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Masteriza [31]
2 years ago
9

25. IF THE ROADWAY IS WET AND THE CAR STARTS TO SKID, YOU SHOULD: A. Slow down by pumping the brakes quickly and firmly B. Slow

down by shifting to a lower gear C. Slowly ease your foot off the gas pedal
Physics
1 answer:
AnnZ [28]2 years ago
5 0

If the roadway is wet and the cars starts to skid, one should slow down by shifting to lower gear. The correct option is B.

<h3>What are gears?</h3>

A gear is component used to transmit motion. They are used in cars, trucks, machines, bicycles, etc.

For a car to avoid skidding on the wet roads, one must drive the car slowly. This can be done by shifting the gear to the lower one. This will help car to keep balance when brakes are applied on wet roads.

Thus, the correct option is B.

Learn more about gears.

brainly.com/question/14455728

#SPJ1

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A 0.54 kg particle has a speed of 5.0 m/s at point A and kinetic energy of 7.5 J at point B. What is
Wewaii [24]

<u>Answers</u>

(a)  6.75 Joules.

(b)  5.27 m/s

(c) 0.75 Joules


<u>Explanation</u>

Kinetic energy is the energy possessed by a body in motion.

(a) its kinetic energy at A?

K.E = 1/2 mv²

       = 1/2 ×  0.54 × 5²

       = 6.75 Joules.

(b) its speed at point B?

K.E = 1/2 mv²

7.5 = 1/2 × 0.54 × V²

V² = 7.5 ÷ 0.27

     = 27.77778

V = √27.77778

   = 5.27 m/s

(c) the total work done on the particle as it moves from A to B?

Work done = 7.5 - 6.75

                 = 0.75 Joules

4 0
3 years ago
A ball is thrown horizontally from the top of a building 54 m high. The ball strikes the ground at a point 35 m horizontally awa
Ivanshal [37]

Answer:

V=34.2 m/s

Explanation:

Given that

Height , h= 54 m

Horizontal distance , x = 35 m

Given that , the ball is thrown horizontally , therefore the initial vertical velocity will be zero.

In vertical direction :

We know that

V^2_y=U^2_y+2 g h

Now by putting the values in the above equation we got

V^2_y=U^2_y+2 g h

V^2_y=0^2+2\times 9.81 \times 54

Assume g= 9.81 m/s^2

Thus

V^2_y=1059.48

V_y=\sqrt{1059.48}\ m/s

V_y=32.54 m/s

We also know that

V_y=U_y+ g\times t

32.54=9.81\times t

t=\dfrac{32.54}{9.81}=3.31\ s

In horizontal direction :

x=U_x\times t

U_x=\dfrac{35}{3.31}=10.54\ m/s

Thus the resultant velocity

V=\sqrt{V^2_y+U^2_x}

V=\sqrt{32.54^2+10.54^2}=34.2\ m/s

V=34.2 m/s

Therefore the velocity will be 34.2 m/s.

5 0
4 years ago
A ball hangs on the end of a string that is connected to the ceiling so that it swings like a pendulum. You pull the ball up so
saw5 [17]

Answer:

When extra energy is added

Explanation:

When the ball is released from rest and swings back towards your face, it will only pass closer to the end of the nose as per the initial conditions. However, when extra energy is added to the ball, it strikes the nose since its velocity and heights are increased. Therefore, the only condition under which the ball hits your nose is when extra energy is added to the system.

3 0
3 years ago
When is an object moving in uniform circular motion?
Tems11 [23]

Answer: An object undergoing uniform circular motion is moving

Explanation:

4 0
3 years ago
You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m.
Gwar [14]

Answer:

0.95 seconds

Explanation:

t = Time taken

u = Initial velocity = 15 m/s

v = Final velocity

s = Displacement

a = Acceleration = 9.81 m/s² (downward positive, upward negative)

Time taken by the ball to reach the maximum height

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-15}{-9.81}\\\Rightarrow t=1.52\ s

Maximum height

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-15^2}{2\times -9.81}\\\Rightarrow s=11.47\ m

Distance between maximum height of the ball and the branch is 11.47-7 = 4.46 m

So, the distance that will be covered on the way down is 4.46 m

Now

u = 0

s = 4.46

s=ut+\frac{1}{2}at^2\\\Rightarrow 4.46=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4.46\times 2}{9.81}}\\\Rightarrow t=0.95\ s

Time taken by the ball from the maximum height to the tree branch is 0.95 seconds.

Total time taken from the moment the ball is thrown to reach the tree branch is 1.52+0.95 = 2.47 seconds

7 0
3 years ago
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