13.0m/s
1.2m/s
Explanation:
Given parameters:
Initial speed of the body = 7.1m/s
time taken = 2.23s
Acceleration = 2.64m/s²
Unknown:
Final speed = ?
Solution:
Acceleration is the rate of change of velocity with time.
a = 
a = acceleration
V = final speed
U = initial speed
T = time taken
Input the variables and solve for V;
2.64 = 
V - 7.1 = 5.9 expression 1
V = 5.9 + 7.1 = 13.0m/s
B
Using the same parameters, the speed after a uniform deceleration of -2.64m/s², the negative sign implies deceleration;
from expression 1;
V - 7.1 = -5.9
V = -5.9 + 7.1 = 1.2m/s
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Answer:
a) w = 7.27 * 10^-5 rad/s
b) v1 = 463.1 m/s
c) v1 = 440.433 m/s
Explanation:
Given:-
- The radius of the earth, R = 6.37 * 10 ^6 m
- The time period for 1 revolution T = 24 hrs
Find:
What is the earth's angular speed?
What is the speed of a point on the equator?
What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?
Solution:
- The angular speed w of the earth can be related with the Time period T of the earth revolution by:
w = 2π / T
w = 2π / 24*3600
w = 7.27 * 10^-5 rad/s
- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.
v1 = R*w
v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)
v1 = 463.1 m/s
- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.
π/2 ........... s
x ............ 1/5 s
x = π/2*5 = 18°
- The radius of the earth R' at point where θ = 18° from the equator is:
R' = R*cos(18)
R' = (6.37 * 10 ^6)*cos(18)
R' = 6058230.0088 m
- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.
v2 = R'*w
v2 = (6058230.0088)*(7.27 * 10^-5)
v2 = 440.433 m/s
Answer:
Explanation: When the electrons move in another direction, they convert this chemical potential energy to electricity in the circuit, thus discharging the battery. So, the battery is all potential energy.
A vertical polarizing filter is used on the lens of a camera, they block out the light that is horizontally polarized, so they allow all of the vertically polarized<span> light to pass through.</span>
