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Aloiza [94]
3 years ago
13

interval (from lowering the body to his feet taking off) lasts for 0.3 s and the mass of the player is 90 kg. Ignore air resista

nce. Part A Determine the force that the the player exerts on the surface when he jumps. Express your answer to one significant figure and include the appropriate units. Enter positive value if the force is upward and negative value if the force is downward.
Physics
1 answer:
steposvetlana [31]3 years ago
8 0

Answer:

The force applied to the surface is 9 kilo Newton.

Explanation:

While jumping on the surface the player applies the force that is equal to its weight on the surface.

The mass of the player is given as 90 kg.

Force applied by the player = weight of the player

Force applied by the player = m × g

Where m is the mass of the player and g is acceleration due to gravity

Force applied by the player = 90 × 9.8

Force applied by the player = 882 Newton

Expressing your answer to one significant figure, we get

Force applied by the player =0. 9 kilo Newton

The force applied to the surface is 0.9 kilo Newton.

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A particle travels 15 times around a 10-cm radius circle in 42 seconds. what is the average speed (in m/s) of the particle?
nikklg [1K]
<span>Circumference = 2 * pi * r 62.8318 = 2 * 3.14159 * 10 cm 62.8318 * 15 rotations / 42 seconds = 22.44 cm / second 22.44 cm / 100 cm per meter = .2244 m / s</span>
7 0
3 years ago
A 49.6-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.584 and 0.399,
NISA [10]

Answer:

(a) Force must be grater than 283.87 N

(B) Force will be equal to 193.945 N      

Explanation:

We have given mass of the crate m = 49.6 kg

Acceleration due to gravity g=9.8m/sec^2

Coefficient of static friction \mu _s=0.584

Coefficient of kinetic friction \mu _k=0.399

(a) Static friction force is given by F_S=\mu _smg=0.584\times 49.6\times 9.8=283.8707N

So to just start the crate moving we have to apply more force than 283.87 N

(B) This force will be equal to kinetic friction force

We know that kinetic friction force is given by F_k=\mu _kmg=0.399\times 49.6\times 9.8=193.945N

3 0
3 years ago
A flowerpot falls from the ledge of an apartment building. A person in an apartment below, coincidentally holding a stopwatch, n
valkas [14]

Answer

given,

height of window = 4 m

time taken to travel = 1 s

acceleration due to gravity = 9.8 m/s²

s = ut + \dfrac{1}{2}at^2

u = \dfrac{s - \dfrac{1}{2}at^2}{t}

u = \dfrac{4 - \dfrac{1}{2}\times 9.8\times 1^2}{1}

u = -0.905 m/s

initial velocity of ledge v = 0

now,

v² = u² + 2 a s

(-0.905)² = 0 + 2 × 9.8 ×s

s = 0.042 m

6 0
3 years ago
A car is traveling at 7.0 m/s when the driver applies the brakes. The car moves 1.5 m before it comes to a complete stop. If the
viva [34]

Answer:

d. 6.0 m

Explanation:

Given;

initial velocity of the car, u = 7.0 m/s

distance traveled by the car, d = 1.5 m

Assuming the car to be decelerating at a constant rate when the brakes were applied;

v² = u² + 2(-a)s

v² = u² - 2as

where;

v is the final velocity of the car when it stops

0 = u² - 2as

2as = u²

a = u² / 2s

a = (7)² / (2 x 1.5)

a = 16.333 m/s

When the velocity is 14 m/s

v² = u² - 2as

0 = u² - 2as

2as = u²

s = u² / 2a

s = (14)² / (2 x 16.333)

s = 6.0 m

Therefore, If the car had been moving at 14 m/s, it would have traveled 6.0 m before stopping.

The correct option is d

4 0
3 years ago
A bicyclist of mass 112 kg rides in a circle at a speed of 8.9 m/s. If the radius of the circle is 15.5 m, what is the centripet
kogti [31]
The centripetal force, Fc, is calculated through the equation, 
                                    Fc = mv²/r
where m is the mass,v is the velocity, and r is the radius. 
Substituting the known values,
                                     Fc = (112 kg)(8.9 m/s)² / (15.5 m)
                                         = 572.36 N
Therefore, the centripetal force of the bicyclist is approximately 572.36 N. 
4 0
3 years ago
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