Q = m.s.Δt
Δt = Q /m.s
Here, Q = 30,000 J
m = 390 g
s = 3.9 J/g C
Substitute their values,
= 30,000 / 390 *3.9
= 30,000 / 1521
= -20 C (approx.) [-ve sign 'cause temperature is decreasing ]
In short, Your Answer would be Option D
Hope this helps!
Answer:
The answer to your question is below
Explanation:
Data 1
mass 1 = 250
mass 2 = 250 kg
gravity constant = 6.67 x 10⁻¹¹ Nm²/kg²
distance = 8 m
Formula
Substitution
Result
F = 0.000000065 N
Data 2
mass 1 = 1000 kg
mass 2 = 1000 kg
distance = 5 m
Substitution
Result
F = 0.000002667 N
Answer:
Explanation:
from the question we have the following:
distance between Sacramento and los angles = 400 miles
speed of car A = 60 mph
start time of car A = 11 am
speed of car B = 75 mph
start time of car B = 12 pm
distance of Fresno from Los Angeles = 150 miles
- To start off let's allow car A to travel for one hour (from 11 am to 12 pm), during which it would have covered a distance of 60 miles.
- Now the time would be 12 pm and the distance between the two cars would be 400 - 60 (distance traveled by car A within 11 am to 12 pm) = 340 miles
- From 12 pm to the time both cars will meet, the distance covered by car A + distance covered by car B would be equal to 340 miles. Therefore
- Distance covered by car A = speed x time(t) = 60 x t = 60t
- Distance covered by car B = speed x time(t) = 75 x t = 75t
- 60t + 75t = 340 miles
- 135t = 340
- t = 2.51 hours
- Recall that at their meeting point, the distance covered by car B = 75t = 75 x 2.62 = 188.89 miles
- Since Fresno is 150 miles from Los Angeles, car B which is 188.89 miles from Los Angeles at their meeting point would be 188.89 - 150 = 38.89 miles from Fresno
- 38.89 miles would also be the distance of car A from Fresno since that is their meeting point.