Answer:
1.25 m
Explanation:
From the question given above, the following data were obtained:
Force ratio = 2.5
Distance of load from the fulcrum = 0.5 m
Distance of effort =.?
The distance of the effort from the fulcrum can be obtained as illustrated below:
Force ratio = Distance of effort / Distance of load
2.5 = Distance of effort / 0.5
Cross multiply
Distance of effort = 2.5 × 0.5
Distance of effort = 1.25 m
Therefore, the distance of the effort from the fulcrum is 1.25 m
Answer:
50 W
Explanation:
<h3>
<u>Given :</u></h3>
- Force applied = 100 N
- Distance covered = 5 metres
- Time = 10 seconds
<h3>
<u>To find :</u></h3>
Power
<h3>
<u>Solution :</u></h3>
For calculating power, we first need to know about the work done.

Now, substituting values in the above formula;
Work = 100 × 5
= 500 Nm or 500 J
We know that,

Substituting values in above formula;
Power = 500/ 10
= 50 Nm/s or 50 W
Hence, power = 50 W .
He has 1 1/30 gallons, or 31/30 gallons, you can find this by setting all the fractions to a common denominator and adding them
Answer:
857.5 m
2.8583×10⁻⁶ seconds
Explanation:
Time taken by the sound of the thunder to reach the student = 2.5 s
Speed of sound in air is 343 m/s
Speed of light is 3×10⁸ m/s
Distance travelled by the sound = Time taken by the sound × Speed of sound in air
⇒Distance travelled by the sound = 2.5×343 = 857.5 m
⇒Distance travelled by the sound = 857.5 m
Time taken by light = Distance the light travelled / Speed of light

Time taken by light = 2.8583×10⁻⁶ seconds
Answer:
A) F=-20.16×10⁹N
B) if the distance doubles, force is 4 times smaller.
Explanation:
q1=-28C
q2=5mC=0.005C
d=25cm=0.25m
Electrostatic force between charges: F=k×q1×q2/d², where k is a coefficient that has the value k=9 × 10⁹ N⋅m²⋅C^(-2) for air.
Thus:
F=9×10⁹×(-28)×0.005/0.25²
F=-20.16×10⁹N
The minus sign indicates attraction.
If distance doubles, d1=2×d, then we have 4d² at the denominator and the force is 4 times smaller.