Answer:
e. All of the above are projectile
Explanation:
A projectile is an object with motion, aka a non-zero speed. A cannonball throwing straight up, rolling down a slope, rolling off the edge of a tale, thrown through the air have motion. They all have speed and kinetic energy. Therefore they can all be considered a projectile.
Answer:
Explanation:
The centripetal acceleration requirement must equal gravity at the top of the circle
mg = mv²/R
v = √Rg
v = √(1.0(9.8))
v = 3.1304951...
v = 3.1 m/s
Answer:
V(average)=6.37 V
Explanation:
Given Data
Peak Voltage=10V
Frequency=10 kHZ
To Find
Average Voltage
Solution
For this first we need to find Voltage peak to peak
So
Voltage (peak to peak)= 2× voltage peak
Voltage (peak to peak)= 2×10
Voltage (peak to peak)= 20 V
Now from Voltage (peak to peak) formula we can find the Average Voltage
So
Voltage (peak to peak)=π×V(average)
V(average)=Voltage (peak to peak)/π
V(average)=20/3.14
V(average)=6.37 V
The point P lies at a linear distance 1 m from the corner. Magnetic field due to current in wire Magnetic field due to current in wire. Applying Right hand thumb rule, the magnetic field at P is perpendicular to xy plane and into the plane of drawing (i.e. negative z-direction).
Answer:
b) G.P.E = Mgh
300j = M x 10 m/s² x 15 m
300 j/ 10 m/s² x 15 m = M
300j/ 150 s² = M
2kg = M
c) K.E = 1/2 m v²
K.E = 1/2 (50) (50)²
K.E = 1/2 (50) (2500)
K.E= 125000/2
K.E = 625 000 J
