Answer:
75
Explanation:
<em>The velocity would be 75 and the wavenumber would be 0.2!</em>
<em />
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Answer:
From the figure,
The free-body diagrams for and are shown in the figures below. The only forces on the blocks are the upward tension and the downward gravitational forces and . Applying Newton’s second law, we obtain:
which can be solved to yield
Substituting the result back, we have
(a) With and , the acceleration becomes
(b) Similarly, the tension in the cord is
Answer:
The magnitude of the average drag force is 2412.34 N.
Explanation:
Given that,
Mass of car 
Velocity v = 25.8 m/s
Distance 
Speed of car = 13.1 m/s
Height = 12.5 m
We need to calculate the magnitude of the average drag force
Using equation kinetic energy


Where,
= initial velocity
= final velocity
h = height
g = acceleration due to gravity
=drag force
m = mass of the car
d = distance
Put the value into the formula




Hence, The magnitude of the average drag force is 2412.34 N.
Answer:
–2.25 m/s²
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 15 m/s
Distance travelled (s) = 50 m
Final velocity (v) = 0 m/s
Deceleration (a) =?
v² = u² + 2as
0² = 15² + (2 × a × 50)
0 = 225 + 100a
Collect like terms
0 – 225 = 100a
– 225 = 100a
Divide both side by 100
a = –225/100
a = –2.25 m/s²
Thus, the deceleration of the vehicle is –2.25 m/s²
Answer:
The object distance is 20cm
Explanation:
Given
Magnification = -1.5
Image distance = 30 cm.
Required
Object Distance
We can calculate the object's distance using magnification formula;
M = -V/U
Where M = Magnification = -1.50
V = Image Distance = 30cm
U = Object Distance
Substitute the above parameters in the formula above.
-1.50 = -30/U
Multiply both sides by -1
-1.50 * -1 = -30/U * -1
1.50 = 30/U
Multiply both sides by U
1.50 * U = 30/U * U
1.50U = 30
Divide through by 1.50
1.50U/1.50 = 30/1.50
U = 30/1.50
U = 20cm
Recall that U represented the object distance.
Hence, the object distance is 20cm