Answer:
V
I and II
III and IV
Explanation:
The impulse is equal to the change in momentum of the object involved, so we can calculate the change in momentum in each situation and compare them all.
Taking always east as positive direction, and labelling
u the initial velocity
v the final velocity
m = 1000 kg the mass (which is always equal)
We find:
(i)
u = 25 m/s
v = 0
(II)
u = 25 m/s
v = 0
(III)
In this case,
F = 2000 N is the force
is the time
So the magnitude of the impulse is
(IV)
F = 2000 N is the force
is the time
So the magnitude of the impulse is
(V)
u = 25 m/s
v = -25 m/s
So the ranking from largest to smallest is:
V
I and II
III and IV
Answer:34 cm
Explanation:
Given
mass of meter stick m=80 gm
stick is balanced when support is placed at 51 cm mark
Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark
balancing torque
Answer:
43.58 m
Explanation:
If you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees
Using trigonometry ratio
Sin 5 = opposite/hypothenus
Where the hypothenus = 500m
Opposite = height h
Sin 5 = h/500
Cross multiply
500 × sin 5 = h
h = 500 × 0.08715
h = 43.58m
Therefore, the height above the starting point is equal to 43.58m
Answer:
correct is d) a ’= g / 2
Explanation:
For this exercise let's use the kinematics equations
On earth
v = v₀ - a t
a = (v₀- v) / T
On planet X
v = v₀ - a' t’
a ’= (v₀-v) / 2T
Let's substitute the land values in plot X
a’= a / 2
Now let's use Newton's second law
W = ma
m g = m a
a = g
We substitute
a ’= g / 2
So we see that on planet X the acceleration is half the acceleration of Earth's gravity
