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melisa1 [442]
3 years ago
12

Which statements about braking a car are true? A)The greater the kinetic energy of a car, the longer it takes for the car to sto

p. B) The lower the kinetic energy of a car, the longer it takes for the car to stop. C)If the speed of a car doubles, the car’s kinetic energy and braking distance quadruple. D)If the speed of a car doubles, the car’s kinetic energy and braking distance remain the same.
Physics
2 answers:
lianna [129]3 years ago
4 0

Answer:

I believe the correct option is <em>C</em>.

Thank You

NikAS [45]3 years ago
3 0

Answer:

option c  

Explanation:

Kinetic energy is due to the speed of a body.

K.E = \frac{1}{2}mv^2

When speed is doubled, the kinetic energy  is quadruple.

From third equation of motion, braking distance is also proportional to square of speed. Thus, when speed is doubled, the braking distance is quadruple.

Thus, option c is correct.

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-7 2/3 or -23 / 3

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8 0
3 years ago
Read 2 more answers
A ball is dropped from rest at point O . It passes a window with height 3.8 m in time interval tAB = 0.02 s.Identify the correct
Taya2010 [7]

Answer:

VB − VA = g tAB   &   (VA + VB)/2 = h / tAB

Explanation:

s = h = Displacement

tAB = t = Time taken

VA = u = Initial velocity

VB = v = Final velocity

a = g = Acceleration due to gravity = 9.8 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{3.8-\frac{1}{2}\times 9.8\times 0.02^2}{0.02}\\\Rightarrow u=189.902\ m/s

v=u+at\\\Rightarrow v=189.902+9.8\times 0.02\\\Rightarrow v=190.098\ m/s

\frac{v+u}{2}=\frac{190.098+189.902}{2}=190\ s

\frac{h}{t}=\frac{3.8}{0.02}=190\ s

Hence, the equations VB − VA = g tAB   &   (VA + VB)/2 = h / tAB will be used

3 0
3 years ago
An athlete whose mass is 87.0 kg is performing weight-lifting exercises. Starting from the rest position, he lifts, with constan
expeople1 [14]

Answer: X = 52,314.12 N

Explanation: Let X be the force the feet of the athlete exerts on the floor.

According to newton's third law of motion the floor gives an upward reaction based on the weight of the athlete and the barbell which is known as the normal reaction ( based on the mass of the athlete and the barbell)

Mass of athlete = 87kg, mass of barbell = 600/ hence total normal reaction from the floor = 87* 61.22/ 9.8 *9.8 = 52,200N.

The athlete lifts the barbell from rest thus making it initial velocity u=0, distance covered = S = 0.65m and the time taken = 1.3s

The acceleration of the barbell is gotten by using the equation of constant acceleration motion

S= ut + 1/2at²

But u = 0

S = 1/2at²

0.65 = 1/2 *a (1.3)²

0.65 = 1.69 * a/2

0.65 * 2 = 1.69 * a

a = 0.65 * 2/ 1.69

a = 0.77m/s²

According to newton's second law of motion

Resultant force = mass * acceleration

And resultant force in this case is

X - 52,200 = (87 + 61.22) * 0.77

X - 52,200 = 148.22 * 0.77

X - 52, 200 = 114.132

X = 114.132 + 52,200

X = 52,314.12 N

6 0
3 years ago
In the diagram, q1= -2.60*10^-9 C and
Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

E_{net}=90.37\: N/c

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}

Where:

  • k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
  • q1 is the first charge
  • d1 is the distance from q1 to P

|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}

|E_{1}|=80.84\: N/C

And E2 will be:

|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}

|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}

|E_{2}|=40.39\: N/C

Finally, we need to use the  Pythagoras theorem to find the magnitude of the net electric field.

E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}

E_{net}=\sqrt{80.84^{2}+40.39^{2}}

E_{net}=90.37\: N/c

I hope it helps you!

7 0
3 years ago
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