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3 years ago

Guys answer with a clear explanation and plzz don't spam.

Physics

1 answer:

timama [110]3 years ago

3 0

Answer:

20.7N

Explanation:

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A big box of sausages (30 kg) are lifted from the ground to the top shelf of the freezer. If the box is lifted at a constant spe

morpeh [17]

Answer:

The work done by gravity moving the box up is -514.5 Joules.

Explanation:

The gravity acts in the direction of the displacement, which in this case is negative (ground - top shelf), and so the resulting work is negative. The work is calculated as:

$W_g = mg\Delta h = 30kg\cdot 9.8\frac{m}{s^2}\cdot(0-1.75)m=-514.5J$

The work done by gravity moving the box up is -514.5 Joules.

4 0

3 years ago

Find the gravitational potential energy of a light that has a mass of 13.0 kg and is 4.8 m above the ground

Lynna [10]

Answer:

611.52 <---- Is your answer

Explanation:

If you need explanation mark me as brain list

8 0

3 years ago

1. Quais motivos fazem com que o Croquet Golf seja a variação da

KiRa [710]

Probablemente porque es un deporte tranquilo y relajante si de eso estás hablando

Espero que esto ayude

3 0

3 years ago

A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm

Airida [17]

Answer:

126.99115 g

Explanation:

50 g at 90 cm

Stick balances at 61.3 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

$mgl_1=Mgl_2\\\Rightarrow M=\dfrac{ml_1}{l_2}\\\Rightarrow M=\dfrac{50\times (61.3-90)}{50-61.3}\\\Rightarrow M=126.99115\ g$

The mass of the meter stick is 126.99115 g

6 0

3 years ago

Read 2 more answers

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

Nostrana [21]

Answer:

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

$E_1 = E_2$

Let the position where net field is zero will lie at distance "r" from q1

$\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}$

now we will have

$\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}$

now square root both sides

$\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}$

now we have

$\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1$

so we have

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

8 0

3 years ago