Answer:
Explanation:
<u>Free Fall Motion</u>
A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.
The speed vf of the object when a time t has passed is given by:
Where 
Similarly, the distance y the object has traveled is calculated as follows:
If we know the height h from which the object was dropped, we can solve the above equation for t:
The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:
The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:
Therefore, it has traveled down a distance:
Thus, the height of the pen is:
Answer:
T₂ = 111.57 °C
Explanation:
Given that
Initial pressure P₁ = 9.8 atm
T₁ = 32°C = 273 + 32 =305 K
The final pressure P₂ = 11.2 atm
Lets take the final temperature = T₂
We know that ,the ideal gas equation
If the volume of the gas is constant ,then we can say that
Now by putting the values in the above equation ,we get
T₂ = 384.57 - 273 °C
T₂ = 111.57 °C
Answer:
19.99 kg m²/s
Explanation:
Angular Momentum (L) is defined as the product of the moment of Inertia (I) and angular velocity (w)
L = m r × v.
r and v are perpendicular to each other,
where r = lsinθ.
l = 2.4 m
θ= 34°
g = 9.8 m/s² and m = 5 kg
resolving using newtons second law in the vertical and horizontal components.
T cos θ − m g = 0
T sin θ − mw² lsin θ = 0
where T is the force with which the wire acts on the bob
w = √g / lcosθ
= √ 9.8 / 2.4 ×cos 34
= 2.2193 rad/s
the angular momentum L = mr× v
= mw (lsin θ)²
= 5 × 2.2193 (2.4 ×sin 34°)²
=19.99 kg m²/s
Answer:
Magnetic force, F = 0.24 N
Explanation:
It is given that,
Current flowing in the wire, I = 4 A
Length of the wire, L = 20 cm = 0.2 m
Magnetic field, B = 0.6 T
Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :
F = 0.24 N
So, the force on the wire at an angle of 30° with respect to the field is 0.24 N. Hence, this is the required solution.
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