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mina [271]
3 years ago
7

The potential energy of a watermelon is 15.0 J. The watermelon is 3.0 m high. What is the mass of the watermelon?

Physics
1 answer:
maks197457 [2]3 years ago
5 0

Answer:

m = 0.51[kg]

Explanation:

Potential energy is defined as the product of mass by gravity by height.

E_{pot}=m*g*h

where:

Epot = potential energy = 15 [J]

m = mass [kg]

g = gravity acceleration = 9.8 [m/s²]

h = elevation = 3 [m]

Now replacing:

E_{pot}=m*g*h\\15=m*9.8*3\\m = 0.51[kg]

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Gnoma [55]

D. When a substance reacts with another substance, it shows its chemical property

3 0
3 years ago
When the saw slices wood, the wood exerts a 104-N force on the blade, 0.128 m from the blade’s axis of rotation. If that force i
Soloha48 [4]

Answer:

Explanation:

When saw slices wood by exerting a force on the wood , wood also exerts a reaction force on the saw in opposite direction which is equal to the force of action that is 104 N.

So torque exerted by wood on the blade

= force x perpendicular distance from the axis of rotation

= 104 x .128

=13.312 Nm.

Since this torque opposes the movement of blade , it turns the blade slower.

5 0
3 years ago
In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w
Daniel [21]

Answer:

v_2=-133.17m/s, the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before (p_i) and after (p_f) the explosion. We will take the east direction as positive.

Before the explosion we have p_i=m_iv_i=Mv_i.

After the explosion we have pieces 1 and 2, so p_f=m_1v_1+m_2v_2.

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

Mv_i=m_1v_1+m_2v_2

Which means (since we want v_2 and M=m_1+m_2):

v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}

So for our values we have:

v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s

5 0
3 years ago
How high would the level be in a gasoline barometer at normal atmospheric pressure?
Sergio [31]

Answer:

h = 13.06 m

Explanation:

Given:

- Specific gravity of gasoline S.G = 0.739

- Density of water p_w = 997 kg/m^3

- The atmosphere pressure P_o = 101.325 KPa

- The change in height of the liquid is h m

Find:

How high would the level be in a gasoline barometer at normal atmospheric pressure?

Solution:

- When we consider a barometer setup. We dip the open mouth of an inverted test tube into a pool of fluid. Due to the pressure acting on the free surface of the pool, the fluid starts to rise into the test-tube to a height h.

- The relation with the pressure acting on the free surface and the height to which the fluid travels depends on the density of the fluid and gravitational acceleration as follows:

                                         P = S.G*p_w*g*h

Where,                              h = P / S.G*p_w*g

- Input the values given:

                                         h = 101.325 KPa / 0.739*9.81*997

                                         h = 13.06 m

- Hence, the gasoline will rise up to the height of 13.06 m under normal atmospheric conditions at sea level.

7 0
3 years ago
Which element's atoms have the greatest average number of neutrons ?
Mkey [24]
I am pretty sure this is uranium. it has 140 neutrons. 
6 0
3 years ago
Read 2 more answers
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