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aniked [119]
3 years ago
10

the winner of a 10km road race took half an hour to complete the race. Calculate the average speed. Give your answer in metres p

er second.
Physics
1 answer:
leonid [27]3 years ago
4 0
Hello there!

For this:

1). Convert 10km to meters!
2). Convert the 30 minutes into seconds!

3). Use the following formula to solve for speed!

speed= distance/time

Note: The units should automatically work out to m/s. :)

My goal is to make sure you understand the problem, which is why I won't be giving you the answer. It'll be more work now, but less work in the future! :)

Hope this helped!

-------------


DISCLAIMER: I am not a professional tutor or have any professional background in your subject. Please do not copy my work down, as that will only make things harder for you in the long run. Take the time to really understand this, and it'll make future problems easier. I am human, and may make mistakes, despite my best efforts. Again, I possess no professional background in your subject, so anything you do with my help will be your responsibility. Thank you for reading this, and have a wonderful day/night!
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A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration?
Scrat [10]

Answer:

{ \boxed{ \bold{ \sf{Acceleration \: ( \: a) = 8 \: m/ {s \: }^{2} }}}}

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

♨ Question :

  • A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration ?

♨ \underbrace{ \sf{Required \: Answer : }}

☄ Given :

  • Initial velocity ( u ) = 0
  • Final velocity ( v ) = 60.0 m/s
  • Time ( t ) = 7.50 s

☄ To find :

  • Acceleration ( a )

✒ We know ,

\boxed{ \underline{ \bold{ \sf{Acceleration \: ( \: a) =  \frac{Final velocity ( v )  - Initial velocity ( u)}{t} }}}}

Substitute the values and solve for a.

➛ \sf{a =  \frac{60.0 - 0}{7.50}}

➛ \sf{a =  \frac{60.0}{7.50}}

➛ \boxed{ \boxed{ \sf{a = 8 \: m/ {s \: }^{2} }}}

---------------------------------------------------------------

✑ Additional Info :

  • When a certain object comes in motion from rest , in the case , initial velocity ( u ) = 0
  • When a moving object comes in rest , in the case , final velocity ( v ) = 0
  • If the object is moving with uniform velocity , in the case , u = v.
  • If any object is thrown vertically upwards in the case , a = -g
  • When an object is falling from certain height , in the case , final velocity at maximum height ( v ) = 0.

Hope I helped!

Have a wonderful time ツ

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4 0
3 years ago
What happens when the crests of two waves overlap
Sergio [31]
There are two<span> main types of </span>wave<span> interference: constructive interference and destructive interference. Constructive interference </span>happens<span> when the amplitude of the combined </span>waves<span> is larger than the amplitudes of the single </span>waves<span>. This can occur when the </span>crests of two<span> transverse </span><span>waves overlap.

Hope this helps!!! :D

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4 0
3 years ago
Which describes radioactive decay of a substance?
Margarita [4]
Radioactive decay is the loss of elementary particles from an unstable nucleus, ultimately changing the unstable element into another more stable element. There are five types of radioactive decay: alpha emission, beta emission, positron emission, electron capture, and gamma emission.
4 0
3 years ago
If R is the total resistance of three resistors, connected in parallel, with resistances R1, R2, R3, then 1 R = 1 R1 + 1 R2 + 1
rjkz [21]

Answer:

maximum error is 0.03333

Explanation:

given data

R1 = 100 Ω,

R2 = 25 Ω,

R3 = 10 Ω

1/ R = 1/ R1 + 1/ R2 + 1 /R3

possible error = 0.5%

to find out

maximum error

solution

we know

1/ R = 1/ R1 + 1/ R2 + 1 /R3

put all value R1, R2 and R3

1/ R = 1/ 100 + 1/ 25 + 1 /10

R = 20/3

now take derivative  

dR/dR(i) = R²/R(i)² for i = 1, 2, 3

we have given error 0.005

so dR(i) = 0.005×R(i)  for the i = 1,2,3

so the equation will be

dR = dR/dR(1) ×dR(1) +dR/dR(2) ×dR(2) + dR/dR(3) ×dR(3)

dR = R²/R²(1) ×dR(1) + R²/R²(2) ×dR(2) +  R²/R²(3) ×dR(3)

put the value dR(1) and dR(2) and dR(3) and R

dR =  (20/3)²/R²(1) ×0.005×R(1) +  (20/3)²/R²(2) ×0.005×R(2) +   (20/3)²/R²(3) ×0.005×R(3)

dR =  (20/3)²/R(1) ×0.005 +  (20/3)²/R(2) ×0.005 +   (20/3)²/R(3) ×0.005

dR =  (20/3)²/100 ×0.005 +  (20/3)²/20 ×0.005 +   (20/3)²/10 ×0.005

dR =  (20/3)² ( 0.005/100 + 0.005/25 + 0.005/10)

dR = 0.033333

maximum error is 0.03333

8 0
3 years ago
One long wire lies along an x axis and carries a current of 43 A in the positive x direction. A second long wire is perpendicula
qwelly [4]

Answer:

Bnet=1.006*10^-6T

Explanation:

One long wire lies along an x axis and carries a current of 43 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 5.9 m, 0), and carries a current of 41 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0, 1.7 m, 0)?

the magnetic field Bnet=\sqrt{b1^2+b2^2}

the magnetic field due this long wire is given by

B1=∨I1/(2\pi *R1)..............................1

B2=∨I2/(2\pi *R2)............................2

Bnet=\sqrt{(vI1/2*pi*R1)^2+(vI2/2*pi*R2)^2}.......................3

Bnet=v/2*pi\sqrt{(I1/R1)^2+(i2/R2)^2}

Bnet=4*pi*10^-7/(2\pi)\sqrt{(43/1.7)^2+(41/29.5)^2}

Bnet=0.0000002*(641.72)^.5

Bnet=1.006*10^-6T

8 0
4 years ago
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