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Flauer [41]
2 years ago
8

A 2000 kg car experiences a constant braking force

Physics
1 answer:
Marat540 [252]2 years ago
4 0
In that case, there are three possible scenarios:

-- If the braking force is less than the force delivered by the engine,
then the car will continue to accelerate, and the brakes will eventually
overheat and erupt in flame.

-- If the braking force is exactly equal to the force delivered by the engine,
then the car will continue moving at a constant speed, and the brakes will
eventually overheat and erupt in flame.

-- If the braking force is greater than the force delivered by the engine,
then the car will slow down and eventually stop.  If it stops soon enough,
then the absorption of kinetic energy by the brakes will end before the
brakes overheat and erupt in flame.  Even if the engine is still delivering
force, the brakes can be kept locked in order to keep the car stopped ...
They do not absorb and dissipate any energy when the car is motionless.
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27. Improvements in space technology have greatly helped scientists better understand the stars, planets, and other
stellarik [79]

Answer:

B

Explanation:

Technological advancements in astronomy have led to the development of thousands of products such as

new materials, medical devices, and communications satellites.

For example, the production of instruments like telescopes, microscopes, and other measuring devices now enables us to identify things that we couldn't examine with the naked eye.

Similarly, the development of communication satellites has led to the emergence and improvements in the mobile network industry.

8 0
3 years ago
What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?
Stolb23 [73]

Answer:

1.1\cdot 10^{-10}C

Explanation:

The electric field produced by a single point charge is given by:

E=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C

8 0
2 years ago
How many nanoseconds are there in 1.90 yr ?<br> Express your answer using three significant figures.
shepuryov [24]

       (1.9 yr) x (365.24 day/yr) x (86,400 sec/day) x (10⁹ nsec/sec)

   =  (1.9 x 365.24 x 86,400 x 10⁹) nanosec

   =  6.00 x 10¹⁶ nanoseconds

5 0
3 years ago
When this circuit is closed, which way do the electrons flow?
gizmo_the_mogwai [7]

Answer:

from the positive end of the battery through the capacitor through the resistors to the negative end...

Current flows from higher potential (+) to lower potential (-)..

3 0
3 years ago
Read 2 more answers
What is the maximum value of the magnetic field at a distance of 2.5 m from a light bulb that radiates 100 W of single-frequency
Anvisha [2.4K]

Answer:

1.04\times 10^{-7} T

Explanation:

IP  = Power of the bulb = 100 W

r  = distance from the bulb = 2.5 m

I = Intensity of light at the location

Intensity of the light at the location is given as

I = \frac{P}{4\pi r^{2}}

I = \frac{100}{4(3.14) (2.5)^{2}}

I = 1.28 W/m²

B_{o} = maximum magnetic field

Intensity is given as

I = \frac{B_{o}^{2}c}{2\mu _{o}}

1.28 = \frac{B_{o}^{2}(3\times 10^{8})}{2(12.56\times 10^{-7})}

B_{o} = 1.04\times 10^{-7} T

7 0
3 years ago
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