Acceleration = (change in speed) / (time for the change)
Change in speed = (speed after the change) - (speed before the change)
Change in speed = (65 m/s) - (35 m/s) = 30 m/s
Acceleration = (30 m/s) / (5 s)
<em>Acceleration = 6 m/s²</em>
Answer:
a = = 37.2V
b = 13.39MJ
Explanation:
Given that
L = 170 × 10³
r = d/2
= 10cm / 2 = 5 cm
current I = 100A
we are to find the potential drop across the cable
so, we can use ohm’s law
V = IR = I (ρL/A)
ρ = resistivity of the copper
= 1.72 × 10⁻⁸ Ω.m
A = πr²
V = I(ρL/πr²)
= 100 (( 1.72 × 10⁻⁸ * 170 × 10³) / ( π * 0.05²))
= 37.2V
(b)
Energy (loss) = Pt
Enery (loss) = IVt
3600s per hour
= (100A)(37.2V)(3600s)
= 13.39MJ
Given:
Object in circular motion 25 m/s
1 second to go quarter circle
Required:
Centripetal acceleration:
Solution:
Acceleration = v2/r
Where v is the velocity and r is
the radian
Substituting the values into the
equation,
Acceleration = v2/r = (25
m/s)2/(4*pi/180) = 8952.47 m2/s2
Answer: 1100 W
Explanation:
Input power = 220(5) = 1100 W
The transformer will step up/down voltage, but will also step down/up current.
Neglecting hysteresis and other minor losses, the power will remain the same.
Explanation:
Precision represents that how close the different measurements of the sample one take are to one another.
- One can increase the precision in lab by paying attention to each and every detail.
- Usage of the equipment properly and also increasing the sample size.
-
Ensuring that the equipment is calibrated properly. They should be clean and functioning. Using equipment which is not functioning correctly can cause results to swing wildly and also bits of the debris stuck to the equipment can influence the measurements of the mass and the volume.
- Each measurement must be taken multiple times, especially if experiments in which combining of the substances in specific amounts is involved.
