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Setler [38]
2 years ago
13

Bolt starts the race not moving, but then increased his speed until he reaches a top speed. Once Bolt reaches his top speed he m

aintains that speed for a few seconds to:
a) move at a constant velocity
b) accelerating by moving faster
c) accelerating by moving slower
Is it a, b, or c?
Physics
1 answer:
ludmilkaskok [199]2 years ago
6 0

Answer:

a) move at a constant velocity

Explanation:

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A manometer is used to measure the air pressure in a tank. The fluid used has a specific gravity of 1.25, and the differential h
Sloan [31]

Answer:

77.88 lbm/ft³

Explanation:

Given,

Specific gravity, SG = 1.25

Density of water, ρ = 62.30 lbm/ft³

density of the fluid =

   = S.G x ρ_{water}

   = 62.30 x 1.25

   = 77.88 lbm/ft³

Density of the fluid is equal to 77.88 lbm/ft³

3 0
3 years ago
The path the moon's umbra traces across earth's surface is the path of totality.What would you see if you were standing in the p
ivann1987 [24]
You would see the darkest part of the moon
4 0
3 years ago
A driver in a 2290-kg car car traveling at 42.7 m/s slams on the brakes and skids to a stop. If the coefficient of friction betw
8_murik_8 [283]

Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

4 0
2 years ago
As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, afte
Klio2033 [76]

Answer:

 r^2 = \frac{ 2 k  \ Ze^2}{ 2m}

Explanation:

For this exercise we must use the principle of conservation of energy

starting point. The proton very far from the nucleus

          Em₀ = K = ½ m v²

final point. The point where the proton is stopped (v = 0)

          Em_f = U = q V

where the potential is

          V = k Ze / r²

Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching

Energy is conserved

          Em₀ = Em_f

           ½ m v² = e (k \frac{Ze}{r^2})

          r^2 = \frac{ 2 k  \ Ze^2}{ 2m}

with this expression we can find the closest approach distance (r)

3 0
3 years ago
A plane leaves Seattle, flies 76.0 mi at 22.0 ∘ north of east, and then changes direction to 51.0 ∘ south of east. After flying
mote1985 [20]
This can be solve by using a triangle, because the path of the plane formed a triangle. first solve the angle form by the second direction
angle = 180 - 51 - 22 = 107 degrees
then using the cosine law
c^2 = a^2 + b^2 - 2ab cos C
c^2 = 76^2 + 123^2 - 2 ( 76) ( 123) cos ( 107)
c = 162.4 mi <span>the crew fly to go directly to the field
</span>
5 0
3 years ago
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