Answer:
38,437.5
Explanation:
Density(d)= 102.5g/ml
Volume (v)=375ml
Mass(m) = ?
D =m/v
102.5= m/375
102.5*375=m
38,437.5=m
therefore Mass = 38,437.5g/ml.
The total resistance of an electric circuit with resistors widener series in the sum of the individual resistances:
Each resistor in a series circuit has a same amount of current flowing through it.
Each resistor in a parallel circuit has the same for voltage of the source applied to it.
When was this is are connected in parallel, the supply current is equal to the sum of the current through each resistor. In other words the currents in the branches of a parallel circuit add up to the supply current. When resistors are connected in parallel they have the same potential differences across them.
Ice sheets water vapor rivers
I think you want to determine the exit speed?
You have to determine how much velocity was decreased by calculating it from the kinetic energy.
KE = (1/2)mv²
1.4 x 10^5 = (1/2)*(1100)v²
v² = 254.55
v =15.95 m/s
So the velocity reduces by 15.95 m/s. Subtracting this to the initial velocity: 22 - 15.95 = 6.05 m/s.
So, the final speed was 6.05 m/s.
I hope I was able to help :)
This question is incomplete, the complete question is;
A particle is directed along the axis of the instrument in the figure below. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by d = 2.0 mm and the value of the magnetic field is B = 0.60T.
Calculate the potential difference, between the capacitor plates, required to allow a particle with speed v = 5.0 × 10⁵ m/s to pass straight through without deflection.
<em>Hint </em>: ΔV = Ed <em>
</em>
Answer:
the required potential difference, between the capacitor plates is 600 V
Explanation:
Given the data in the question;
B = 0.60 T
d = 2.0 mm = 0.002 m
v = 5.0 × 10⁵ m/s.
since particle pass straight through without deflection.
F
= 0
so, F
= F
qE = qvB
divide both sides by q
E = vB
we substitute
E = (5.0 × 10⁵) × 0.6
E = 300000 N/C
given that; potential difference ΔV = Ed
we substitute
ΔV = 300000 × 0.002
ΔV = 600 V
Therefore, the required potential difference, between the capacitor plates is 600 V