Answer:
The total distance is equal to the change in total position of the person, but the displacement of the person is zero.
Explanation:
- The distance is the total change in the positions of the toy robot. it is having only magnitude and it is a scalar quantity.
- The displacement is the difference between the final and initial position of the toy robot. it is a vector quantity having magnitude and direction
- Since the distance is the change in positions, so the magnitude of the distance will be equal to the up and down distance covered.
- If the toy robot travels in a straight line path and returns back to its original location, the magnitude of the distance and displacement doesn't depend on the speed of the toy robot.
Answer:
Explanation:
From the question we are told that
Radius 
Magnetic field
Generally the equation for area of circular path is mathematically given by
Generally the equation for Magnetic flux is mathematically given by
Answer:
total distance = 1868.478 m
Explanation:
given data
accelerate = 1.68 m/s²
time = 14.2 s
constant time = 68 s
speed = 3.70 m/s²
to find out
total distance
solution
we know train start at rest so final velocity will be after 14 .2 s is
velocity final = acceleration × time ..............1
final velocity = 1.68 × 14.2
final velocity = 23.856 m/s²
and for stop train we need time that is
final velocity = u + at
23.856 = 0 + 3.70(t)
t = 6.44 s
and
distance = ut + 1/2 × at² ...........2
here u is initial velocity and t is time for 14.2 sec
distance 1 = 0 + 1/2 × 1.68 (14.2)²
distance 1 = 169.37 m
and
distance for 68 sec
distance 2= final velocity × time
distance 2= 23.856 × 68
distance 2 = 1622.208 m
and
distance for 6.44 sec
distance 3 = ut + 1/2 × at²
distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²
distance 3 = 76.90 m
so
total distance = distance 1 + distance 2 + distance 3
total distance = 169.37 + 1622.208 + 76.90
total distance = 1868.478 m
Answer:
a. 17 μC b. 14.45 J
Explanation:
Here is the complete question
A pair of 10μF capacitors in a high-power laser are charged to 1.7 kV.
a. What charge is stored in each capacitor?
b. How much energy is stored in each capacitor?
a. The charge Q stored in a capacitor is Q = CV where C = capacitance and V = voltage. Here, C = 10μF = 10 × 10⁻⁶ F and V = 1.7 kV = 1.7 × 10³ V.
So, Q = CV = 10 × 10⁻⁶ × 1.7 × 10³ = 1.7 × 10⁻² C = 17 × 10⁻³ C = 17 μC
b. The energy stored in a capacitor is given by W = 1/2CV² = 1/2 × 10 × 10⁻⁶ F × (1.7 × 10³ V)² = 1/2 × 10 × 10⁻⁶ F × 2.89 × 10⁶ V² = 1/2 × 10 × 2.89 × 10⁶ × 10⁻⁶ J = 14.45 J
v
Convert the given temperatures from celsius to kelvin since we are dealing with gas.
To convert to kelvin, add 273.15 to the temperature in celsius.
T1 = 22 + 273.15 = 295.15 k
T2 = 4 + 273.15 = 277.15 k
V1 = 0.5 L
Let's find the final volume (V2).
To solve for V2 apply Charles Law formula below:
