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polet [3.4K]
3 years ago
13

(a) calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff

with an initial velocity of 8.00 m/s. (b) how long would it take to reach the ground if it is thrown straight down with the same speed?
Physics
1 answer:
erastovalidia [21]3 years ago
3 0
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Calculate the required rate of return for Climax Inc., assuming that (1) investors expect a 4.0% rate of inflation in the future
Westkost [7]

Answer:

Required rate of return = 18.5 %

Explanation:

given,                            

rate of inflection = 4 %

risk free rate = 3 %                      

market risk premium = 5 %                    

firm has a beta  = 2.30                                  

rate of return has averaged 15.0% over the last 5 years

now,                                                                      

Nominal risk free rate = risk free rate  + inflation

                                    = 3%  +  4%

                                   = 7%

Required rate of return = Nominal risk free rate + β (RPM)

                                       = 7%       +          2.3 x 5.0%

Required rate of return = 18.5 %

5 0
3 years ago
What was the greatest contribution of the monasteries? A. Illuminated manuscripts. B. Cloisters. C. Sculpture.d. Transepts
REY [17]
I am not entirely sure but i believe the answer is C scupture

3 0
4 years ago
Read 2 more answers
A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
Antonina throws a coin straight up from a height of
vichka [17]

Answer:

s=vt-\frac{1}{2}gt^2

Explanation:

We could use the following suvat equation:

s=vt-\frac{1}{2}gt^2

where

s is the vertical displacement of the coin

v is its final velocity, when it hits the water

t is the time

g is the acceleration of gravity

Taking upward as positive direction, in this problem we have:

s = -1.2 m

g=-9.8 m/s^2

And the coin reaches the water when

t = 1.3 s

Substituting these data, we can find v:

v=\frac{s}{t}+\frac{1}{2}gt=-\frac{1.2}{1.3}+\frac{1}{2}(-9.8)(1.3)=-7.3 m/s

where the negative sign means the direction is downward.

5 0
3 years ago
Hi, I am having some difficulties in solving this question. Could someone please explain this question to me in detail. The thin
aniked [119]

Answer:

a) 3.0×10⁸ m

b) 0 m

Explanation:

Displacement is the distance from the starting position to the final position.

a) In half a year, the Earth travels from one point on the circle to the point on the exact opposite side of the circle (from 0° to 180°).  The distance between the points is the diameter of the circle.

x = 2r

x = 2 (1.5×10⁸ m)

x = 3.0×10⁸ m

b) In a full year, the Earth travels one full revolution, so it ends up back where it started.  The displacement is therefore 0 m.

8 0
3 years ago
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