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PIT_PIT [208]
3 years ago
9

Suppose that a 1000 kg car is traveling at 25 m/s. Its brakes can apply a force of 5000N. What is the minimum distance required

for the car to stop?
Physics
1 answer:
stealth61 [152]3 years ago
6 0
The car's kinetic energy is

                             (1/2) (mass) (speed)²

                       =    (1/2) (1,000 kg) (25 m/s)²

                       =      (500 kg)  (625 m²/s²)

                       =        312,500 joules .

THAT's the work the brakes have to do in order to stop the car.
They have to absorb that kinetic energy and send it somewhere.

Work done by the brakes = (force) x (distance)

             312,500 joules  =  (5,000 N) x (distance)

                     Distance  =  (312,500 joules) / (5,000 N)

                                   =     62.5 meters .

The brakes soak up the car's kinetic energy, turn it to heat,
and let it blow away in the wind.

Too bad you paid good money to buy that energy in gasoline,
and it ended up blowing away in the wind.  You could have 
stayed home, and just opened some windows and let some
money blow away while you ate chips and watched TV.
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What is the magnitude acceleration of a body in a projectile motion at the highest point​
MrRa [10]

Answer:

Zero

Explanation:

At a projectile's highest point, its velocity is zero. At a projectile's highest point, its acceleration is zero

6 0
3 years ago
A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi
vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

V=\frac {C_2V_2-C_1V_1}{C_1+C_2}

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}

Therefore

V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437

Charge, Q is given by CV hence for the first capacitor charge will be Q_1=C_1V

Here, Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C

8 0
3 years ago
Question 4 (20 points)
irinina [24]

im pretty sure this is right 4. true 5. force

5 0
3 years ago
Read 2 more answers
an average net force of 31.6 N is used to accelerate a 15.0 kg object uniformly from rest to 10.0 m/s leftward. What is the chan
pickupchik [31]

Answer:

-150 kg m/s

Explanation:

The change of momentum is calculated as ;

Δp= m*Δv where

where

Δp= change in momentum

Δv = vf-vi

m= mass of the object

Change in momentum is also calculated when using the formula;

Δp = F * Δt   when F is the net force applied and Δt is the time of action.

In this case;

m= 15 kg

vf=  -10 m/s

vi= 0 m/s

Taking rightward direction to be positive, then leftward will be negative.

Applying the formula as;

Δp = m*Δv

Δp = 15 * { -10 -0} = 15*-10 = -150 kg m/s

6 0
3 years ago
A certain car going from Atlanta to Philadelphia averages 24.0 m/s over the course of the trip. Determine how long it takes the
Dennis_Churaev [7]

Answer:

14.5 hours

Explanation:

The relationship between speed, time and distance for a uniform motion is

v=\frac{d}{t}

where

v is the speed

d is the distance

t is the time taken

Here we know:

v = 24.0 m/s is the average speed

d=1250 km = 1.25\cdot 10^6 m is the distance

Solving the equation, we find the time taken:

t=\frac{d}{v}=\frac{1.25\cdot 10^6 m}{24.0 m/s}=52083 s

And since

1 hour = 3600 s

The time in hours is

t=52083 s \cdot \frac{1}{3600 s}=14.5 h

6 0
4 years ago
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