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Llana [10]
3 years ago
7

Pleaseeee help!!!!!!!!

Physics
1 answer:
kenny6666 [7]3 years ago
6 0
Most likely B. Will erode, if not it will grow weeds
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A 75kg hockey player is skating across the ice at a speed of 6.0m/s. What is the magnitude of the average force required to stop
liq [111]

Answer:

692.31 N

Explanation:

Applying,

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

Where F = Average force required to stop the player, m = mass of the player, a = acceleration of the player

But,

a = (v-u)/t............ Equation 2

Where v = final velocity, u = initial velocity, t = time.

Substitute equation 2 into equation 1

F = m(v-u)/t............ Equation 3

From the question,

Given: m = 75 kg, u = 6.0 m/s, v = 0 m/s (to stop), t = 0.65 s

Substitute these values into equation 3

F = 75(0-6)/0.65

F = -692.31 N

Hence the average force required to stop the player is 692.31 N

6 0
2 years ago
Which is an example of radiation? Check all 
vampirchik [111]

Answer:

Explanation:

Examples are;

Ultraviolet light from sun.

Heat from a stove burner.

X-ray from an x-ray machine.

Alpha particle emit from a radio active decay of uranium.

Sound waves from your stereo.

Microwave from micro oven.

ultraviolet light from a black light.

Gamma radiations from a supernova.

AND MANY MORE.

8 0
3 years ago
Read 2 more answers
A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend.
MatroZZZ [7]

Answer: Part(a)=0.041 secs, Part(b)=0.041 secs

Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction

now we know that

a=-9.81 m/s^2  ( negative because it is pulling the player downwards)

we also know that

s=76 cm= 0.76 m ( maximum s)

using kinetic equation

v^2=u^2+2as

where v is final velocity which is zero at max height and u is it initial

hence

u^2=-2(-9.81)*0.76

u=3.8615 m/s\\

now we can find time in the 15 cm ascent

s=ut+0.5at^2

0.15=3.861*t+0.5*9.81t^2\\

using quadratic formula

t=\frac{-3.861+\sqrt{3.86^2-4*0.5*9.81(-0.15)} }{2*0.5*9.81}

t=0.0409 sec

the answer for the part b will be the same

To find the answer for the part b we can find the velocity at 15 cm height similarly using

v^2=u^2+2as

where s=0.76-0.15

as the player has traveled the above distance to reach 15cm to the bottom

v^2=0^2 +2*(9.81)*(0.76-0.15)

v=3.4595

when the player reaches the bottom it has the same velocity with which it started which is 3.861

hence the time required to reach the bottom 15cm is

t=\frac{3.861-3.4595}{9.81}

t=0.0409

8 0
3 years ago
What mass has a rest energy of 100J?
alexandr1967 [171]

Answer:

option a is correct

Explanation:

<h2>I hope it's help you ❣️❣️</h2>
6 0
3 years ago
A deep space probe travels in a straight line at a constant speed of over 16,000 m/s. Assuming there is no friction in space, if
diamong [38]

Answer:

I believe that the answer is d.

Explanation:

Because there is nothing to make the aircraft accelerate or decelerate, it is going to stay in constant motion with no acceleration.

3 0
3 years ago
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