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tatyana61 [14]
3 years ago
7

when a human cannon ball shoots in the air for 9 seconds, she travels 37 meters before ut lands. Caculate the speed​

Physics
1 answer:
max2010maxim [7]3 years ago
7 0

Answer:

9.1 mph

Explanation:

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A person, with a mass of 50.0 kg, stands on a weighing scale in a lift which is moving
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Answer:

D

Explanation:

Because I just had that answer

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3 years ago
Jessica has a mass of 55kg she sleds down a hill that has a slope of 32 degrees. what is the component of her weight that is alo
Arturiano [62]

Answer:

W = 285.62 N

Explanation:

It is given that,

Mass of Jessica is 55 kg

Slope of the hill is 32 degrees

We need to find the component of her weight that is along her direction of motion.

The component along her direction of motion is shown in attached figure. It means

W_y=mg\sin\theta\\\\W_y=55\times 9.8\times \sin(32)\\\\W_y=285.62\ N

So, the component of her weight that is along her direction of motion is 285.62 N.

7 0
3 years ago
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In a first order decomposition in which the rate constant is 0.204 sec-1, how long will it take (in seconds) until 0.399 mol/L o
DENIUS [597]

Answer:

2.902 to 3 decimal places

5 0
3 years ago
A car traveling 34 mi/h accelerates uniformly for 4 s, covering 615 ft in this time. What was its acceleration? Round your answe
Contact [7]

Answer:

51.94 ft/s²

257.63 ft/s

Explanation:

t = Time taken = 4 s

u = Initial velocity = 34 mi/h

v = Final velocity

s = Displacement = 615 ft

a = Acceleration

Converting velocity to ft/s

34\ mi/h=\frac{34\times 5280}{3600}=49.87\ ft/s

Equation of motion

s=ut+\frac{1}{2}at^2\\\Rightarrow a=2\frac{s-ut}{t^2}\\\Rightarrow a=2\left(\frac{615-49.87\times 4}{4^2}\right)\\\Rightarrow a=51.94\ ft/s^2

Acceleration is 51.94 ft/s²

v=u+at\\\Rightarrow v=49.87+51.94\times 4\\\Rightarrow v=257.63\ ft/s

Final velocity at this time is 257.63 ft/s

5 0
3 years ago
A bumper cart has a mass of 200 kg and has a protective bumper around it that behaves like a spring. The spring constant is 5000
34kurt
Part A:
For this part we’re assuming all the kinetic energy of the moving bumper car is converted into elastic potential energy in the spring since the car is brought to rest. Therefore you can find the total kinetic energy to get your answer:

KE = ½ mv^2
KE = ½ (200)(8)^2
KE = 6400 J

Part B:
Now you can use Hooke’s law to find the force:

F = kx
F = (5000)(0.2)
F = 1000 N
4 0
3 years ago
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