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notka56 [123]
3 years ago
15

A boy of mass 40kg eats bananas contains of 980 joule. If this energy is used to lift him up from ground,the height to which he

can climb is
Physics
1 answer:
Oksanka [162]3 years ago
7 0

Answer:

h = 2.49 [m]

Explanation:

In order to solve this problem we must use the definition of potential energy, which tells us that energy is equal to the product of mass by gravity by height.

The potential energy can be calculated by means of this equation:

Ep = m*g*h

where:

Ep = potential energy = 980 [J]

m = mass = 40 [kg]

g = gravity acceleration = 9.81 [m/s^2]

h = elevation [m]

Now replacing:

980 = 40*9.81*h

h = 2.49 [m]

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A 240 V circuit has 20 A flowing through it. How much power is it using?
Delvig [45]

Explanation:

P = IV

V = 240

I = 20

P = ?

P = 240 × 20

P = 4800 watts

6 0
3 years ago
This one too dude help me
Mrac [35]

Answer:

Vas happenin!!

atomic number: 5

Atomic mass: 10.81

Number of protons: 5

Number of neutrons: 5

Number of electrons: 5

Hope this helps!

-Zayn Malik

Explanation:

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3 years ago
The force of gravity acting on an object is the object's ______. A. acceleration B. mass C. weight D. matter
ryzh [129]

Answer:

d

Explanation:

3 0
2 years ago
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A 10,000kg space ship is orbiting the moon with a radius of 25km from the ship to the center of the moon. It's tangential speed
galina1969 [7]

Answer:

False

Explanation:

ac = v^2/r

acceleration is not dependent on the mass of the orbiting object.

3 0
2 years ago
A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each
ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

E_{1}+E_{3}-E_{2}=0

We know that the electric field is:

E=k\frac{q}{r^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge
  • r is the distance from the charge to the point

So, we have:

k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0

Let's solve it for r(3).

\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0

r_{3}=0.0743\:  

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

 

3 0
3 years ago
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