Answer:
18 radians
Explanation:
The computation is shown below:
As we know that
Torque = Force × Moment arm
= 1N × 1M
= 1N-M
Torque =
Now
Here t = 1 minutes = 60 seconds
6) 100 ml of water is initially at 20°C. 30,000 J of heat is added to the water. What is temperature change for the water?
1 answer:
<h2>Δt = 71.67 °C</h2>
The temperature change of water is equal to 71.67 °C
<h3>Explanation:</h3>
Given:
Amount of transferred energy = 30,000 K J
Mass of water = 100 ml
Initial temperature = 20°C
To find the change in temperature of water.
Formula for Heat capacity is given by
Q = m×c×Δt ........................................(1)
where:
Q = Heat capacity of the substance (in J)
m=mass of the substance being heated in grams(g)
c = the specific heat of the substance in J/(g.°C)
Δt = Change in temperature (in °C)
Δt = (Final temperature - Initial temperature) = T(f) - T(i)
Q = 30,000 J
Mass of water = m = 100 ml
1 ml = 1 g ................................................(2)
Therefore m = 100 ml = 100 g
Specific heat of water is c = 4.186 J /g.
Δt = ?
Substituting these in equation (1), we get
Q = m×c×Δt
Rearranging the terms for Δt,
Δt =
Δt =
Δt = 71.67 °C
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Answer:
the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J
Explanation:
The torque is given by :
where ;
m = 0.160 A.m²
B = 0.0800 T
θ = 35°
So the magnitude of the torque N = mBsinθ
N = (0.160)(0.0800)(sin 35°)
N = 0.007341
N = 7.34×10⁻³ Nm
Hence, the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
b) The potential energy
U = -mBcosθ
U = (- 0.160)(0.0800)(cos 45)
U = -0.010485
U = -1.0485 ×10⁻² J
Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J