There are several information's of immense importance that are already given in the question. Based on those information's the answer to the question can be very easily determined.
Initial velocity of the object (V1) = 65 m/s
Final velocity of the object (V2) = 98 m/s
Time taken by the object = 12 seconds
Then
Acceleration of the object = (V2 - V1)/Time taken
= (98 - 65)/12 m/s^2
= 33/12 m/s^2
= 2 9/12 m/s^2
= 2.75 m/s^2
So the acceleration of the object can be written as 2 9/12 meter per square second or 2.75 meter per square second.
3.00 is my correct answer this is what my teacher told me.thanks
Answer:
<em>Increase in wave frequency only</em>
Answer:
A) ΔU = 3.9 × 10^(10) J
B) v = 8420.75 m/s
Explanation:
We are given;
Potential Difference; V = 1.3 × 10^(9) V
Charge; Q = 30 C
A) Formula for change in energy of transferred charge is given as;
ΔU = QV
Plugging in the relevant values gives;
ΔU = 30 × 1.3 × 10^(9)
ΔU = 3.9 × 10^(10) J
B) We are told that this energy gotten above is used to accelerate a 1100 kg car from rest.
This means that the initial potential energy will be equal to the final kinetic energy since all the potential energy will be converted to kinetic energy.
Thus;
P.E = K.E
ΔU = ½mv²
Where v is final velocity.
Plugging in the relevant values;
3.9 × 10^(10) = ½ × 1100 × v²
v² = [7.8 × 10^(8)]/11
v² = 70909090.9090909
v = √70909090.9090909
v = 8420.75 m/s