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3 years ago

I need help on this please

Physics

1 answer:

MrMuchimi3 years ago

4 0

C because it has a small amplitude and long wavelength

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Consider an opaque horizontal plate that is well insulated on its back side. The irradiation on the plate is 2500 W/m2, of which

GuDViN [60]

Answer:

i. 0.34

ii. 0.4

iii. 1700 w/m²

iv. 2211.36 w/m²

Explanation:

Given that

Irradiation of the plate, G = 2500 w/m²

Reflected rays, p = 500 w/m²

Emissive power, E = 1200 w/m²

See attachment for calculations

8 0

3 years ago

How long would it take for a car to travel 200 km if it has an average speed of 55 km hr?

Rainbow [258]

Answer:

3.63 hours or 3 and 37.5 minutes

Explanation:

200/55

Hope this helps :)

6 0

3 years ago

A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the

Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

==> a = 900 ÷ 0.24 = 3750 m/sec2

Now we know;

F = ma

F= -Kx

==> ma= -kx

==> 20 × 3750 = -K × 0.12

==> k = 625000 N/ m

5 0

2 years ago

How long does it take light from the sun to reach earth?

storchak [24]

It takes approximately 8 minutes and 20 seconds.

5 0

3 years ago

Read 2 more answers

How much equal charge should be placed on the earth and the moon so that the electrical repulsion balances the gravitational for

kumpel [21]

As we know that electrostatic force between two charges is given as

$F = \frac{kq_1q_2}{r^2}$

here we know that electrostatic repulsion force is balanced by the gravitational force between them

so here force of attraction due to gravitation is given as

$F_g = 1.98 \times 10^{20} N$

here we can assume that both will have equal charge of magnitude "q"

now we have

$1.98 \times 10^{20} = \frac{kq^2}{r^2}$

$1.98 \times 10^{20} = \frac{(9\times 10^9)(q^2)}{(3.84 \times 10^8)^2}$

$1.98 \times 10^{20} = (6.10 \times 10^{-8}) q^2$

now we have

$q = 5.7 \times 10^{13} C$

6 0

3 years ago