Answer:
The answer to your question is
Explanation:
Data
mass = 0.5kg
T1 = 35
T2 = ?
Q = - 6.3 x 10⁴ J = - 63000 J
Cp = 4184 J / kg°C
Formula
Q = mCp(T2 - T1)
T2 = T1 + Q/mCp
Substitution
T2 = 35 - 63000/(0.5 x 4184)
T2 = 35 - 63000/2092
T2 = 35 - 30.1
T2 = 4.9 °C
Answer:
Distance travel by go-cart = 500 meter
Explanation:
Given:
Speed of go cart = 25 m/s
Time travel = 20 seconds
Find:
Distance travel by go-cart
Computation:
Distance = Speed x time
Distance travel by go-cart = Speed of go cart x Time travel
Distance travel by go-cart = 25 x 20
Distance travel by go-cart = 500 meter
Answer:
a) The current density ,J = 2.05×10^-5
b) The drift velocity Vd= 1.51×10^-15
Explanation:
The equation for the current density and drift velocity is given by:
J = i/A = (ne)×Vd
Where i= current
A = Are
Vd = drift velocity
e = charge ,q= 1.602 ×10^-19C
n = volume
Given: i = 5.8×10^-10A
Raduis,r = 3mm= 3.0×10^-3m
n = 8.49×10^28m^3
a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]
J = (5.8×10^-10) /(2.83×10^-5)
J = 2.05 ×10^-5
b) Drift velocity, Vd = J/ (ne)
Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)
Vd = (2.05×10^-5)/(1.36 ×10^10)
Vd = 1.51× 10^-5
The magnitude of the magnetic force acting on the charge is 2.34×10⁻³ N.
<h3>What is magnetic force?</h3>
A magnetic force is the force that act in a magnetic field.
To calculate the magnetic force, we use the formula below.
Formula:
- F = qvB.........Equation 1
Where:
- F = magnetic force
- q = point charge
- v = Velocity of the the charge
- B = Field strength
From the question,
Given:
- q = 5.0×10⁻⁷ C
- v = 2.6×10⁵ m/s
- B = 1.8×10⁻² T
Substitute these values into equation 2
- F = (5.0×10⁻⁷)(2.6×10⁵)(1.8×10⁻²)
- F = 23.4×10⁻⁴
- F = 2.34×10⁻³ N
Hence, the magnitude of the magnetic force acting on the charge is 2.34×10⁻³ N.
Learn more about magnetic force here: brainly.com/question/2279150
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