Answer:
9.98 × 10⁻⁹ C
Explanation:
mass, m = 1.00 × 10⁻¹¹ kg
Velocity, v = 23.0 m/s
Length of plates D₀ = 1.80 cm = 0.018 m
Magnitude of electric field, E = 8.20 × 10⁴ N/C
drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m
density of the ink drop = 1000 kg/m^3
Now,
Time =
or
Time =
or
Time = 6.9 × 10⁻⁴ s
Now, force due to the electric field, F = q × E
where, q is the charge
Also, Force = Mass × acceleration
q × E = 1.00 × 10⁻¹¹ × a
or
a =
Now from the Newton's equation of motion
where,
d is the distance
u is the initial speed
a is the acceleration
t is the time
or
or
q = 9.98 × 10⁻⁹ C
Answer:
Hello your question is incomplete below is the complete question
Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000
answer : V = 1.624* 10^-5 m/s
Explanation:
First we have to calculate the value of a
a = 93 * 10^6 mile/m * 1609.344 m
= 149.668 * 10^8 m
next we will express the distance between the earth and the sun
--------- (1)
a = 149.668 * 10^8
E (eccentricity ) = ( 1/60 )^2
= 90°
input the given values into equation 1 above
r = 149.626 * 10^9 m
next calculate the Earths velocity of approach towards the sun using this equation
------ (2)
Note :
Rc = 149.626 * 10^9 m
equation 2 becomes
(
therefore : V = 1.624* 10^-5 m/s
Answer:
1, When Jane brakes, the brakes slow the car wheels turning and the road surface exerts a backwards force on the tires, causing the car to decelerate. The pocket book tends to continue on in a straight line (Newton's first law). If she brakes hard enough that the friction between the book and the car seat is insufficient to decelerate the book as fast as the car is decelerating, the book will slide off the seat, and gravity pulls it to the floor
2.
When the diver uses his / her force to depress the springboard, the springboard pushes him back with equal force
3.Newton's Second Law (F=ma)
4. 5 N
5. 19.5 N
65kg * 0.3 m/s^2
6.0.2 N/s
10kg divided by 2N
7.-Walking then pushing the moving forward
-Dribbling
-Basketball is pushed but bounces back
Explanation:
Answer:
b. 0.20 m/s.
Explanation:
Given;
initial mass, m = 0.2 kg
maximum speed, v = 0.3 m/s
The total energy of the spring at the given maximum speed is calculated as;
K.E = ¹/₂mv²
K.E = 0.5 x 0.2 x 0.3²
K.E = 0.009 J
If the mass is changed to 0.4 kg
¹/₂mv² = K.E
mv² = 2K.E

Therefore, the maximum speed is 0.20 m/s