Answer:
1.045 m from 120 kg
Explanation:
m1 = 120 kg
m2 = 420 kg
m = 51 kg
d = 3 m
Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.
By use of the gravitational force
Force on m due to m1 is equal to the force on m due to m2.
3 - y = 1.87 y
3 = 2.87 y
y = 1.045 m
Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.
Answer:
fog and mist are the examples
Answer:
55.128kg
Explanation:
P= m×v
or, m=P/v = 430/7.8 = 55.128 kg
No but the sun could be a white dwarf stellar remnant.
Answer:
378 KWh
Explanation:
We'll begin by converting 1.2×10³ W to KW. This can be obtained as follow:
10³ W = 1 KW
Therefore,
1.2×10³ W = 1.2×10³ W × 1 KW / 10³ W
1.2×10³ W = 1.2 KW
Next, we shall convert 6.3×10² mins to hours (h). This can be obtained as follow:
60 mins = 1 h
Therefore,
6.3×10² mins = 6.3×10² mins × 1 h / 60 mins
6.3×10² mins = 10.5 h
Finally, we shall determine the electrical energy in KWh used for 1 month (i.e 30 days). This can be obtained as follow:
Power (P) = 1.2 KW
Time (t) for 1 month (30 days) = 10.5 h × 30
= 315 h
Energy (E) =?
E = Pt
E = 1.2 × 315
E = 378 KWh
Thus, the electrical energy used for 1 month (i.e 30 days) is 378 KWh.
