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cluponka [151]
4 years ago
11

The atomic mass of an element is A. the sum of the protons and electrons in one atom of the element. B. twice the number of prot

ons in one atom of the element. C. a ratio based on the mass of a carbon-12 atom. D. a weighted average of the masses of an element’s isotopes.
Physics
1 answer:
sveticcg [70]4 years ago
3 0

Answer:imagine Bohr coming up to you and saying i have 2 atoms; it sticks together and becomes a compound.

Explanation:

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What is the momentum of a 45-kg quarterback moving eastward at 15<br> m/s?
puteri [66]

Answer:

Given

mass (m) =45kg

velocity (v) =15m/s

momentum (p) =?

Form

p=mv

=45x 15

p=675kg.m/s

the momentum =675kg.m/s

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3 years ago
A 5kg object is moving at a height of 2 m. The potential energy of the object is closest to ___ j
elena55 [62]

Answer:

In this case, a body of mass 5 kg kept at a height of 10 m. So the potential energy is given as 5 * 10 *10 = 500 J.

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3 years ago
James Joule (after whom the unit of energy is named) claimed that the water at the bottom of Niagara Falls should be warmer than
Molodets [167]

Answer:

0.12 K

Explanation:

height, h = 51 m

let the mass of water is m.

Specific heat of water, c = 4190 J/kg K

According to the transformation of energy

Potential energy of water = thermal energy of water

m x g x h = m x c x ΔT

Where, ΔT is the rise in temperature

g x h =  c x ΔT

9.8 x 51 = 4190 x ΔT

ΔT = 0.12 K

Thus, the rise in temperature is 0.12 K.

7 0
3 years ago
.Need and actual answer not a guess
cricket20 [7]

Answer:

it's C (this is not a guess)

8 0
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Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 10
Tema [17]

Answer:

Explanation:

(a) It is given that Joseph jogs on a straight road of 300m in a time interval of 2 minutes and 30 seconds, which is equal to 150seconds. Therefore, when Joseph jogs from point A to point B, he covers a distance of 300m in time of 150seconds. Hence, his average speed is 300m/150s=2ms^−1. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m.

Hence, his average velocity is 300m/150s=2ms^−1

(b) Then it is given that he turns back and points B and jogs on the same road but in the opposite direction for a time interval for 1 minute and covers a distance of 100m.If we consider the whole motion of Joseph, i.e. from point A to point C, then he covers a total distance of 300m+100m=400m. And he covers this total distance in a time interval of 2.5min+1min=3.5min=210s.

Therefore, his average speed for this journey is 400m210s=1.9ms−1.

For the same journey is displacement is equal to the distance between the points A and C,i.e. 300m−100m=200m.

Hence, his average velocity for this case is 200m/210s=0.95ms^−1

7 0
3 years ago
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