To complete a task in a short amount of time/get from A to B in the quickest amount of time
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The distance covered by the body is 114.3 m
Explanation:
The work done by a force exerted on an object is given by
where
F is the magnitude of the force
d is the displacement of the object
is the angle between the direction of the force and of the displacement
For the object in this problem, we have
F = 350 N is the force applied
is the work done
if we assume that the force is applied parallel to the motion of the object
Solving for d, we find the distance covered by the object:
Learn more about work:
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Answer:
232 J/K
Explanation:
The amount of heat gained by the air = the amount of heat lost by the tea.
q_air = -q_tea
q = -mCΔT
q = -(0.250 kg) (4184 J/kg/ºC) (20.0ºC − 85.0ºC)
q = 68,000 J
The change in entropy is:
dS = dQ/T
Since the room temperature is constant (isothermal):
ΔS = ΔQ/T
Plug in values (remember to use absolute temperature):
ΔS = (68,000 J) / (293 K)
ΔS = 232 J/K
Formula for Electric Field strength , E = V/d
where V = Voltage in volts, and d = distance of separation in meters.
d = 0.5 cm = 0.005 m, V = 12 V
E = V/d = 12 / 0.005
E = 2400
Electric Field Strength = 2400 Volts/meter
Answer:
T=1.384×10⁶seconds
Explanation:
Given data
p (Intensity)=1.30 kw/m²
E (Energy)=1.8×10⁹ J
A (Area)=1.00 m²
T (Time required)=?
Solution
E=PT ................eq(i)
where E is energy
P is radiation power
T is time
Radiating Power is given as
P=pA
Where p is intensity
A is Area
Put P=pA in eq(i) we get
E=pAT
T=E/pA