Answer: 
Explanation:
According to Newton's 2nd Law of motion the force
is proportional to the mass
and acceleration
:
(1)
On the other hand, the equation for the Centripetal force is:
(2)
Where:
is the velocity
is the radius of the circular motion
Making (1) and (2) equal:
(3)
Hence:
This is the expression for the centripetal acceleration
It should be noted, this acceleration is directed toward the center of the circumference of the circular motion (that's why it's called centripetal acceleration).
Answer:
313.92w
Explanation:
Formula for power:
P=W/∆t = Fv
Givens:
m=20kg
∆y=4.0m
∆t=2.5s
a=9.81m/s²
In order to find power, we first need to solve for work.
W=Fd (force*displacement), f=mg
W=mg∆y
W=(20kg)(9.81m/s²)(4.0m)
W=784.8J
P=W/∆t
P=784.8J/2.5s
P=313.92 watts
<span>There is six horizen.
1. O Horizon - The top, organic layer of soil,
2. A Horizon - The layer called topsoil;
3. E Horizon - This layer is beneath the A Horizon and above the
B Horizon. It is made up mostly of sand.
4. B Horizon - Also called the subsoil - this layer is beneath the E
Horizon and above the C Horizon.
5. C Horizon - it's called regolith: the layer beneath the B Horizon
and above the R Horizon.
6 R Horizon - this is last and the unweathered rock layer that is
beneath all the other layers.</span>
Answer:
The magnitude of the applied torque is 
(e) is correct option.
Explanation:
Given that,
Mass of object = 3 kg
Radius of gyration = 0.2 m
Angular acceleration = 0.5 rad/s²
We need to calculate the applied torque
Using formula of torque

Here, I = mk²

Put the value into the formula



Hence, The magnitude of the applied torque is 
Answer:
R = 35.27 Ohms
Explanation:
Given the following data;
Voltage = 230V
Power = 1500W
To find the resistance, R;
Power = V²/R
Where:
V is the voltage measured in volts.
R is the resistance measured in ohms.
Substituting into the equation, we have;
1500 = 230²/R
Cross-multiplying, we have;
1500R = 52900
R = 52900/1500
R = 35.27 Ohms.
Therefore, the resistance which the heating element needs to have is 35.27 Ohms.