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julsineya [31]
3 years ago
12

A race car driver must average 200 km/hr for four laps to qualify for a race. because of engine trouble, the car averages only 1

70 km/hr over the first two laps. what average speed must be maintained for the last two laps?
Physics
1 answer:
AnnZ [28]3 years ago
6 0
To know the average, let us assign values for each of the four laps as x, y, v and w. The average of the first two is 170. So, we can determine the sum of x and y:

(x+y)/2 = 170
x = y = 340

Next, we have to find the value for (v+w):
(x+y+v+w)/4 = 200
[340 + (v+w)]/4 = 200
v+w = 460

So, we can now determine the average of the last two laps:
Average = (v+w)/2 = 460/2 = 230 km/h
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Explanation:

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A 56-N net force acting on a cart accelerates it at a rate of 0.5 m/s/s. What is the mass of the cart
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3 years ago
The volume of a gas is reduced from 4 l to 0.5 l while the temperature is held constant. How does the gas pressure change?
shusha [124]

Answer:

The pressure increases by a factor 8

Explanation:

For a gas held at constant temperature, Boyle's law can be applied. It states that the product of the gas pressure and the gas volume is constant, so we can write:

p_1 V_1 = p_2 V_2

where

p_1 is the initial pressure

p_2 is the final pressure

V_1 is the initial volume

V_2 is the final volume

For the gas in this problem, the volume is reduced from

V_1 = 4 L

to

V_2 = 0.5 L

so we can rewrite the equation as

\frac{p_2}{p_1}=\frac{V_1}{V_2}=\frac{4 L}{0.5 L}=8

this means that the pressure of the gas will increase by a factor 8.

5 0
4 years ago
To test a slide at an amusement park, a block of wood with mass 3.00 kgkg is released at the top of the slide and slides down to
dem82 [27]

Answer:

511.1 J

Explanation:

We are given that

Mass of wood block=m=3 kg

Vertical distance,h=23 m

Horizontal distance =x=30 m

Distance traveled in downward direction y=40 m

Initial velocity,u=0

y=ut+\frac{1}{2} gt^2

Where g=9.8 m/s^2

40=0+\frac{1}{2}(9.8)t^2=4.9t^2

t^2=\frac{40}{4.9}

t=\sqrt{\frac{40}{4.9}}=2.86 s

x=v_x\times t

30=v_x(2.86)

v=v_x=\frac{30}{2.86}=10.49 m/s

By work energy theorem

Change in kinetic energy=Work done= mgh-W

\frac{1}{2}mv^2-\frac{1}{2}mu^2=3\times 9.8\times 23-W

\frac{1}{2}(3)(10.49)^2-0=676.2-W

W=676.2-\frac{1}{2}(3)(10.49)^2=511.1 J

Hence, the work done due to friction on the block as it slides down the ramp=511.1 J

6 0
3 years ago
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