Answer:
38.87 m/s
Explanation:
Given that the ball is dropped from a height = 77 m
u = 0 m/s
s = 77 m
a = g = 9.81 m/s²
Applying the expression as:
Applying values as:
<u>The speed with which the ball hit the ground = 38.87 m/s</u>
Answer:
C
Explanation:
vectors are any quantities that have both magnitude and direction
therefore both inertia and weight are both vectors
The speed of the vehicle after the push is 2.7 m/s.
<h3>What is velocity?</h3>
velocity is the rate of change of displacement.
To calculate the velocity of the vehicle, we use the formula below.
Formula:
- v = √[2(W-W')/m]............... Equation 1
Where:
- W = work done by the pusher
- W' = Work done by friction
- m = mass of the car
From the question,
Given:
- W = 6500 J
- W' = 2000 J
- m = 1200 kg
Substitute these values into equation 1
- v = √[2(6500-2000)/1200]
- v = √(7.5)
- v = 2.7 m/s
Hence, the speed of the vehicle after the push is 2.7 m/s.
Learn more about velocity here:
#SPJ1
Answer:
<em>the mass of the original unstable particle is</em><em> 1115.08 MeV/c²</em>
Explanation:
The momentum of a particle is determined by:
p = e B R
where
- B is the magnetic field
- R is the radius of curvature
- e is the energy of the particle
Therefore,
p = e B R kg · m/s
We can transform the units to MeV/c and we do that by taking:
e = 0.511 MeV and
c = 3 × 10⁸ m/s
Therefore,
p = 300 B R MeV/c
p = 300(0.250 T)(1.33 m) MeV/c
p = 99.75 MeV/c
The energy of the unstable decayed particle is determined as:
E = √ [m²c⁴ + p²c²]
where
- m is the mass of the particle
- c is the speed of light
- p is the particle's momentum
Therefore,
E = E_p + E_(π⁻)
E = √[ (938.3)² + (99.75)² ] + √[ (139.5)² + (99.75)² ]
E = 1115.08 MeV
Since the particle was initially at rest, its energy is only rest-mass energy so its <em>mass will be 1115.08 MeV/c²</em>