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Ne4ueva [31]
3 years ago
15

What is 4and 4/5ths divided by 3 1/4

Physics
1 answer:
Julli [10]3 years ago
4 0
8.05 that is the answer lllllllll
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Which type of light ray can produce identical but reversed
barxatty [35]

Answer: Identical rays are refracted upon entering

Explanation:

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3 years ago
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A 51.0 kg cheetah accelerates from rest to its top speed of 31.7 m/s. HINT (a) How much net work (in J) is required for the chee
andrey2020 [161]

(a) 2.56\cdot 10^4 J

The work-energy theorem states that the work done on the cheetah is equal to its change in kinetic energy:

W= \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

where

m = 51.0 kg is the mass of the cheetah

u = 0 is the initial speed of the cheetah (zero because it starts from rest)

u = 31.7 m/s is the final speed

Substituting, we find

W=\frac{1}{2}(51.0 kg)(31.7 m/s)^2 - \frac{1}{2}(51.0 kg)(0)^2=2.56\cdot 10^4 J

(b) 6.1 cal

The conversion between calories and Joules is

1 cal = 4186 J

Here the energy the cheetah needs is

E=2.56\cdot 10^4 J

Therefore we can set up a simple proportion

1 cal : 4186 J = x : 2.56\cdot 10^4 J

to find the equivalent energy in calories:

x=\frac{(1 cal)(2.56\cdot 10^4 J)}{4186 J}=6.1 cal

3 0
3 years ago
What does the solar system consist of?
scZoUnD [109]
Planets, Sun, stars, plasma, and black matter
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4 years ago
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A ski jumper travels down a slope and leaves
Serhud [2]

Question seems to be missing. Found it on google:

a) How long is the ski jumper airborne?

b) Where does the ski jumper land on the incline?

a) 4.15 s

We start by noticing that:

- The horizontal motion of the skier is a uniform motion, with constant velocity

v_x = 28 m/s

and the distance covered along the horizontal direction in a time t is

d_x = v_x t

- The vertical motion of the skier is a uniformly accelerated motion, with initial velocity u_y = 0 and constant acceleration g=9.8 m/s^2 (where we take the downward direction as positive direction). Therefore, the vertical distance covered in a time t is

d_y = \frac{1}{2}gt^2

The time t at which the skier lands is the time at which the skier reaches the incline, whose slope is

\theta = 36^{\circ} below the horizontal

This happens when:

tan \theta = \frac{d_y}{d_x}

Substituting and solving for t, we find:

tan \theta = \frac{\frac{1}{2}gt^2}{v_x t}= \frac{gt}{2v_x}\\t = \frac{2v_x}{g}tan \theta = \frac{2(28)}{9.8} tan 36^{\circ} =4.15 s

b) 143.6 m

Here we want to find the distance covered along the slope of the incline, so we need to find the horizontal and vertical components of the displacement first:

d_x = v_x t = (28)(4.15)=116.2 m

d_y = \frac{1}{2}gt^2 = \frac{1}{2}(9.8)(4.15)^2=84.4 m

The distance covered along the slope is just the magnitude of the resultant displacement, so we can use Pythagorean's theorem:

d=\sqrt{d_x^2+d_y^2}=\sqrt{(116.2)^+(84.4)^2}=143.6 m

7 0
3 years ago
Gravity is a <br><br> A.pushing force<br> B. not a force at all<br> C. pulling force
melamori03 [73]

Answer:

C

Explanation:

gravity is a pulling force according to Newton

4 0
3 years ago
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