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Kazeer [188]
3 years ago
13

Distant galaxies appear to be much larger than those nearby. true or false

Physics
1 answer:
jonny [76]3 years ago
3 0

Answer:True

Explanation: Distant galaxies appear to be larger than those close by because of cosmic expansion. light from nearer galaxies travel farther and makes the other to be billions of years away. The farther it takes to get to another galaxy, the bigger it appears

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А masd<br>Of 500kg a raised to a height of 6m In 30s<br>Find (a) Workdone .​
Mazyrski [523]

Answer:

Work done is 882000joule.

power is 29400watt.

Explanation:

given,

Mass(m)=500kg

Acceleration due to gravity(g)=9.8m/s²

Height(h)=6m

Time taken(t)=30s

Workdone=?

Power=?

now,

workdone=force*displaxement

= m*g*h

=500*9.8*6

=8,82,000joule

so, the work done by the man is 8,82,000joule.

then,

power=workdone/time taken

=8,82,000/30

=29,400watt

so, the required power to lift a load is 29,400watt.

5 0
3 years ago
How do good and bad ozone forms
sasho [114]
Good are free zones bad is car pollution ext.
3 0
3 years ago
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How long does it take the lava bomb to reach its maximum height? Answer with three significant digits and the correct unit. A sm
wlad13 [49]

Answer:

The time taken to reach the maximum height is 3.20 seconds

Explanation:

The given parameters are;

The initial height from which the volcano erupts the lava bomb = 64.4 m

The initial upward velocity of the lava bomb = 31.4 m/s

The acceleration due to gravity, g = 9.8 m/s²

The time it takes the lava bomb to reach its maximum height, t, is given by the following kinematic equation as follows;

v = u - g·t

Where;

v = The final velocity  = 0 m/s at maximum height

u = The initial velocity = 31.4 m/s

g = The acceleration due to gravity = 9.8 m/s²

t = The time taken to reach the maximum height

Substituting the values gives;

0 = 31.4 - 9.8 × t

∴ 31.4 = 9.8 × t

t = 31.4/9.8  ≈ 3.204

The time taken to reach the maximum height rounded to three significant figures = t ≈ 3.20 seconds

4 0
3 years ago
A current of 9 A flows through an electric device with a resistance of 43 Ω. What must be the applied voltage in this particular
lubasha [3.4K]

Voltage, V = IR

Where I is current in Ampere, R is Resistance in Ohms.

V = 9A * 43 Ω

V = 387 V

8 0
3 years ago
A 1.50 cm high diamond ring is placed 20.0 cm from a concave mirror with radius of curvature 30.00 cm. The magnification is ____
Rama09 [41]

Answer:

Magnification, m = -0.42

Explanation:

It is given that,

Height of diamond ring, h = 1.5 cm

Object distance, u = -20 cm

Radius of curvature of concave mirror, R = 30 cm

Focal length of mirror, f = R/2 = -15 cm (focal length is negative for concave mirror)

Using mirror's formula :

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}, f = focal length of the mirror

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}

\dfrac{1}{v}=\dfrac{1}{-15}+\dfrac{1}{-20}

v = -8.57 cm

The magnification of a mirror is given by,

m=\dfrac{-v}{u}

m=\dfrac{-(-8.57)}{-20}

m = -0.42

So, the magnification of the concave mirror is 0.42. Thew negative sign shows that the image is inverted.

5 0
3 years ago
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