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
Answer:
The automobile's acceleration in that time interval is -2 m/s^2
Explanation:
The acceleration is defined as the rate of change of the velocity.
The average acceleration in a given lapse of time is calculated as:
A = (final velocity - initial velocity)/time.
In this case, we have:
initial velocity = 31 m/s
final velocity = 15 m/s
time = 8 seconds.
Then the average acceleration is:
A = (15m/s - 31m/s)/8s = -2 m/s^2
Answer:
Explanation:
Distance between plates d = 2 x 10⁻³m
Potential diff applied = 5 x 10³ V
Electric field = Potential diff applied / d
= 5 x 10³ / 2 x 10⁻³
= 2.5 x 10⁶ V/m
This is less than breakdown strength for air 3.0×10⁶ V/m
b ) Let the plates be at a separation of d .so
5 x 10³ / d = 3.0×10⁶ ( break down voltage )
d = 5 x 10³ / 3.0×10⁶
= 1.67 x 10⁻³ m
= 1.67 mm.
Answer:
Earth is nearest the Sun in July and farthest away in July.
Explanation:
