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kicyunya [14]
3 years ago
10

An archer defending a castle is on a 15.5 m high wall. He shoots an arrow straight down at 22.8 m/s. How much time does it take

to reach the ground?
Physics
1 answer:
vazorg [7]3 years ago
3 0

Answer: 1.907

Explanation:

I did the math

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A car increased its velocity to 62m/s in 10s starting from rest. Calculate the distance it covers during this time?
Roman55 [17]

Answer:

distance= velocity ×time

distance= 62×10

distance=620m

hope it helps you mate please mark me as brainliast

6 0
3 years ago
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You place ammonium nitrate crystals in water and stirred. As you do so, the container becomes cold to the touch. This an example
ankoles [38]

The process of flask becoming cold is due to endothermic reaction.

Answer: Option B

<u>Explanation:</u>

So two kinds of heat transfer can be possible in any chemical reaction. If the sample is considered as system and the sample container is considered as the surrounding, then heat transfer can occur between them.

If the heat is transferred from the surrounding to the system , then it is an endothermic reaction. And in those cases, the sample holder will be becoming colder. This is because the heat from the surrounding that is the container will be utilized to complete the reaction.

While when there is transfer of heat from the system to surrounding , it will be exothermic reaction and the beaker will be getting hot in this process. So in the present case, the container is becoming cold because of occurrence of endothermic process.

8 0
3 years ago
PLEASE HELP! Can you do the write about it! Will thank and try to mark brainiest!
goblinko [34]
Move the objects faster to get more friction.
8 0
3 years ago
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In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1
asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, f=5\times 10^{14}\ Hz

Separation between slits, d=2.2\times 10^{-5}\ m

(a) The condition for maxima is given by :

d\ sin\theta=n\lambda

For third maxima,

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{nc}{fd})  

\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})  

\theta=4.69^{\circ}

(b) For second dark fringe, n = 2

d\ sin\theta=(n+1/2)\lambda

\theta=sin^{-1}(\dfrac{5\lambda}{2d})

\theta=sin^{-1}(\dfrac{5c}{2df})

\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})

\theta=3.90^{\circ}

Hence, this is the required solution.

8 0
2 years ago
Please need in hurry
gulaghasi [49]

Explanation:

i) center of gravity (or mass)

ii) m = W/g = (160 N)/(9.8 m/s^2)

= 16.3 kg

3 0
2 years ago
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