A kangaroo hops at an angle of 25° to the horizontal with a velocity of 20 m/s. What is the vertical component of the velocity?
2 answers:
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Answer:
Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. What is the speed v(final) of the electron when it is 10.0 cm from
The answer to the question is
The speed of the electron when it is 10.0 cm from charge Q₁
= 7.53×10⁶ m/s
Explanation:
To solve the question we have
Q₁ = 3.45 nC = 3.45 × 10⁻⁹C
Q₂ = 1.85 nC = 1.85 × 10⁻⁹ C
2·d = 50.0 cm
a = 10.0 cm
q = -1.6×10⁻¹⁹C
Also initial kinetic energy = 0 and
Initial electric potential energy =
Final kinetic energy due to motion = 0.5·m·v²
Final electric potential energy =
From the energy conservation principle we have
Solving for v gives
where k = 9.0×10⁹ and m = 9.109×10⁻³¹ kg
gives v =7528188.32769 m/s or 7.53×10⁶ m/s
= 7.53×10⁶ m/s
Answer:
Truck will be at a distance of 720 m
Explanation:
Speed of the truck is given as 90 m/sec
Time for which truck travel t = 10 seconds
We have to find the distance that how much truck travel in 8 sec with speed of 90 m/sec
We know that distance is given by
So distance traveled by truck in 8 sec will be equal to
So truck will be at a distance of 720 m