Answer:
If the force applied is larger than 185.2 N, yes.
Explanation:
In order to move the table, the pushing force must be larger than the frictional force. The frictional force is given by:
where
is the coefficient of static friction
is the mass of the table
is the gravitational acceleration
Substituting,
So, we are able to move the table if we push with a force larger than 185.2 N.
Answer:
Answer:
Explanation:
Given that
K=8.98755×10^9Nm²/C²
Q=0.00011C
Radius of the sphere = 5.2m
g=9.8m/s²
1. The electric field inside a conductor is zero
εΦ=qenc
εEA=qenc
net charge qenc is the algebraic sum of all the enclosed positive and negative charges, and it can be positive, negative, or zero
This surface encloses no charge, and thus qenc=0. Gauss’ law.
Since it is inside the conductor
E=0N/C
2. Since the entire charge us inside the surface, then the electric field at a distance r (5.2m) away form the surface is given as
F=kq1/r²
F=kQ/r²
F=8.98755E9×0.00011/5.2²
F=36561.78N/C
The electric field at the surface of the conductor is 36561N/C
Since the charge is positive the it is outward field
3. Given that a test charge is at 12.6m away,
Then Electric field is given as,
E=kQ/r²
E=8.98755E9 ×0.00011/12.6²
E=6227.34N/C
<span>The three types of seismic waves produced by an earthquake are primary, secondary, and (D.) surface.</span>
Answer:
F = 120 N
Explanation:
Force x distance = energy
The bike has energy 1/2 . 80 . 6^2 = 1440 J
You are looking at an example of not reading the question properly.
Impulse = Force . time = change in momentum
F . 4 = 80 .6
F = 120 N
(A) We can solve the problem by using Ohm's law, which states:
where
V is the potential difference across the electrical device
I is the current through the device
R is its resistance
For the heater coil in the problem, we know
and
, therefore we can rearrange Ohm's law to find the current through the device:
(B) The resistance of a conductive wire depends on three factors. In fact, it is given by:
where
is the resistivity of the material of the wire
L is the length of the wire
A is the cross-sectional area of the wire
Basically, we see that the longer the wire, the larger its resistance; and the larger the section of the wire, the smaller its resistance.
