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Valentin [98]
3 years ago
7

A runner can run a 400.0 meter lap in 50.0 seconds.   He finishes at the same spot as he began.  What is his average velocity?

Physics
2 answers:
cestrela7 [59]3 years ago
8 0
8 meters per second. To find velocity is to divide distance by total time. so 400/50.
Tom [10]3 years ago
5 0
Velocity: displacement/time 
as he's back at the same spot, displacement=0
0/50= 0 
hope this helps! 
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Gekata [30.6K]
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It should also be noted that the additional elements that combine with the silicon-oxygen tetrahedron are involved with the other elements commonly found in the Earth's crust and mantle. They are aluminum, calcium, iron, magnesium, potassium and sodium.
8 0
3 years ago
Read 2 more answers
A 1560-kilogram truck moving with a speed of 28.0 m/s runs into the rear end of a 1070-kilogram stationary car. If the collision
Nimfa-mama [501]

Answer:

Δ KE =  249158.6 kJ  

Explanation:

given data

Truck mass  M =  1560 Kg

Truck initial speed, u = 28 m/s

mass of car m = 1070 Kg

initial speed of car u1 = 0 m/s

solution

first we get here final speed by using conservation of momentum  that is express as

Mu = (M+m) V     .......................1

put here  value we get

1560 × 28 = (1560 + 1070 ) V

solve it we get

final speed V = 16.60 m/s

and

Change in kinetic energy  will be here

Δ KE =   \frac{1}{2} Mu^2 - \frac{1}{2}(M+m)V^2         .................2

put here value and we get

 Δ KE = \frac{1}{2}\times 1560\times 28^2 - \frac{1}{2}\times (1560 + 1070)\times 16.60^2  

solve it we get

Δ KE =  249158.6 kJ  

6 0
3 years ago
4A. How high is a 12 kg monkey in a tree if it has 509 J of gravitational potential Energy?
True [87]

4A. PE = MxGxH. (You can consider g as 9.8 / 10m/s as well)

509 J = 12x10xH

509 J = 120xH

H = 509/120

H = 4.24 m

Hope u got the answer....pls rate the answer if it is helpful for u....and I'm sorry I could not understand B part so I didn't do it.

Thank you

8 0
3 years ago
Sometimes i wish i wasnt alive....
pshichka [43]

Answer:

Swim, jogging, weight lifting are some activities to do. This will help the overall health of the heart and mind.

Explanation:

5 0
3 years ago
Read 2 more answers
Ch 27 HW Exercise 27.12 10 of 20 Constants A horizontal rectangular surface has dimensions 2.80 cm by 3.15 cm and is in a unifor
deff fn [24]

Answer:

Magnetic field, B = 0.88 T

Explanation:

It is given that,

The dimension of rectangular surface is 2.80 cm by 3.15 cm. The area of rectangular surface is, A=8.82\ cm^2=0.000882\ m^2

Angle between the uniform magnetic field and the horizontal, \theta=31

Magnetic flux, \phi=4\times 10^{-4}\ Wb

Let B is the magnitude of magnetic field in which the rectangular surface is placed. It is given by :

\phi=BA\ cos\theta

\theta is the angle between magnetic field and the area

Here, \theta=90-31=59^{\circ}

B=\dfrac{\phi}{A\ cos\theta}

B=\dfrac{4\times 10^{-4}}{0.000882\times cos(59)}

B = 0.88 T

So, the magnitude of magnetic field is 0.88 T. Hence, this is the required solution.

7 0
3 years ago
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