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3 years ago

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A charge gains 2.77 J of energy when it moves between the terminals of a battery with a potential difference of 9.00 V What is t

he amount of the charge? (Unit = C)

1 answer:

Dmitry_Shevchenko [17]3 years ago

8 0

Answer: 0.308

Explanation:

acellus

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Please help Need good grade

ivann1987 [24]

Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

Speed = 5.0

We need to calculate the range

Using formula of range

$R=\dfrac{v^2\sin(2\theta)}{g}$ ...(I)

The range for the other angle is

$R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}$ ....(II)

Here, distance and speed are same

On comparing both range

$\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}$

$\sin(2\theta)=\sin(2\times(\alpha-\theta))$

$\sin120=\sin2(\alpha-60)$

$120=2\alpha-120$

$\alpha=\dfrac{120+120}{2}$

$\alpha=120^{\circ}$

Hence, The other angle is 120°

6 0

4 years ago

What happens to the kinetic energy of water when it changes from the liquid state to the solid state

d1i1m1o1n [39]

k e is energy of movement. When thing solidifies, atoms stop moving ke is less.

8 0

3 years ago

A heavy copper ball of mass 2 kg is dropped from a fiftieth-floor apartment window. Another one with mass 1 kg is dropped immedi

STALIN [3.7K]

Answer:

C

Explanation:

Because everything on Earth falls at the same speed, the masses of the balls do not matter. Since the acceleration due to gravity is constant, their speeds will both be increasing at the same rate, and therefore the difference in speeds would remain constant until they hit the ground. Hope this helps!

5 0

3 years ago

If a rocket transfers 2,500,000 MJ of energy when it takes off, how much work is done?

erastovalidia [21]

Answer:

2,500,000 MJ ( $2.5\cdot 10^6 J$ )

Explanation:

According to the work-energy theorem:

The work done on an object is equal to the amount of energy transferred

In this problem, the energy transferred by the rocket is

$E=2,500,000 MJ=2.5\cdot 10^6 J$

Therefore, the work done must be equal to the energy transferred, therefore:

$W=E=2.5\cdot 10^6 J$

6 0

3 years ago

An experimenter using a gas thermometer found the pressure at the triple point of water (0.01°C) to be 4.80 × 10⁴ Pa and the pre

irakobra [83]

Answer:

$T = -282.33^o C$

Explanation:

As we know that the relation between temperature and pressure is a linear relation

so we have

$P - P_o = \frac{P_1 - P_o}{T_1 - T_o} (T - T_o)$

here we know that

$P_1 = 6.50 \times 10^4$

$P_o = 4.80 \times 10^4$

$T_1 = 100^o C$

$T_o = 0.01^o C$

now we will have

$P - 4.80 \times 10^4 = \frac{(6.50 - 4.80)\times 10^4}{100 - 0.01}(T - 0.01)$

$P = 4.80 \times 10^4 + 170.02(T - 0.01)$

now if P = 0

then we will have

$0 = 4.80 \times 10^4 + 170.02(T - 0.01)$

$T = -282.33^o C$

7 0

3 years ago