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2 years ago

14

How many million kilometers is one astronomical unit.

1 answer:

Andrews [41]2 years ago

6 0

Answer:

150 million kilometres

Explanation:

The astronomical unit (symbol: au, or AU or AU) is a unit of length, roughly the distance from Earth to the Sun and equal to 150 million kilometres (93 million miles) or 8.3 light minutes.

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A scientist is trying to determine whether a newly observed phenomenon can be described as a wave. Which of the following questi

Step2247 [10]

Answer:

B. does it transfer energy

A wave is a disturbance in a medium. Particles vibrate and TRANSFER their ENERGY to their neighbors.

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3 years ago

How far does a 1.2g bullet with kinetic energy of 1.2 j go in 2 seconds ?<br><br><br>

almond37 [142]

Answer:

To calculate the energy in joules, simply enter the mass of ammunition (in grams) that you use, and the fps that you've read from your Chrono unit.

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2 years ago

Which two optical devices use convex lenses to focus and enlarge images?

Andreas93 [3]

It is a reflecting telescope and a compound microscope. I know this for sure

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2 years ago

A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of

Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

$h=(u\sin60)\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2$

$h=41.6\ m$

(b). We need to calculate the maximum height of the ball

Using formula of height

$h_{max}=\dfrac{(u\sin\theta)^2}{2g}$

Put the value into the formula

$h=\dfrac{(33\sin60)^2}{2\times 9.8}$

$h=41.67\ m$

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

$v=u-gt$

$v=u\sin\theta-gt$

Put the value into the formula

$v_{y}=33\times\sin 60-9.8\times3.0$

$v_{y}=-0.82\ m/s$

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=33\times\cos60$

$v_{x}=16.5\ m/s$

We need to calculate the ball's impact speed

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(16.5)^2+(-0.82)^2}$

$v=16.52\ m/s$

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

3 0

2 years ago

You have 1.2kg of (AU)gold (that's quite a bit of money at $1839/ounce). Using a heat source you apply 3096 J to the gold and re

melamori03 [73]

Answer:

129 J/Kg°C

Explanation:

Given :

Mass of gold, m = 1.2kg

Quantity of heat applied, Q = 3096 J

Temperature, t2 = 40°C

Temperature, t1 = 20°C

Change in temperature, dt = (40-20)°C = 20°C

Using the relation :

Q = mcdt

Where, C = specific heat capacity of gold

3096 = 1.2kg * C * 20°C

3096 J = 24kg°C * C

C = 3096 J / 24 kg°C

C = 129 J/Kg°C

7 0

2 years ago