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Xelga [282]
3 years ago
12

Light travels 300,000,000m/sec, and one year has approximately 32,000,000 seconds. A light year is the distance light travels in

one year. Based on this data: a. state the speed in m/sec using scientific notation b. state the speed in km/sec using scientific notation c. state the number of seconds in one year using scientific notation d. calculate the distance in meters of one light year e. state this distance in centimeters
Physics
1 answer:
dem82 [27]3 years ago
3 0
A) 3 x 10 ^ 8
b) 3 x 10 ^ 5
c) 3.2 x 10 ^ 7
d) 9.6 x 10 ^ 15 m
e) 9.6 x 10 ^ 17 cm
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How is a projectile different from an object in free fall
aksik [14]

Answer:

Explanation:

Projectile Motion. Projectile motion is different than free fall: it involves two dimensions instead of one. ... Balls traveling in two dimensions, only one of which experiences acceleration, require two sets of equations: one set for the x-direction and the other for the y-direction.

7 0
3 years ago
The diagram shows a ballistic pendulum. A 200 g bullet is fired into the suspended 4 kg block of wood and remains embedded insid
adoni [48]

Question

What was the initial momentum of the bullet before collision?

Answer:

10 Kg.m/s

Explanation:

Momentum is a product of velocity of an object in m/s and its mass in kgs hence numerically expressed as p=mv where p is momentum, v is velocity and m is mass. Substituting m for 0.2 kg and v for 50 m/s then p=0.2*50=10 kg.m/s

5 0
3 years ago
What velocity must a car with a mass of 1110 kg have in order to have the same momentum as a 2280 kg pickup truck traveling at 2
shutvik [7]

Answer:

49.3 m/s

Explanation:

The momentum is defined as the product of the object velocity and its mass.

So the momentum of the truck is

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For the car to have the same momentum, its speed must be

v_c = P_t/m_c = 54720 / 1110 = 49.3 m/s

6 0
3 years ago
How much thermal energy is required to raise the temperature of 5 kg of water 50℃? The specific heat of water is 4168 J/kg*C.
sasho [114]

Answer/solution:

Given :

Mass =5kg

T 1 =20 C,T 2  =100 ∘C

ΔT=100−20=80 ∘C

Q=m×C×ΔT

where C= specific heat capacity of water

=4200J/(kgK)

Q=5×4200×80

=1680000 Joule.

=1680KJ

3 0
2 years ago
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grin007 [14]
C. A little backward, 100 or 110 degrees
6 0
3 years ago
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