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gogolik [260]
3 years ago
14

In a ballistics test, a 28-g bullet pierces a sand bag that is 30cm thick. If the initial bullet velocity was 55 m/s and it emer

ged from the sandbag moving at 18m/s, what was the magnitude of the friction force (assuming it to be constant and the only force present) that the bullet experienced while it traveled through the bag
Physics
1 answer:
slega [8]3 years ago
6 0

Answer:

Explanation:

Given that,

The mass of the bullet is

M= 28g = 0.028kg.

Thickness oof sack bag

d = 30cm = 0.3m

Initial velocity of bullet Vi = 55m/s

Velocity at which the bullet emerges the sack bag is Vf = 18m/s

We want to calculate the frictional force present in the sack bag.

Using conservation of energy.

Work done by friction in the sack is equal to change in kinetic energy.

Work done by friction is given as

Wf = Fr × d

The distance is the thickness of the bag

Wf = Fr × 0.3

Wf = 0.3 • Fr

Change in kinetic energy is calculated by

∆K.E = ½mVf² — ½ mVi²

∆K.E = ½m(Vf² — Vi²)

∆K.E = ½ × 0.028 ( 18²-55²)

∆K.E = 0.014 × -2701

∆K.E = -37.814 J

Since

Work done by friction = ∆K.E

0.3 Fr = -37.814

Fr = -37.814/0.3

Fr = -126.05 N

The, negative sign show that frictional force is opposing the motion.

Fr = 126.05 N

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Answer:

m = B²qR² / 2 V

Explanation:

If v be the velocity after acceleration under potential difference of V

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from (1) and (2)

B²q²R² / m²  = 2 Vq / m

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m = B²qR² / 2 V

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3 years ago
Please help me,its urgent!!
GREYUIT [131]

Test:

Performing a Litmus Test

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3 years ago
A football is place kicked with a velocity having a vertical component of 12 m/s and a horizontal component of 6 m/s. Find the r
SSSSS [86.1K]

The velocity is given by:

V = √(Vx²+Vy²)

V = velocity, Vx = horizontal velocity, Vy = vertical velocity

Given values:

Vx = 6m/s, Vy = 12m/s

Plug in and solve for V:

V = √(6²+12²)

V = 13.42m/s

Now find the direction:

θ = tan⁻¹(Vy/Vx)

θ = angle of velocity off horizontal, Vy = vertical velocity, Vx = horizontal velocity

Given values:

Vx = 6m/s, Vy = 12m/s

Plug in and solve for θ:

θ = tan⁻¹(12/6)

θ = 63.4°

The resultant velocity is 13.42m/s at an angle of 63.4° off the horizontal.

6 0
3 years ago
The carbon atom with 12 protons and 12 electrons has a charge of:
olga55 [171]

Answer:

Isotopes are the elements having the same atomic number i.e same number of protons which in turn is equal to same number of electrons. Number of protons and electrons are equal as the atom is electrically neutral. So, the only option is A.

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2 years ago
When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius
Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

\tau = I\alpha

where

I = Moment of inertia \rightarrow \frac{1}{2}mr^2 For a disk

\alpha = Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

2 \alpha \theta = \omega_f^2-\omega_i^2

Where

\omega_{f,i} = Final and Initial Angular velocity

\alpha = Angular acceleration

\theta = Angular displacement

Our values are given as

\omega_i = 0 rad/s

\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})

\omega_f = 47.12rad/s

\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad

r = 7cm = 7*10^{-2}m

m = 17g = 17*10^{-3}kg

Using the expression of angular acceleration we can find the to then find the torque, that is,

2\alpha\theta=\omega_f^2-\omega_i^2

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}

\alpha = \frac{47.12^2-0^2}{2*6\pi}

\alpha = 58.89rad/s^2

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

\tau = I\alpha

\tau = (\frac{1}{2}mr^2)\alpha

\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)

\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m

Therefore the torque exerted on it is 2.45*10^{-3}N\cdot m

3 0
3 years ago
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