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suter [353]
3 years ago
13

How do we gather most of the information we get about the universe around us?

Physics
1 answer:
8090 [49]3 years ago
6 0

Answer:

B IS THE ONLY LOGICAL ANSWER

Explanation:

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standing side by side, you and a friend step off a bridge and fall for 1.6s to the water below. your friend goes first, and you
Andrew [12]

(a) The distance will be more than 2.0 meters.

In fact, you starts your fall after your friend has already fallen 2.0 meters. This means that your friend has already accelerated for a while, therefore his velocity will be greater than yours. But this statement will be actually true for the entire fall, since you has some delay, therefore when your friend will hit the water, the separation between you and him will be greater than the initial separation of 2.0 meters.


b) First of all we need to calculate the height of the bridge with respect to the water. We know that you take 1.6 s to fall down, therefore we can use the following equation:

S=\frac{1}{2}gt^2=\frac{1}{2}(9.81 m/s^2)(1.6s)^2=12.56 m

We know that your friend will take 1.6 s to falls down. Instead, you start your jump after he has already fallen 2.0 m, therefore after a time given by the equation:

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

Using S=2.0 m,

t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(2.0 m)}{9.81 m/s^2}}=0.64 s

So we know that you start your fall 0.64 s after your friend. Therefore, now we can find how much did you fall between the moment you started your fall (0.64 s) and the moment your friend hits the water (1.6 s). Using

t=1.6 s-0.64 s=0.96 s

we find

S=\frac{1}{2}gt^2=\frac{1}{2}(9.81 m/s^2)(0.96 s)^2 =4.52 m

So, when your friend hits the water, you just covered 4.52 m, while he already covered 12.56 m. Therefore, the separation between you and your friend is more than 2 meters.

8 0
3 years ago
A ball is thrown 20.0 m/s at an angle of 40.0° with the horizontal. Assume the ball is thrown at ground level.
TEA [102]
The ball's horizontal component of velocity (ie it's horizontal speed) is 20 cos 40degrees. Without knowing the distance of the ball to the wall it's difficult to go further ...
8 0
3 years ago
Astronauts often undergo special training in which they are subjected to extremely high centripetal accelerations. One device ha
egoroff_w [7]

The centripetal acceleration of an object is given by the relation,

Ac =V^2/R

where Ac = centripetal acceleration = 98 m/s^2

R = radius of rotation = 15 m

V = speed of astronaut

Hence, \frac{V^2}{15} =98

solving this we get, V = 38.34 m/s

3 0
3 years ago
Read 2 more answers
What is the current
Art [367]

Answer:

2.5mA

Explanation:

In the picture.

I think it's a clear.

8 0
2 years ago
A garden hose having with an internal diameter of 1.1 cm is connected to a (stationary) lawn sprinkler that consists merely of a
trapecia [35]

Answer:

Water leaves the sprinkler at a speed of 2.322 m/sec

Explanation:

We have given internal diameter of the garden d_1=1.1cm

Speed of water in the hose is v_1=0.95m/sec

Number of holes n = 22

Diameter of each holes d_2=15cm

According to continuity equation A_1v_1=A_2v_2

d_1^2\times v_1=22\times d_2^2v_2

1.1^2\times 0.95=22\times 0.15^2\times v_2

v_2=2.322m/sec

So water leaves the sprinkler at a speed of 2.322 m/sec

6 0
3 years ago
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