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
Alex_Xolod [135]
3 years ago
5

Suppose you are in a moving car and the motor stops running. You step on the brakes and slow the car to half speed. If you relea

se your foot from the brakes, will the car speed up a bit, or will it continue at half speed and slow due to friction?
Physics
1 answer:
In-s [12.5K]3 years ago
4 0

Answer:

See the answer below

Explanation:

If you step on the brake of a car while driving, the frictional force between the tires of the car and the surface of the road increases in opposition to the motion of the car. Consequently, the car slows down.

If you release your foot from the brake pedal when the car is still at half speed, the frictional force reduces and the car speeds up a bit even without pressing the throttle. Eventually, the frictional force will slow down and stop the car if the throttle is not pressed.

You might be interested in
You ride on an elevator that is moving with constant upward acceleration while standing on a bathroom scale. the reading on the
Blababa [14]

The reading on the scale is greater than your actual weight.

4 0
3 years ago
A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre
vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}

v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

And the velocity achieved in part 1:

v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

Then, to get y_2, we do:

2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

3 0
3 years ago
a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien
Law Incorporation [45]

Answer:

<u>The magnitude of the friction force is 8197.60 N</u>

Explanation:

Using the definition of the centripetal force we have:

\Sigma F=ma_{c}=m\frac{v^{2}}{R}

Where:

  • m is the mass of the car
  • v is the speed
  • R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

F_{f}=m\frac{v^{2}}{R}

F_{f}=2100\frac{18^{2}}{83}

F_{f}=8197.60 \: N

<u>Therefore the magnitude of the friction force is 8197.60 N</u>

I hope it helps you!

7 0
3 years ago
You are driving at 50 miles per hour. If you decrease the time it takes you to travel 1 mile by 8 seconds, what is your new spee
Dmitriy789 [7]

Answer:

The new speed is 56.25 miles/hour.

Explanation:

Since speed = distance/time;

time = distance/speed.

While driving at 50 miles/hour, time taken for one to complete 1 mile is (1/50) hour

(1/50) hour = (1/50) × 3600s = 72 seconds.

So, if this time to complete 1 mile (72 seconds) is reduced by 8 seconds,

New time to complete 1 mile will be = 72 - 8 = 64 seconds = (64/3600) hour = 0.0178 hour

New speed would be = (1 mile/64 seconds) = (1 mile/0.0178 hour) = 56.25 miles/hour.

Hope this Helps!!!

7 0
3 years ago
how will resistance change when the voltage increases in an electric circuit if the current remains constant
Yakvenalex [24]
V=IR
Potential Difference (v)= Current (A) * Resistance (Ω)
As V increases, R also increases.
3 0
3 years ago
Other questions:
  • To investigate seed growth in different levels of light, Jarrod covers seeds with soil in three different containers. He places
    7·2 answers
  • Drag the tiles to the correct boxes to complete the pairs.
    12·2 answers
  • What’s the difference between the speed and velocity of an object?
    8·1 answer
  • Most papers do not reflect light very well because its surface is somewhat rough. State True or False.
    11·1 answer
  • A slender rod of length L has a varying mass-per-unit-length from the left end (x=0) according to dm/dx=Cx where C has units kg/
    13·1 answer
  • Describe the core-mantle-crust structures of the terrestrial worlds. What is differentiation? What do we mean by the lithosphere
    15·1 answer
  • A foam ball of mass 0.150 g carries a charge of -2.00 nC. The ball is placed inside a uniform electric field, and is suspended a
    5·1 answer
  • Which of the following instrument measure time most accurately​
    7·2 answers
  • A compressed spring does not have elastic potential energy.
    10·1 answer
  • 5. Did objects have to touch to interact? What causes this?
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!