Answer:
Kinetic energy of the shot is 400 Joules
Explanation:
It is given that,
Mass of the shot, m = 8 kg
Speed of thrown, v = 10 m/s
We have to find the kinetic energy of the shot. It is given by the product of mass of an object and the square of velocity divided by 2 i.e.
KE = 400 Joules
Hence, the kinetic energy of the shot is 400 joules.
b. 9.8 meters per minute
c. 8.9 meters per minute squared
d. 9.8 meters per second squared
Answer:
9.8 meters per second squared
Explanation:
exportation.
civil war.
dictatorships.
b) By what factor does the current change if the side dimensions of the membrane portion is halved? The other values do no change
increase by factor of 2
decrease by factor of 8
decrease by factor of 2
decrease by a factor of 4
increase by factor of 4
Answer:
Part a: The current is 1.49x10⁻¹² A
Part b: The current is decreased by a factor of 4.
Explanation:
Part a
The area is given as
A =(1.4*10⁻⁶)² =1.96*10⁻¹² m²
The resistance is given as where resistivity of the membrane material is 1.30 x 10⁷ ohms*m
R=pL/A =(1.3*10⁷)(7.5*10⁻⁹)/(1.96*10⁻¹²)
R=4.97x10¹⁰ ohms
So the resistance is 4.97x10¹⁰ ohms.
I=V/R =86.0x10⁻³/5.77x10¹⁰
I=1.49x10^⁻¹² A
So the current is 1.49x10⁻¹² A
b)
S=So/2=1.4/2 =0.7μm
A=(0.7*10^-6)^2=4.9*10⁻¹³ m²
R=pL/A =(1.3*10⁷)(7.5*10⁻⁹)/(4.9*10⁻¹³)
R=1.98x10¹¹ ohms
So the resistance is 1.98x10¹¹ ohms.
I=V/R =86.0x10⁻³/1.98x10¹¹
I2=4.52x10^⁻¹³ A
So the ratio of the new current vs the old current is as
I2/I=4.52x10^⁻¹³/1.49x10⁻¹²=0.25
So the current is decreased by a factor of 4.
B. Iron core
C. Rocky crust
D. Thin atmosphere
E. Dense atmosphere
F. Relatively high density