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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.

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The national high magnetic field laboratory holds the world record for creating the strongest magnetic field. for brief periods
max2010maxim [7]

Newton’s 2nd law states that Force is equal to the product of mass (m) and acceleration (a):

F = m a                                  ---> 1

While in magnetic forces, force can also be expressed as:

F = q v B                               ---> 2

where,

q = total charge

v = velocity = 45 cm / s = 0.45 m / s

B = the magnetic field = 85 T

First we solve for the total charge, q:

q = 3.8 × 10^-23 g (1 mol / 23 g) (6.022 × 10^23 electrons / mol) (1.602 × 10^-19 C / electron)

q = 1.594 × 10^-19 C

 

We equate equations 1 and 2 then solve for acceleration a:

m a = q v B

a = q v B / m

a = [1.594 × 10^-19 C * 0.45 m / s * 85 T] / 3.8 × 10-26 kg

a = 160,437,862.2 m/s^2

 

Therefore the maximum acceleration of Na ions is about 160 × 10^6 m/s^2.

5 0
3 years ago
Read 2 more answers
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
Charra [1.4K]

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

5 0
3 years ago
Non-metals can form cations when combined with metals and anions when combined with other non-metals.
exis [7]
Well i think it is false caus ebunch of products these day have metals incorparded in the product
4 0
3 years ago
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The orbit of the planets in our solar system is generally due to which characteristic of the Sun?
larisa [96]
The answer to your question is C. <span> the Sun's strong gravitational field . This is correct because i took the test :D</span>
6 0
3 years ago
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A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift i
WINSTONCH [101]

Answer:

Total work done = = 29811.60 J

Explanation:

Since the person is being moved upward, the person’s potential and kinetic energy are increasing.

To determine the increase in potential energy, the following equation is used;

∆ PE = m * g * ∆ h

m = 78.0 Kg, g = 9.8 m/s, ∆ h = 13.0 m

∆ PE = 78.0 * 9.8 * 13.0 = 9937.20 J

Increase in kinetic energy is given by the following equation;

∆ KE = ½ * m * (vf² – vi²)

vf = 3.4, vi = 0

∆ KE = ½ * 78.0 * 3.4² = 450.84 J

Total work = 9937.20 + 450.84 = 10388.04 J

(b) He is then lifted at the constant speed of 3.40 m/s

Velocity is constant, therefore, there is no increase in kinetic energy. The only work done is the increase of potential energy

∆ PE = 78.0 * 9.8 * 13.0

∆ PE = 9937.20 J

(c) He is then decelerated to zero speed.

Since his velocity decreased from 3.40 m/s to 0 m/s, his kinetic energy decreased.

∆ KE = ½ * 78.0 * (0² - 3.4²)

∆ KE = - 450.84 J

∆ PE = 78.0 * 9.8 * 13.0 = 9937.20 J

Total work = 9937.20 - 450.84 = 9486.36 J

Sum of works = 10388.04 + 9937.20 + 9486.36 = 29811.60 J

8 0
2 years ago
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