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2 years ago

The Steamboat Geyser in Yellowstone National Park shoots water into the air at 48.0 m/s. How

Physics

2 answers:

qwelly [4]2 years ago

7 0

Answer:

The maximum height reached by the water is 117.55 m.

Explanation:

Given;

initial velocity of the water, u = 48 m/s

at maximum height the final velocity will be zero, v = 0

the water is going upwards, i.e in the negative direction of gravity, g = -9.8 m/s².

The maximum height reached by the water is calculated as follows;

v² = u² + 2gh

where;

h is the maximum height reached by the water

0 = u² + 2gh

0 = (48)² + ( 2 x -9.8 x h)

0 = 2304 - 19.6h

19.6h = 2304

h = 2304 / 19.6

h = 117.55 m

Therefore, the maximum height reached by the water is 117.55 m.

Anon25 [30]2 years ago

4 0

Answer:

The maximum height reached by the water is 117.55 m.

Explanation:

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Two cars cover the same distance in a straight line. Car A covers the distance at a constant velocity. Car B starts from rest an

SpyIntel [72]

Answer:

a) 2.43 m/s

b) 4.83 m/s

c) 0.023 m/s²

Explanation:

a) Both cars cover a distance of 510 m in 210 s. Since car A has no acceleration

Speed = Distance / Time

$\text{Speed}=\frac{510}{210}=2.43\ m/s$

Velocity of car A is 2.43 m/s

t = Time taken = 210 seconds

u = Initial velocity

v = Final velocity

s = Displacement = 510 m

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 510=0\times 210+\frac{1}{2}\times a\times 210^2\\\Rightarrow a=\frac{510\times 2}{210^2}\\\Rightarrow a=0.023\ m/s^2$

Acceleration of car B is 0.023 m/s²

$v=u+at\\\Rightarrow v=0+0.023\times 210\\\Rightarrow v=4.83\ m/s$

Final velocity of car B is 4.83 m/s

6 0

3 years ago

Ms. Stafford wants to try racing with a push start. A student pushes her at 5 m/s before the rocket kicks in. The rocket still o

Gnom [1K]

The initial velocity of Ms. Stafford is $v_0 = 5 m/s$ , while her acceleration is
$a=4 m/s^2$
This is a uniform accelerated motion, so we can calculate the total distance travelled by Ms. Stafford in a time of $t=7.0 s$ using the law of motion for a uniform accelerated motion:
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2 years ago

Computers A and B implement the same ISA. Computer A has a clock cycle time of 200 ps and an effective CPI of 1.5 for some progr

lesya692 [45]

Answer:

Computer A is 1.41 times faster than the Computer B

Explanation:

Assume that number of instruction in the program is 1

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