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4 years ago

A rocket is fired at 100 m/s at an angle of 37, how many seconds is it in the air?

Physics

1 answer:

Sonbull [250]4 years ago

7 0

Answer:

12.3 seconds

Explanation:

time in air = 2(v)/g

2(100sin(37))/9.8

=12.28 seconds

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stira [4]

It’s D. The pitch would be lower. I’m 100% sure. I just took the test. I hope this helps you!! <3

5 0

3 years ago

Where is the deepest cave in the world? How far down is it located?

Sergeeva-Olga [200]

Answer:

Geology Notes

1 Veryovkina Cave 2212[1] 13.5 km (8.4 mi)[1] Abkhazia / Georgia 43°23′52″N 40°21′37″E.

6 0

3 years ago

A frog falls from its rainforest tree. If we ignore wind resistance, (a) how much time does it take the frog to fall a distance

gtnhenbr [62]

Answer:

Explanation:

a) Using the equation of motion

S = ut + 1/2gt²

S is the distance of fall

g is the acceleration due to gravity

t is the time taken

Given S = 12.0m, g = 9.81m/s^2, un= 0m/s

12 = 0+1/2(9.81)t²

12 = 4.905t²²²

t² = 12/4.905

t² = 2.446

t = √2.446

t = 1.56secs

b) To determine how fast is the frog falling at this point, we need to calculate the speed of the frog. Using the equaton v = u+gt

v = 0+9.81(1.56)

v = 15.34m/s

Hence the frog is falling at the rate of 15.34m/s

5 0

3 years ago

A 95kg clock initially at rest on a horizontal floor requires a 560N horizontal force to set it in motion. After the clock is in

miv72 [106K]

Complete Question

A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.

Answer:

The value for static friction is $\mu_s = 0.60$

The value for static friction is $\mu_k = 0.70$

Explanation:

From the question we are told that

The mass of the clock is $m = 95 \ kg$

The first horizontal force is $F _s = 560 \ N$

The second horizontal force is $F _k = 650 \ N$

Generally the static frictional force is equal to the first horizontal force

$F _s = m * g * \mu_s$

=> $560 = 95 * 9.8 * \mu_s$

=> $\mu_s = 0.60$

Generally the kinetic frictional force is equal to the second horizontal force

$F _k = m * g * \mu_k$

$650 = 95 * 9.8 * \mu_k$

$\mu_k = 0.70$

3 0

3 years ago

A sports car is advertised to be able to stop in a distance of 50.0 m from a speed is 99.0km/h. What is its acceleration in m/s2

Shtirlitz [24]

99.0km/h =27.5m/s (this is the initial speed)
The final speed is zero
The distance is 50.0m
Therefore you use the formula:
vfinal²=vinitial²+2ad
a=(vfinal²-vinitial²)/2d
= (0²-27.5²)/(2x50.0)
=-7.5625 or in correct sigdigs -7.56m/s²
Hope this helps!

4 0

3 years ago