
Actually Welcome to the Concept of the kinematics of a body.
Since, we know that Velocity = Distance / time
hence, V = 20/5 = 4 m/s
hence the velocity of the RC car is 4 m/s westwards direction.
The Image distance and Magnification of The Image will be 30 cm and 3.
<h3>What is focal length?</h3>
The focal length of the lens, which is often expressed in millimeters, is the distance between the lens and the image sensor when the subject is in focus.
Given data;
Focal length,f=?
Image distance,v=?
Object distance,u= 10 cm
Magnification,m= 2.85
The focal length is half of the radius;
f=R/2
f=30 Cm/2
f= 15 Cm
The mirror equation is found as;

The magnification of the lens is found as;

Hence, the image distance and magnification of The image will be 30 cm and 3.
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Answer:
mass of the neutron star =3.45185×10^26 Kg
Explanation:
When the neutron star rotates rapidly, a material on its surface to remain in place, the magnitude of the gravitational acceleration on the central material must be equal to magnitude of the centripetal acc. of the rotating star.
That is

M_ns = mass odf the netron star.
G= gravitational constant = 6.67×10^{-11}
R= radius of the star = 18×10^3 m
ω = 10 rev/sec = 20π rads/sec
therefore,

= 3.45185... E26 Kg
= 3.45185×10^26 Kg
Answer:
Option D
670 Kg.m/s
Explanation:
Initial momentum is given by mv=82*5.6=459.2 Kg.m/s (taking eastward as positive)
Final momentum is also mv but v being westward direction, we take it negative
Final momentum=82*-2.5= -205 Kg.m/s
Change in momentum=Final momentum-Initial momentum=-205-459.2=-664.2 Kg.m/s
Impulse=change in momentum=664.2 Kg.m/s rounded off as 670 Kg.m/s
As capacitor was discharging, The charge on the plate got reversed and the motion of charge is opposite to the flow of current.
The charging contemporary asymptotically processes 0 as the capacitor becomes charged up to the battery voltage.
The capacitor is completely charged when the voltage of the electricity supply is equal to that at the capacitor terminals. that is referred to as capacitor charging; and the charging segment is over when modern-day stops flowing thru the electrical circuit.
A capacitor can be slowly charged to the important voltage and then discharged quick to provide the power wanted. it's far even viable to charge several capacitors to a positive voltage and then discharge them in any such way as to get extra voltage out of the gadget than became installed.
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