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marusya05 [52]
3 years ago
9

During a race, a runner runs at a speed of 6 m/s. 2 seconds later, she is running at a speed of 10 m/s. What is the runner’s acc

eleration? Show your work.
Physics
1 answer:
Lapatulllka [165]3 years ago
5 0
Let's calculate the average acceleration. It is the rate of changing speeds. Hence, we need to calculate the difference of speeds. 10-6=4 m/s. The rate is now \frac{4m/s}{2s} =2m/s^2.
In general, the formula for the mean acceleration between two times 1 and 2 is given by:
\frac{u_2-u_1}{T} where v1 and v2 are the speeds at the respective points and T is the time interval between them.
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When a 70 kg man sits on the stool, by what percent does the length of the legs decrease? Assume, for simplicity, that the stool
allochka39001 [22]

The diameter of one leg of the stool is missing and it's 2cm.

Answer:

(ΔL/L) = 0.00729%

Explanation:

If the Weight of the man is W, the weight will be distributed equally on the 3 legs and so the reactions for each leg will be W/3 or F/3.

Now, Youngs modulus(Y) of douglas fir wood is about 1.3 x 10^(10) N/m^2. Gotten from youngs modulus of common materials.

Now, weight of man is 70kg.

Now diameter of one leg is 2cm.so radius of one leg = 2/2 = 1cm = 1 x 10^(-2)m

Area for one leg is; π( 1 x 10^(-2)m)^2 = 3.14 x 10^(-4)m

Now as stated earlier, the force on one leg is; F/3.

Now F = mg = 70 x 9.81 = 686.7N

So, force on one leg = 686.7/3 = 228. 9N

Now we know youngs modulus(Y) = Stress/Strain.

Stress = F/A while Strain = ΔL/L

Therefore Y = (F/A) / (ΔL/L)

And therefore, (ΔL/L) = F/(AY)

So (ΔL/L) = 228.9/(3.14 x 10^(-4))x(1.3 x 10^(10)) = 7. 29 x 10^(-5)

When expressed in percentage, it becomes 0.00729%

7 0
3 years ago
At what point does the external energy enter the system?
Phoenix [80]
The correct answer as the first one above !
8 0
3 years ago
Coasting due west on your bicycle at 8 m/s, you encounter a sandy patch of road 7.2m
jeka57 [31]

Answer:

V = (v1 + v2) / 2 = (8 + 6.5) / 2 = 7.25 m/s     average speed

t = 7.2 / 7.25 = .993 sec      time to cross patch

a = (v2 - v1) / t = (6.5 - 8) / .993 = -1.51 m/s^2     or 1.5 m/s^2

8 0
2 years ago
A ball is dropped from some height. It bounces off the floor and rebounds with a speed that is one-half the speed it had just be
Arada [10]

Answer:

The correct option is C

Explanation:

According to third equation of motion, v

2

=u

2

+2ax

Here, u=0 m/s

a=−g and x=−h

Negative sign indicates downward direction. Displacement and acceleration both are downwards.

So,v=±

2(−g)(−h)

​

We take minus sign because it is downwards.

v=−

2gh

​

After bouncing. velocity becomes 80% of v, i.e.,

v

′

=+0.8

2gh

​

 

(positive sign because the direction of ball has reversed after bouncing and is upwards.

Applying third equation of motion again, for u=v

′

, v=0 and a=−g

v

2

=u

2

+2×a×x

Thus,

0=0.64(2gh)+2(−g)x

or

x=0.64h

3 0
2 years ago
Suppose a tidal basin is 6 m above the ocean at low tide and that the area of the basin is 5×107 m2. Estimate the gravitational
Neko [114]

Answer:

Potential energy will be 176.58\times 10^{11}j          

Explanation:

We have given the height of the basin is h = 6 m

Area of the basin A=5\times 10^7m^2

Volume V=area \times height=5\times 10^7\times 6=30\times 10^7m^3

Density \rho =1000kg/m^3

We know that mass is given by m=\rho V=1000\times 30\times 10^7=3\times 10^{11}kg

We know that potential energy is given by E=mgh=3\times 10^{11}\times 9.81\times 6=176.58\times 10^{11}j

3 0
3 years ago
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