Complete question is:
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.
Answer:
(V_A) = 31.32 m/s
Explanation:
We are given;
car's mass, m = 1200 kg
h_A = 100 m
h_B = 150 m
v_B = 0 m/s
From law of conservation of energy,
the distance from point A to B is;
h = 150m - 100 m = 50 m
From Newton's equations of motion;
v² = u² + 2gh
Thus;
(V_B)² = (V_A)² + (-2gh)
(negative next to g because it's going against gravity)
Thus;
(V_B)² = (V_A)² - (2gh)
Plugging in the relevant values;
0² = (V_A)² - 2(9.81 × 50)
(V_A) = √981
(V_A) = 31.32 m/s
Answer:
kinetic energy will change by a factor of 1/2
Option C) 1/2 is the correct answer
Explanation:
Given the data in the question;
we know that;
Kinetic energy = 1/2.mv²
given that mass of the object is doubled; m1 = 2m
speed is halved; v1 = V/2
Now, New kinetic energy will be; 1/2.m1v1²
we substitute
Kinetic Energy = 1/2 × 2m × (v/2)²
Kinetic Energy = 1/2 × 2m × (v²/4)
Kinetic Energy = 1/2 × m × (v²/2)
Kinetic Energy = 1/2 [ 1/2mv² ]
Kinetic Energy = 1/2 [ KE ]
Therefore; kinetic energy will change by a factor of 1/2
Option C) 1/2 is the correct answer
Answer:
X and Z
Explanation:
Conduction occurs through direct physical contact. Heat transferred from the pot to the handle, and from the handle to the hand, are both examples of conduction.
Answer:
Δx = 2.76 x 10⁻³ m = 2.76 mm
Explanation:
The distance between any two consecutive dark or consecutive bright fringes is given by:
Δx = λL/d
where,
Δx = distance between first and second dark fringe = ?
λ = wavelength of light = 546.1 nm = 546.1 x 10⁻⁹ m
L = distance between slits = 1.35 m
d = slit separation = 0.267 mm = 2.67 x 10⁻⁴ m
Therefore,
Δx = (546.1 x 10⁻⁹ m)(1.35 m)/(2.67 x 10⁻⁴ m)
<u>Δx = 2.76 x 10⁻³ m = 2.76 mm</u>
