It takes a minimum distance of 48.96 m to stop a car moving at 12.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.

Answers

Answer 1

Answer:

102 m

Explanation:

Given that It takes a minimum distance of 48.96 m to stop a car moving at 12.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.

Let the stopping distance be equal to S.

According to the definition of speed,

Speed = distance / time.

make time the subject of the formula

Time = distance / speed

then, the equivalent time is:

48.96 / 12 = S / 25

Cross multiply

12S = 48.96 x 25

12S = 1224

S = 1224 / 12

S = 102 m

Therefore, the stopping distance is 102 m


Related Questions

If it takes you 5 minutes to dry your hair using a 1200-W hairdryer plugged into a 120-V power outlet, how many Coulombs of charge pass through your hair dryer

Answers

Answer:

The charge pass through your hair dryer is 3000 C.

Explanation:

Given that,

Power = 1200 W

Voltage = 120 V

Flow time = 5 min

We need to calculate the current

Using formula of power

[tex]P=VI[/tex]

[tex]I=\dfrac{P}{V}[/tex]

Put the value into the formula

[tex]I=\dfrac{1200}{120}[/tex]

[tex]I=10\ A[/tex]

We need to calculate the charge pass through your hair dryer

Using formula of current

[tex]I=\dfrac{Q}{t}[/tex]

[tex]Q=It[/tex]

Put the value into the formula

[tex]Q=10\times5\times60[/tex]

[tex]Q=3000\ C[/tex]

Hence, The charge pass through your hair dryer is 3000 C.

While making some observations at the top of the 66 m tall Astronomy tower, Ron
accidently knocks a 0.5 kg stone over the edge. How long will a student at the bottom
have to get out of the way before being hit?

Answers

Analysing the question:

Since the stone was dropped, there was no initial velocity applied on it and hence it's initial velocity of the stone is 0 m/s

We are given:

height of the tower (h) = 66 m

mass of the stone (m) = 0.5 kg

initial velocity of the stone (u) = 0 m/s

time taken by the stone to reach the ground (t) = t seconds

acceleration due to gravity = 10 m/s²

** Neglecting air resistance**

Finding the time taken by the stone to reach the ground:

from the second equation of motion

h = ut + 1/2at²

replacing the variables

66 = (0)(t) + 1/2 (10)(t)²

66 = 5t²

t² = 13.2

t = 3.6 seconds

I initially wanted to subtract the height of the student from the height of the tower since the time i calculated is the time taken by the stone to reach the ground and that means that the stone has already hit the student before 3.6 seconds

but since we were NOT given the height of a student, the person who posed this question wants the time taken by the stone to reach the ground and that is what we solved

what is the force produced on a body of 30kg mass when a body moving with the velocity of 26km/hr is acceleted to gain the velocity of 54 km/hr in 4 sec​

Answers

Answer:

F = 58.35 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that force is equal to the product of mass by acceleration. But first we must use the following equation of kinematics.

We have to convert speeds from kilometers per hour to meters per second

[tex]\frac{26km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=\frac{7.22m}{s} \\\frac{54km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=15\frac{m}{s}[/tex]

[tex]v_{f}=v_{o}+(a*t) \\[/tex]

where:

Vf = final velocity = 15 [m/s]

Vi = initial velocity = 7.22 [m/s]

a = acceleration [m/s^2]

t = time = 4 [s]

Note: the positive sign of the above equation is because the car increases its speed

15 = 7.22 + (a*4)

a = 1.945 [m/s^2]

Now we can use the Newton's second law:

F = m*a

F = 30*1.945

F = 58.35 [N]

If you weigh 660 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 20.0 km? Take the mass of the sun to be 1.99×10^30, the gravitational constant to be G = 6.67×10^−11Nm^2/kg^2, and the acceleration due to gravity at the earth's surface to be g = 9.810 m/s^2.p

Answers

Answer:

8.93*10^13 N.

Explanation:

Assuming that in this case, the weight is just the the force exerted on you by the mass of the star, due to gravity, we can apply the Universal Law of Gravitation:

       [tex]F_{g}= \frac{G*m_{1}*m_{s}}{r_{s}^{2} }[/tex]

where, m1 = mass of  the man = 660 N / 9.81 m/s^2 = 67.3 kg, ms = mass of the star = 1.99*10^30 kg, G= Universal Constant of Gravitation, and rs= radius of the star = 10.0 km. = 10^4 m.Replacing by the values, we get:

       [tex]F_{g}= \frac{6.67e-11Nm^2/kg^2*1.99e30 kg*67.3 kg}{10e4m^2} = 8.93e13 N[/tex]

Fg = 8.93*10^13 N.

21. Prediction: If you were to measure the current at points A, B and C, how do you think the values would compare? Why? 22. Prediction: If you were to measure the potential differences across these bulbs (what the voltmeter measures) how do you think the values will compare to each other and to the potential difference across the battery pack or the power supply? Why?

Answers

Answer:

hello your question is incomplete attached below is the complete question

21) The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C

22) The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)

hence the voltage in the battery will be equal to the voltage across each bulb

Explanation:

The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C

The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)

hence the voltage in the battery will be equal to the voltage across each bulb

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