Answer:
V = 4.63 m/s
V = 11.31 m/s
Explanation:
Given,
The distance traveled by the bus, towards north, d = 2.5 km
= 2500 m
The time taken by the trip is, t = 9 min
= 540 s
The velocity of the bus,
V = d / t
= 2500 / 540
= 4.63 m/s
At another point, the bus travels at a constant speed of v = 18 m/s
Therefore the velocity becomes
V = (4.63 + 18)/2
= 11.31 m/s
Hence, the velocity of the bus, V = 11.31 m/s
Answer:
b) The star is moving away from us.
Explanation:
If an object moves toward us, the light waves it emits are compressed - the wavelength of the light will be shorter, making the light bluer. On the other hand, if an object moves away from us, the light waves are stretched, making it redder. If from laboratory measurements we know that a specific hydrogen spectral line appears at the wavelength of 121.6 nanometers (nm) and the spectrum of a particular star shows the same hydrogen line appearing at the wavelength of 121.8 nm, we can conclude that the star is moving away from npos, since the wavelength related to that star is more expanded.
Answer:
This means that the kinetic energy of second object is 48times that of the first object
Explanation:
Kinetic energy is the energy possessed by a body by virtue of its motion e.g motion of an accelerating car. Mathematically,
Kinetic energy = 1/2mv² where;
m is the mass of the object
v is the velocity of the object
If Object 1 of mass m moves with speed v in the positive direction, its kinetic energy will be expressed as;
K1 = 1/2mv²
For Object 2 of mass 3m moving with speed 4v in the negative x-direction, its kinetic energy can be expressed as;
K2 = 1/2(3m)(4v)²
K2 = 1/2(3m)(16v²)
K2 = (3m)(8v²)
K2 = 24mv²
To compare the kinetic energy of both bodies, we will take the ratio of K2:K1 to have;
K2/K1 = 24mv²/(1/2)mv²
K2/K1 = 24/(1/2)
K2/K1 = 48
K2 = 48K1
This means that the kinetic energy of second object is 48times that of the first object and moving in the negative x direction since the body of mass 3m initially moves in the negative x direction.
Answer:
Explanation:
From the question we are told that
Generally the equation for momentum is mathematically given by
Therefore
T-Joe momentum 
