The correct answer to the question is 130.4 N.
CALCULATION;
The mass of the bullet is given as m = 28 gram = 0. 028 kg.
The initial velocity of the bullet u = 55 m/s
The final velocity of the bullet v = 18 m/s.
The distance covered by the bullet through the sand bag s = 29 cm.
= 0.29 m
Let the acceleration of the bullet is a .
From equation of kinematics, we know that-
⇒ 
The negative sign is used due to the fact that force is opposing in nature. Its velocity is decreasing with time.
From Newton's second law of motion, we know that net force on a body is equal to the product of mass with acceleration.
Mathematically F = ma.
Hence, the frictional force exerted on the bullet is calculated as -
F = m × a
= 0.028 × (-4656.897) N
= -130.4 N [ANS]
Here, N ( newton) stands for the unit of force.
Answer:2800000j
Explanation:
For us to know the kinetic energy of the vehicle,
Where m is the mass
And v is the velocity
Then, K.E=1/2mv^2
While, K.E=1/2×3500×40^2
Therefore, our answer will now be
K.E=2800000j
Answer:
3.70242 nm
Explanation:
Using Compton effect formula
Δλ = ( h / mec) ( 1 - cosθ)
where h is planck constant = 6.62607 × 10 ⁻³⁴ m²kg/s
me, mass of an electron = 9.11 × 10⁻³¹ kg
c is the speed of light = 3 × 10⁸ m/s
Δλ = 6.62607 × 10 ⁻³⁴ m²kg/s / (9.11 × 10⁻³¹ kg × 3 × 10⁸ m/s ) ( 1 - cos 90°) = 0.242 × 10 ⁻¹¹ m = 2.42 × 10⁻¹² m = 0.00242 nm
modified wavelength = 3.7 nm + 0.00242 nm = 3.70242 nm
Answer:
A. The object falls a distance of 250 m
Explanation:
Hi there!
In the question, you have forgotten the acceleration due to gravity. However, looking on the web I´ve found a very similar problem in which the acceleration due to gravity was as twice as much as it is on Earth.
The equation of height of a falling object is the following:
y = y0 + v0 · t + 1/2 · g · t²
Where:
y = height of the object after a time t.
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (on Earth: ≅ -10 m/s² considering the upward direction as positive).
Let´s place the origin of the system of reference at the point where the object is released so that y0 = 0. Since the object falls from rest, v0 = 0.
Then, the height of the object after 5 s will be :
y = 1/2 · 2 · g · t² (notice that the acceleration due to gravity is 2 · g)
y = g · t²
y = -10 m/s² · (5 s)²
y = -250 m
The object falls a distance of 250 m.
