What happens when light rays pass through a converging lens (convex lens)?

5. What is the amount of force required to accelerate a 20 kg object at a rate of 5 m/sz?

GenaCL600 [577]

Force required is 100 N

Given that;

Rate of acceleration = 5 m/s²

Mass of object = 20kg

Find:

Force required

Computation:

Force = Mass × Acceleration

Force required = Rate of acceleration × Mass of object

Force required = 20 × 5

Force required = 100 N

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3 0

2 years ago

During which moon phase do spring tides occur?

Murrr4er [49]

Answer:

The Full Moon and New Moon

Explanation:

6 0

3 years ago

One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the flo

yuradex [85]

Answer:

μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

W₁ x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

cos 60 = x / L

where L is the length of the ladder

x = L cos 60

sin 60 = y / L

y = L sin60

the horizontal distance of man is

cos 60 = x2 / 7.0

x2 = 7 cos 60

we substitute

m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

fr = (735 + 2450) / 8.66

fr = 367.78 N

the friction force has the expression

fr = μ N

write the translational equilibrium equation

N - W₁ -W₂ = 0

N = m₁ g + W₂

N = 30 9.8 + 700

N = 994 N

we clear the friction force from the eucacion

μ = fr / N

μ = 367.78 / 994

μ = 0.37

3 0

3 years ago

A

True [87]

Answer:

$\Delta T=3.615^{\circ}C$ is the drop in the water temperature.

Explanation:

Given:

mass of ice, $m_i=14.7\ g=0.0147\ kg$

mass of water, $m_w=324\ g=0.324\ kg$

Assuming the initial temperature of the ice to be 0° C.

Apply the conservation of energy:

Heat absorbed by the ice for melting is equal to the heat lost from water to melt ice.

Now from the heat equation:

$Q_i=Q_w$

$m_i.L=m_w.c_w.\Delta T$ ......................(1)

where:

$L=$ latent heat of fusion of ice $=333.55\ J.g^{-1}$

$c_w=$ specific heat of water $=4.186\ J.g^{-1}.^{\circ}C^{-1}$

$\Delta T=$ change in temperature

Putting values in eq. (1):

$14.7 \times 333.55=324\times 4.186\times \Delta T$

$\Delta T=3.615^{\circ}C$ is the drop in the water temperature.

8 0

3 years ago

A bar magnet is held above the center of a conducting ring in the horizontal plane. The magnet is dropped so it falls lengthwise

Alenkinab [10]

Explanation:

Since, it is given that the magnet drops and falls lengthwise towards the canter of the ring. As a result, change in magnetic flux will occur which tends to induce an electric current in the ring.

Therefore, a magnetic field is also produced by the ring itself which will actually oppose or repel the magnet.

Thus, we can conclude that the falling magnet be repelled by the ring due to the magnetic interaction of the magnet and the ring.

7 0

3 years ago