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Reil [10]
3 years ago
10

In a catapult, energy is transferred into useful kinetic energy stores. What store is the energy in when it is the catapult?

Physics
1 answer:
yawa3891 [41]3 years ago
6 0

Answer:

actually I was just wondering what you are thinking

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An experiment is being done. Name the independent variable, and the dependent variable for the following question.
Nata [24]

Answer:

Independent variable: Surrounding color

Dependent Variable: Eating Habits

Explanation:

Dependent variables are dependent on Independent variables which means that in this question Eating habits are affected by the Surrounding color.

Hope it Helps!!

3 0
3 years ago
Which of the following types of waves can transmit only through matter? Select all that apply.
Ivan

D) Gamma Rays

Explanation:

Gamma rays can transmit through matter.

3 0
2 years ago
.A car has a steady speed of 10 m/s.<br> How far does the car travel in 10 mins?
aniked [119]

Answer:

i think the far of the car is 30 m/s.

6 0
3 years ago
Read 2 more answers
Benjamin, a transport technician, was punched by a patient in the
Masja [62]
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6 0
3 years ago
A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro
prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope=.2\frac{kg}{m}

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as,W_{1}=Force \times displacement

=\int\limits^{15}_0 {.2x} \, dx ..............(1)

=.1\times 15^2

=22.5 Nm

work done in lifting the sand is given as,W_{2}=Force \times displacement

W_{2}=\int\limits^{15}_0 F \, dx.................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m=\frac{20-18}{15-0}

m=.133

Therefore,

F=.133x+2

Put value of F in equation 2

W_{2}=\int\limits^{15}_0 (.133x+2) \, dx

W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm

Therefore,

work done lifting the bucket (sand and rope) to the top of the building,W=W_{1}+W_{2}

W=22.5 Nm+44.96 Nm

W=67.46 Nm

4 0
3 years ago
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