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cluponka [151]
2 years ago
10

A football player runs 20 meters North of a football field, and then 15 meters East. The total motion lasted 15 seconds. What wa

s his average speed
Physics
1 answer:
vlada-n [284]2 years ago
4 0
Given:
1st run: 20 meters North
2nd run: 15 meters East
time: 15 seconds

Average speed = total distance covered / total time taken
Ave. Speed = (20m + 15m) / 15s
Ave. Speed = 35m / 15s
Ave. Speed = 2 1/3  meters per second
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MrMuchimi

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3 years ago
A 1.5-kg block slides at rest starts sliding down a snow-covered hill Point A, which has an altitude of 10 m. There is no fricti
damaskus [11]

Answer:

the speed of the block when it reaches point B is 14 m/s

Explanation:

Given that:

mass of the block slides = 1.5 - kg

height = 10 m

Force constant  = 200 N/m

distance of rough surface patch = 20 m

coefficient of kinetic friction = 0.15

In order to determine the speed of the block when it reaches point B.

We consider the equation for the energy conservation in the system which can be represented by:

\dfrac{1}{2}mv^2=mgh

\dfrac{1}{2}v^2=gh

v^2=2 \times g \times   h

v^2=2 \times 9.8 \times   10

v=\sqrt{2 \times 9.8 \times   10

v=\sqrt{196

v = 14 m/s

Thus; the speed of the block when it reaches point B is 14 m/s

7 0
3 years ago
A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift i
WINSTONCH [101]

Answer:

Total work done = = 29811.60 J

Explanation:

Since the person is being moved upward, the person’s potential and kinetic energy are increasing.

To determine the increase in potential energy, the following equation is used;

∆ PE = m * g * ∆ h

m = 78.0 Kg, g = 9.8 m/s, ∆ h = 13.0 m

∆ PE = 78.0 * 9.8 * 13.0 = 9937.20 J

Increase in kinetic energy is given by the following equation;

∆ KE = ½ * m * (vf² – vi²)

vf = 3.4, vi = 0

∆ KE = ½ * 78.0 * 3.4² = 450.84 J

Total work = 9937.20 + 450.84 = 10388.04 J

(b) He is then lifted at the constant speed of 3.40 m/s

Velocity is constant, therefore, there is no increase in kinetic energy. The only work done is the increase of potential energy

∆ PE = 78.0 * 9.8 * 13.0

∆ PE = 9937.20 J

(c) He is then decelerated to zero speed.

Since his velocity decreased from 3.40 m/s to 0 m/s, his kinetic energy decreased.

∆ KE = ½ * 78.0 * (0² - 3.4²)

∆ KE = - 450.84 J

∆ PE = 78.0 * 9.8 * 13.0 = 9937.20 J

Total work = 9937.20 - 450.84 = 9486.36 J

Sum of works = 10388.04 + 9937.20 + 9486.36 = 29811.60 J

8 0
2 years ago
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