Why does an object accelerate when it falls toward the earth's surface
2 answers:
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Answer:
6.4m/s
Explanation:
The total mechanical energy of the man is 1780J.
This mechanical energy is the energy due to the motion of the body and it is a form of kinetic energy.
Also, mass = 87kg
Kinetic energy = m v²
m is the mass
v is the velocity
1780 = x 87 x v²
v² = 40.9
v = 6.4m/s
Answer:
just before landing the ground
Explanation:
Let the velocity of projection is u and the angle of projection is 30°.
Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.
initial horizontal component of velocity, ux = u Cos 30
initial vertical component of velocity, uy = u Sin 30
Time of flight is given by
Final horizontal component of velocity, vx = ux = u Cos 30
Let vy is teh final vertical component of velocity.
Use first equation of motion
vy = uy - gT
vy = - u Sin 30
The magnitude of final velocity is given by
v = u
Thus, the velocity is same as it just reaches the ground.
Answer:
the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
Explanation:
Given that;
diameter D = 2.0 mm
current I = 1.0 mA
K.E of each proton is 20 MeV
the number density of the protons in the beam = ?
Now, we make use of the relation between current and drift velocity
I = MeAv ⇒ 1 / eAv
The kinetic energy of protons is given by;
K = v²
v = √( 2K / )
lets relate the cross-sectional area A of the beam to its diameter D;
A = πD²
now, we substitute for v and A
n = I / πeD² ×√( 2K /
)
n = 4I/π eD² × √( / 2K )
so we plug in our values;
n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )
n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵
n = 3.2 × 10¹³ m⁻³
Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³