Theoritically
the body moving with uniform velocity has acceleration zero.
Mathmatically,
u=3m/s
v=3m/s (since body is moving with uniform velocity)
a= v-u/t
3-3/t
0/t
0m/s.s
The law of conservation of momentum tells us that momentum
is conserved, therefore total initial momentum should be equal to total final
momentum. In this case, we can expressed this mathematically as:
mA vA + mB vB = m v
where, m is the mass in kg, v is the velocity in m/s
since m is the total mass, m = mA + mB, we can write the
equation as:
mA vA + mB vB = (mA + mB) v
furthermore, car B was at a stop signal therefore vB = 0,
hence
mA vA + 0 = (mA + mB) v
1800 (vA) = (1800 + 1500) (7.1 m/s)
<span>vA = 13.02 m/s</span>
Answer:
16.8ohms
Explanation:
According to ohm's law which states that the current passing through a metallic conductor at constant temperature is directly proportional to the potential difference across its ends.
Mathematically, V = IRt where;
V is the voltage across the circuit
I is the current
R is the effective resistance
For a series connected circuit, same current but different voltage flows through the resistors.
If the initial current in a circuit is 19.3A,
V = 19.3R... (1)
When additional resistance of 7.4-Ω is added and current drops to 13.4A, our voltage in the circuit becomes;
V = 13.4(7.4+R)... (2)
Note that the initial resistance is added to the additional resistance because they are connected in series.
Equating the two value of the voltages i.e equation 1 and 2 to get the resistance in the original circuit we will have;
19.3R = 13.4(7.4+R)
19.3R = 99.16+13.4R
19.3R-13.4R = 99.16
5.9R = 99.16
R= 99.16/5.9
R = 16.8ohms
The resistance in the original circuit will be 16.8ohms
Answer:
b) 472HZ, 408HZ
Explanation:
To find the frequencies perceived when the bus approaches and the train departs, you use the Doppler's effect formula for both cases:
fo: frequency of the source = 440Hz
vs: speed of sound = 343m/s
vo: speed of the observer = 0m/s (at rest)
v: sped of the train
f: frequency perceived when the train leaves us.
f': frequency when the train is getTing closer.
Thus, by doing f and f' the subjects of the formulas and replacing the values of v, vo, vs and fo you obtain:
hence, the frequencies for before and after tha train has past are
b) 472HZ, 408HZ
