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valentinak56 [21]
4 years ago
7

What are two ways in which you can increase the potential energy of a marble on a ramp?

Physics
1 answer:
valkas [14]4 years ago
8 0

Answer:

As the marble starts rolling down the roller coaster, the amount of potential energy stored in the marble decreases while its kinetic energy increases. Potential energy is also converted into heat energy due to friction.

Explanation:

As the marble rolls down the hill its potential energy is converted to kinetic energy (its height decreases, but its velocity increases). When the marble goes back up the loop its height increases again and its velocity decreases, changing kinetic energy into potential energy.

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Some believe that the positions of the planets at the time
laila [671]

Answer:

1.4007\times 10^{-8}\ N

1.50075\times 10^{-6}\ N

0.000000667\ N

Explanation:

m_1 = Mass of baby = 3 kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Distance between objects

Gravitational force of attraction is given by

F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 70\times 3}{1^2}\\\Rightarrow F=1.4007\times 10^{-8}\ N

The force between baby and obstetrician is 1.4007\times 10^{-8}\ N

F=\dfrac{6.67\times 10^{-11}\times 3\times 2.7\times 10^{27}}{(6\times 10^{11})^2}\\\Rightarrow F=1.50075\times 10^{-6}\ N

The force between the baby and Jupiter is 1.50075\times 10^{-6}\ N

F=\dfrac{6.67\times 10^{-11}\times 3\times 2.7\times 10^{27}}{(9\times 10^{11})^2}\\\Rightarrow F=0.000000667\ N

The force between the baby and Jupiter is 0.000000667\ N

7 0
4 years ago
A parallel-plate vacuum capacitor has 7.72 J of energy stored in it. The separation between the plates is 3.30 mm. If the separa
maxonik [38]

Answer

3.340J

Explanation;

Using the relation. Energy stored in capacitor = U = 7.72 J

U =(1/2)CV^2

C =(eo)A/d

C*d=(eo)A=constant

C2d2=C1d1

C2=C1d1/d2

The separation between the plates is 3.30mm . The separation is decreased to 1.45 mm.

Initial separation between the plates =d1= 3.30mm .

Final separation = d2 = 1.45 mm

(A) if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same

Energy=U =(1/2)q^2/C

U2C2 = U1C1

U2 =U1C1 /C2

U2 =U1d2/d1

Final energy = Uf = initial energy *d2/d1

Final energy = Uf =7.72*1.45/3.30

(A) Final energy = Uf = 3.340J

4 0
3 years ago
The equation shows neutralization of an acid and a base to produce a salt and water.
Masja [62]

The answer is; 2.

To balance a chemical equation, the moles on one side of the equation has to be the same as that on the other side. This ensures that the law of conservation  is observed because matter or energy can't be created or destroyed but can only be transformed from one form to another.

In this equation, putting 2 in front of NaCl ensures that there are 2 moles of Na and CL just as there are 2 moles of Na and CL in the reactants side.


6 0
3 years ago
Read 2 more answers
In an electromagnetic wave in free space, the electric and magnetic fields are A. parallel to one another and perpendicular to t
Leto [7]

Answer:

C. perpendicular to one another and perpendicular to the direction of wave propagation.

Explanation:

An Electromagnetic wave (EM wave) is a wave having both electric and magnetic components in it. These wave radiates electromagnetic energy while propagating through the space. The electric and magnetic field component of the wave have an angle of 90° to each other aming them perpendicular while they both are perpendicular to the direction of wave propagation as well.

Some examples of EM waves are: UV rays, IR radiation, Radio waves etc. These waves propagate at the speed of light in vacuum.

8 0
3 years ago
Water is flowing into a factory in a horizontal pipe with a radius of 0.0183 m at ground level. This pipe is then connected to a
timama [110]

Answer:

0.0168 m^3/s

Explanation:

We are given that

r_1=0.0183 m

h_1=0

r_2=0.0420 m

h_2=12.6 m

Let P_1=P_2=P

By using Bernoulli theorem

P+\frac{1}{2}\rho v^2_1+\rho gh_1=P+\frac{1}{2}\rho v^2_2+\rho gh_2

\frac{1}{2}\rho v^2_1+\rho gh_1=\frac{1}{2}\rho v^2_2+\rho gh_2

v^2_1+2gh_1=v^2_2+2gh_2

A_1v_1=A_2v_2

v_1=\frac{A_2v_2}{A_1}

(\frac{A_2}{A_1})^2v^2_2+2g\times 0=v^2_2+2\times 9.8\times 12.6

(\frac{\pi r^2_2}{\pi r^2_1})^2v^2_2-v^2_2=246.96

v^2_2((\frac{r^2_2}{r^2_1})^2-1)=246.96

v^2_2=246.96\frac{r^4_1}{r^2_4-r^4_1}

v_2=\sqrt{246.96\frac{r^4_1}{r^4_2-r^4_1}}

v_2=\sqrt{246.96\times \frac{(0.0183)^4}{(0.042)^4-(0.0183)^4}}

v_2=3.038 m/s

Volume flow rate =A_2v_2

Volume flow rate =\pi r^2_2v_2=\pi (0.042)^2\times 3.038=0.0168 m^3/s

3 0
3 years ago
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