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Dima020 [189]
3 years ago
8

A sprinter completes a 400m race (one lap around the track). Mr Borst thinks the magnitudes of her distance and displacement are

equalIs he correct?
Physics
1 answer:
GalinKa [24]3 years ago
7 0

Answer:

No

Explanation:

Displacement is how far between your initial and finishing position.

If one lap around the track is 400 m and the sprinter ran 1 lap around the track. Then the sprinter's distance is 400 m and their displacement is 0 m

If the sprinter ran 400 m in a straight line however, then it would be equal.

But since the sprinter ran 1 lap around there is no displacement.

I hope this helped you...  

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Tema [17]
Its similar to the moon orbiting the earth because lets say that the sing is moon and the ball is earth has the "moon" orbits around the "earth" the string ends up tying around the ball till its no more


i think i hope this example helps you somehow srry that i dont know more then that :/
4 0
3 years ago
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A cannon fires a 40.5kg shell toward a target and the shell moves with a velocity of 120 m/s. Calculate the shells momentum
kvv77 [185]

Answer:

4860 kg m/s

Explanation:

P = mv

P = 40.5 x 120

P = 4860 kg m/s

8 0
2 years ago
How do Glaciers (located thousands of miles away) affect Florida's geography?
mr_godi [17]

Answer:

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6 0
3 years ago
A horizontal spring-mass system has low friction, spring stiffness 160 N/m, and mass 0.3 kg. The system is released with an init
anygoal [31]

Answer:

(a) 0.38 m

(b) 2.78 m/s

(c) 0.11 watt

Explanation:

mass, m = 0.3 kg

spring constant, K = 160 N/m

initial compression, d = 12 cm = 01.2 m

initial speed, u = 3 m/s

(a) Let the maximum stretch is y.

Use conservation of energy

Initial potential energy + initial kinetic energy = final potential energy

0.5 x K x d² + 0.5 x m x u² = 0.5 x K x y²

160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 160 x y²

2.304 + 0.00432 = 160 y²

y = 0.38 m

y = 38 cm

(b) Let v is the maximum speed.

The speed is maximum when the stretch in the spring is zero, so by use of conservation of energy

Initial potential energy + initial kinetic energy = final kinetic energy

0.5 x K x d² + 0.5 x m x u² = 0.5 x m x v²

160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 0.3 x v²

2.304 + 0.00432 = 0.3 v²

v = 2.78 m/s

(c) The time period of the spring mass system is given by

T=2\pi\sqrt{\frac{m}{K}}

T=2\pi\sqrt{\frac{0.3}{160}}

T = 0.272 second

Energy dissipated per cycle = 0.03 J

Power, P = 0.03 / 0.272 = 0.11 Watt

5 0
3 years ago
PLEASE HELP ME WITH THIS ONE QUESTION
Kitty [74]

Answer:

Q = 282,000 J

Explanation:

Given that,

The mass of liquid water, m = 125 g

Temperature, T = 100°C

The latent heat of vaporization, Hv = 2258 J/g.

We need to find the amount of heat needed to vaporize 125 g of liquid water. We can find it as follows :

Q=mH_v\\\\Q=125\ g\times 2285\ J/g\\\\Q=282250\ J

or

Q = 282,000 J

So, the required heat is 282,000 J .

6 0
3 years ago
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