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ivann1987 [24]
3 years ago
15

an athlete swims the distance from one end of a 50m pool to the other end in a time of 25 s what is The Athlete's average speed

​
Physics
1 answer:
vredina [299]3 years ago
3 0

Answer:

2.5 N

because Average speed is equal to distance divided by time

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Which one of the following distances is the shortest?
Kitty [74]
I think the correct answer is C
6 0
3 years ago
A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed on 10 m/s. If air
fiasKO [112]

The distance of the rock from the base of the cliff is C) 20 m

Explanation:

The motion of the rock in this problem is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion to find the time of flight of the rock (the time it takes to reach the ground). We can do it by using the suvat equation:

s=u_y t+\frac{1}{2}at^2

where, taking downward as positive direction,

s = 20 m is the vertical displacement of the rock

u_y=0 is the initial vertical velocity

t is the time of flight

a=g=9.8 m/s^2 is the acceleration of gravity

Solving for t,

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s

Now we can analzye the horizontal motion: the rock moves horizontally with a constant velocity of

v_x = 10 m/s

Therefore, the horizontal distance covered after a time t is

d=v_x t

and substituting t = 2.02 s, we find the final distance of the rock from the base of the cliff:

d=(10)(2.02)=20 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0
3 years ago
At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average intensity of approximately
Dmitrij [34]

Intensity of sunlight at given position is defined as power received per unit area

so here we can say

I = 2 kJ/s*m^2

area  on which photons are received is given as

A = 4.80 cm^2 = 4.80 * 10^-4 m^2

now we can find the power received due to sunlight

P = I*A

P = 2* 10^3 * 4.80 * 10^-4

P = 0.96 Watt

now we can say this power is due to photons that strikes on surface of earth

so here we can say

P = N\frac{hc}{\lambda}

given here that

\lambda = 510 nm

0.96 = N\frac{6.6 * 10^{-34}* 3 * 10^8}{510*10^{-9}}

0.96 = N * 3.88 * 10^{-19}

N = \frac{0.96}{3.88*10^{-19}}

N = 2.47 * 10^{18}

so it will strike 2.47 * 10^18 photons on given area per second

3 0
4 years ago
A 10.00-kilogram block slides along a horizontal, frictionless surface at 12.0 meters per second for 6.00 seconds. The magnitude
earnstyle [38]

Answer:

120 kg m/s

Explanation:

The magnitude of the momentum of an object is given by

p=mv

where

m is the mass of the object

v is its speed

For the block in this problem,

m = 10.0 kg (mass of the block)

v = 12.0 m/s (speed of the block)

Therefore the magnitude of the block's momentum is

p=(10.0 kg)(12.0 m/s)=120 kg m/s

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3 years ago
Which examples are a COMPLETE description of motion? Select ALL that apply.
SIZIF [17.4K]
All of them apply because they all are doing some thing to move
8 0
3 years ago
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