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madam [21]
3 years ago
6

a sprinter starts from rest and accelerates at a rate of 0.16 ms ^-2 over a distance of 50.0 meters . How fast is the athlete tr

aveling at the end of the 50.0 meters?
Physics
1 answer:
vlada-n [284]3 years ago
7 0

Answer:

Final velocity v = 4 m/s

Explanation:

Given:

Initial velocity u = 0 m/s

Acceleration a = 0.16 m/s²

Distance s = 50 m

Find:

Final velocity v

Computation:

Using; v² = u² + 2as

v² = (0)² + 2(0.16)(50)

v² = 16 m/s

v = 4 m/s

Final velocity v = 4 m/s

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An observation is received directly or indirectly by
sveta [45]

Answer:

All of the above

Explanation:

because these are all senses of the body and therefore you're receiving signals from all of them all the time

4 0
2 years ago
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This is an object's change in motion per unit time in a specified direction.
timurjin [86]

Answer:

Velocity

Explanation:

Velocity is an object's change in motion per unit time in a specified direction

7 0
2 years ago
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A campground official wants to measure the distance across an irregularly-shaped lake, but can't do so directly. He picks a poin
sleet_krkn [62]

Answer:

149 m

Explanation:

The distances across the lake is forming a triangle.  

let the distance between the point and the left side be 'x'

and the distance between the point and the right be 'y'

and the distance across the lake be 'z' and the angle opposite to 'z' be 'Z' given:

∠Z = 83°

x = 105 m

y = 119 m

Now, applying the Law of Cosines, we get  

z² = x² + y² - 2xycos(Z)  

Substituting the values in the above equation, we get

z² = 105² + 119² - 2×105×119×cos(83°)

or

z = √22140.48

or

z = 148.796 m ≈ 149 m

The point is 149 m across the lake

8 0
2 years ago
Does a volcano contain a medium? Does the energy travel through anything- the transverse waves?
Elza [17]

This molten rock is called magma when it is beneath the surface and lava when it erupts or flows from a volcano. Along with lava, volcanoes also release gases, ash, and rock. ... Volcanoes form at the edges of Earth's tectonic plates



4 0
3 years ago
The suspension system of a 2100 kg automobile "sags" 8.5 cm when the chassis is placed on it. Also, the oscillation amplitude de
Gennadij [26K]

Answer:

Part a)

k = 6.06 \times 10^4 N/m

Part b)

b = 1795.4 kg/s

Explanation:

Part a)

as the mass of the suspension system is given as

m = 2100 kg

also we have

x = 8.5 cm

so now for force balance we have

mg = kx

(525)(9.81) = k(0.085)

k = 6.06 \times 10^4 N/m

Part b)

Now we know that amplitude decreases by 63% in each cycle

so after one cycle the amplitude will become 37% of initial amplitude

so it is given as

A = 0.37 A_o

also we know

A = A_o e^{-bt/2m}

0.37 A_o = A_o e^{-bt/2m}

\frac{bt}{2m} = 1

b = \frac{2m}{t}

here t = time period of one oscillation

so it is

t = 2\pi\sqrt{\frac{m}{k}}

t = 2\pi\sqrt{\frac{525}{6.06 \times 10^4}}

t = 0.58 s

now damping constant is

b = \frac{2(525)}{0.58}

b = 1795.4 kg/s

7 0
3 years ago
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