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Lesechka [4]
2 years ago
10

An athlete executing a long jump leaves the ground at a 28.0º angle and travels 7.80 m. (A) What was the takeoff speed?

Physics
1 answer:
Naddika [18.5K]2 years ago
7 0

Answer:

9.6 m/s

Explanation:

Angle of projection, θ = 28°

Horizontal distance, R = 7.8 m

Let the velocity of projection is given by u.

The formula used to find the velocity of projection is given by

R=\frac{u^{2}Sin2\theta }{g}

7.8=\frac{u^{2}Sin\left (2\times 28  \right ) }{9.8}

7.8=\frac{u^{2}\times 0.829}{9.8}

u = 9.6 m/s

Thus, the velocity of projection is 9.6 m/s.

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A physical quantity, G, is defined by G = (Original mass x time)/(change in mass), what is the S.I. unit of G ?
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The gravitational constant (G) in its base SI units is

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2 years ago
A demented scientist creates a new temperature scale, the "Z scale." He decides to call the boiling point of nitrogen 0°Z and th
slavikrds [6]

Answer:

A) Z = 0.577 C +112.931

Z = 0.577*(100) +112.931=170.631 Z

B) C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C

C) K = C +273.15

K = -22.41 +273.15 =250.739 K

Explanation:

For this case we want to create a function like this:

Z = a C + b

Where Z represent the degrees for the Z scale C the Celsius grades and  tha valus a and b parameters for the model.

The boiling point of nitrogen is -195,8 °C

The melting point of iron is 1538 °C

We know the following equivalences:

-195.8 °C = 0 °Z

1538 °C = 1000 °Z

Let's say that one point its (1538C, 1000 Z) and other one is (-195.8 C, 0Z)

So then we can calculate the slope for the linear model like this:

a = \frac{z_2 -z_1}{c_2 -c_1}= \frac{1000 Z- 0Z}{1538C -(-195.8 C)}=\frac{1000 Z}{1733.8 C}=0.577 \frac{Z}{C}

And now for the slope we can use one point let's use for example (-195.8C, 0Z), and we have this:

0 = 0.577 (-195.8) + b

And if we solve for b we got:

b = 0.577*195.8 =112.931 Z

So then our lineal model would be:

Z = 0.577 C +112.931

Part A

The boiling point of water is 100C so we just need to replace in the model and see what we got:

Z = 0.577*(100) +112.931=170.631 Z

Part B

For this case we have Z =100 and we want to solve for C, so we can do this:

Z-112.931 = 0.577 C

C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C

Part C

For this case we know that K = C +273.15

And we can use the result from part B to solve for K like this:

K = -22.41 +273.15 =250.739 K

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