions that do not change oxidation number or composition during a reaction.
    10·2 answers
  • During the process of cellular respiration water becomes oxygen
    6·2 answers
  • Increased levels of __________ __________ reaching Earth can lead to medical problems such as sunburn and skin cancer. What two
    5·2 answers
  • I need some help with physics ​
    5·2 answers
  • What is the difference between storm spotter and storm chaser?
    10·1 answer
  • A custodian pushes a 119 kg box across a level floor. The box moves at a constant velocity. The custodian pushes horizontally on
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!