How does the mass of an object affect its acceleration?
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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.
Explanation:
A) Draw free body diagrams of both blocks.
Force P is pushing right on block A, which will cause it to move right along the incline. Therefore, friction forces will oppose the motion and point to the left.
There are 5 forces acting on block A:
Applied force P pushing to the right,
Normal force N pushing up and left 10° from the vertical,
Friction force Nμ pushing down and left 10° from the horizontal,
Reaction force Fab pushing down,
and friction force Fab μ pushing left.
There are 2 forces acting on block B:
Reaction force Fab pushing up,
And elastic force kx pushing down.
(There are also horizontal forces on B, but I am ignoring them.)
Sum of forces on A in the x direction:
∑F = ma
P − N sin 10° − Nμ cos 10° − Fab μ = 0
Solve for N:
P − Fab μ = N sin 10° + Nμ cos 10°
P − Fab μ = N (sin 10° + μ cos 10°)
N = (P − Fab μ) / (sin 10° + μ cos 10°)
Sum of forces on A in the y direction:
N cos 10° − Nμ sin 10° − Fab = 0
Solve for N:
N cos 10° − Nμ sin 10° = Fab
N (cos 10° − μ sin 10°) = Fab
N = Fab / (cos 10° − μ sin 10°)
Set the expressions equal:
(P − Fab μ) / (sin 10° + μ cos 10°) = Fab / (cos 10° − μ sin 10°)
Cross multiply:
(P − Fab μ) (cos 10° − μ sin 10°) = Fab (sin 10° + μ cos 10°)
Distribute and solve for Fab:
P (cos 10° − μ sin 10°) − Fab (μ cos 10° − μ² sin 10°) = Fab (sin 10° + μ cos 10°)
P (cos 10° − μ sin 10°) = Fab (sin 10° + 2μ cos 10° − μ² sin 10°)
Fab = P (cos 10° − μ sin 10°) / (sin 10° + 2μ cos 10° − μ² sin 10°)
Sum of forces on B in the y direction:
∑F = ma
Fab − kx = 0
kx = Fab
x = Fab / k
x = P (cos 10° − μ sin 10°) / (k (sin 10° + 2μ cos 10° − μ² sin 10°))
Plug in values and solve.
x = 500 N (cos 10° − 0.4 sin 10°) / (12000 (sin 10° + 0.8 cos 10° − 0.16 sin 10°))
x = 0.0408 m
x = 4.08 cm
B) Draw free body diagrams of both blocks.
Force P is pushing block A to the right relative to the ground C, so friction force points to the left.
Block A moves right relative to block B, so friction force on A will point left. Block B moves left relative to block A, so friction force on B will point right (opposite and equal).
Block B moves up relative to the wall D, so friction force on B will point down.
There are 5 forces acting on block A:
Applied force P pushing to the right,
Normal force Fc pushing up,
Friction force Fc μ₁ pushing left,
Reaction force Fab pushing down and left 15° from the vertical,
and friction force Fab μ₂ pushing up and left 15° from the horizontal.
There are 5 forces acting on block B:
Weight force 750 n pushing down,
Normal force Fd pushing left,
Friction force Fd μ₁ pushing down,
Reaction force Fab pushing up and right 15° from the vertical,
and friction force Fab μ₂ pushing down and right 15° from the horizontal.
Sum of forces on B in the x direction:
∑F = ma
Fab μ₂ cos 15° + Fab sin 10° − Fd = 0
Fd = Fab μ₂ cos 15° + Fab sin 15°
Sum of forces on B in the y direction:
∑F = ma
-Fab μ₂ sin 15° + Fab cos 10° − 750 − Fd μ₁ = 0
Fd μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750
Substitute:
(Fab μ₂ cos 15° + Fab sin 15°) μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750
Fab μ₁ μ₂ cos 15° + Fab μ₁ sin 15° = -Fab μ₂ sin 15° + Fab cos 15° − 750
Fab (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°) = -750
Fab = -750 / (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°)
Sum of forces on A in the y direction:
∑F = ma
Fc + Fab μ₂ sin 15° − Fab cos 15° = 0
Fc = Fab cos 15° − Fab μ₂ sin 15°
Sum of forces on A in the x direction:
∑F = ma
P − Fab sin 15° − Fab μ₂ cos 15° − Fc μ₁ = 0
P = Fab sin 15° + Fab μ₂ cos 15° + Fc μ₁
Substitute:
P = Fab sin 15° + Fab μ₂ cos 15° + (Fab cos 15° − Fab μ₂ sin 15°) μ₁
P = Fab sin 15° + Fab μ₂ cos 15° + Fab μ₁ cos 15° − Fab μ₁ μ₂ sin 15°
P = Fab (sin 15° + (μ₁ + μ₂) cos 15° − μ₁ μ₂ sin 15°)
First, find Fab using the given values.
Fab = -750 / (0.25 × 0.5 cos 15° + 0.25 sin 15° + 0.5 sin 15° − cos 15°)
Fab = 1151.9 N
Now, find P.
P = 1151.9 N (sin 15° + (0.25 + 0.5) cos 15° − 0.25 × 0.5 sin 15°)
P = 1095.4 N
Answer:
L = 130 decibels
Explanation:
The computation of the sound intensity level in decibels is shown below:
According to the question, data provided is as follows
I = sound intensity = 10 W/m^2
I0 = reference level =
Now
Intensity level ( or Loudness)is
Therefore
L = 13 bel
And as we know that
1 bel = 10 decibels
So,
The Sound intensity level is
L = 130 decibels