Answer:
W=1705.2 J
Explanation:
Given that
mass ,m= 60 kg
Acceleration due to gravity ,g= 9.8 m/s²
Height ,h= 2.9 m
As we know that work done by a force given as
W = F . d
F=force
d=Displacement
W=work done by force
Now by putting the values
F= m g (Acting downward )
d= h (Upward)
W= m g h ( work done against the force)
W= 60 x 9.8 x 2.9 J
W=1705.2 J
Therefore the answer will be 1705.2 J.
v
Convert the given temperatures from celsius to kelvin since we are dealing with gas.
To convert to kelvin, add 273.15 to the temperature in celsius.
T1 = 22 + 273.15 = 295.15 k
T2 = 4 + 273.15 = 277.15 k
V1 = 0.5 L
Let's find the final volume (V2).
To solve for V2 apply Charles Law formula below:
Answer:
a) 1450watts
b) 564watts
c) 1.11
Explanation:
Power consumed = IV
I is the current rating
V is the operating voltage
If a blow-dryer and a vacuum cleaner each operate with a voltage of 120 V and the current rating of the blow-dryer is 12 A, while that of the vacuum cleaner is 4.7 A then their individual power rating is calculated thus;
a) For blow-dryer
Operating voltage = 120V
Its current rating = 12A
Power consumed = IV
= 120×12
= 1440watts
b) For vacuum cleaner:
Operating voltage is the same as that of blow dryer = 120V
Its current rating = 4.7A
Power consumed = IV
= 120×4.7
= 564watts
c) Energy used = Power consumed × time taken
Energy used = Power × time
Energy used by blow dryer = 1440×20×60
= 1,728,000Joules
Energy used up by vacuum cleaner = 564×46×60
= 564×2760
= 1,556,640Joules
Ratio of the energy used by the blow-dryer in 20 minutes to the energy used by the vacuum cleaner in 46 minutes will be 1,728,000/1,556,640 = 1.11
Explanation:
It is given that,
Mass of the tackler, m₁ = 120 kg
Velocity of tackler, u₁ = 3 m/s
Mass, m₂ = 91 kg
Velocity, u₂ = -7.5 m/s
We need to find the mutual velocity immediately the collision. It is the case of inelastic collision such that,


v = -1.5 m/s
Hence, their mutual velocity after the collision is 1.5 m/s and it is moving in the same direction as the halfback was moving initially. Hence, this is the required solution.
The strength of the electric field on the point charge at this distance will be 4000 V/m.
<h3>What is the strength of the electric field?</h3>
The strength of the electric field is the ratio of electric force per unit charge.
The given data in the problem is;
Qis the unit charge = 4.0 × 10⁻⁶ C
E is the strength of the electric field
R is the distance from point charge = 3 m
The strength of the electric field is;

Hence, the strength of the electric field on the point charge at this distance will be 4000 V/m.
To learn more about the strength of the electric field refer to the link;
brainly.com/question/15170044
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