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3 years ago

The proper way to brake in an emergency situation with non-ABS brakes is to: (A) apply threshold braking (B) “pump” the brakes

Physics

1 answer:

Makovka662 [10]3 years ago

5 0

Answer:

C. Lock the brakes is the answer

B. Pump the brakes

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# A cheetah can start from rest and attain the velocity 72km/h in 2 seconds. Calculate the acceleration of cheetah

yan [13]

Answer:

Explanation:

Solution,

When a certain object comes in motion from rest, in the case, initial velocity = 0 m/s

Initial velocity ( u ) = 0 m/s

Final velocity ( v ) = 72 km/h ( Given)

We have to convert 72 km /h in m/s

$72 \: km \: per \: hour$

$= 72 \times \frac{1000}{60 \times 60}$

$= 20$ m/s

Final velocity ( v ) = 20 m/s

Time taken ( t ) = 2 seconds

Acceleration (a) = ?

Now,

we have,

$a = \frac{v - u}{t}$

$a = \frac{20 - 0}{2}$

$a = \frac{20}{2}$

$a = 10$ m/s ^2

Hope this helps...

Good luck on your assignment..

7 0

3 years ago

Which of the following is MOST useful to scientists in measuring the size of asteroids?

Alenkasestr [34]

Answer:c-The gravitational effect when spacecraft flies close to the asteriod

Explanation:

Gravitational effect on the spacecraft gives an estimate that how big is the asteroid by experiencing its gravitational pull.

The amount of extra thrust required to maintain the trajectory of the spacecraft during its motion hints at the scientist about the size of the asteroid.

Gravitational pull is directly proportional to the mass of object so greater the mass, greater will be the pull.

5 0

3 years ago

a 5.0 charge is placed at the 0 cm mark of a meterstick and a -4.0 charge is placed at the 50 cm mark. what is the electric fiel

Lesechka [4]

Answer:

-1748*10^N/C

Explanation:

See attached file

8 0

3 years ago

A locomotive accelerates a 25-car train along a level track. Every car has a mass of 7.7 ✕ 104 kg and is subject to a friction f

MaRussiya [10]

To solve the problem it is necessary to apply the concepts related to Force of Friction and Tension between the two bodies.

In this way,

The total mass of the cars would be,

$m_T = 25(7.7*10^4)Kg$

$m_T = 1.925*10^6Kg$

Therefore the friction force at 29Km / h would be,

$f=250v$

$f= 250*29Km/h$

$f = 250*29*(\frac{1000m}{1km})(\frac{1h}{3600s})$

$f = 2013.889N$

In this way the tension exerts between first car and locomotive is,

$T=m_Ta+f$

$T=(1.925*10^6)(0.2)+2013.889$

$T= 3.8701*10^5N$

Therefore the tension in the coupling between the car and the locomotive is $3.87*10^5N$

6 0

3 years ago

A 400.0 ohm resistor has a potential difference of 20.0 volts. What is the magnitude of the power dissipated by the resistor

Sergio [31]

P=IV, where P is power, I is resistance, and V is voltage. Plug in and solve:

P=400(20)

P=8000W

Hope this helps!!

3 0

3 years ago