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Elis [28]
3 years ago
8

a bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s.if the thickness of the

block reduces to one half of the previous value .what will be the emerging velocity of bullet?
Physics
1 answer:
erma4kov [3.2K]3 years ago
3 0

The emerging velocity of the bullet is <u>71 m/s.</u>

The bullet of mass <em>m</em> moving with a velocity <em>u</em>  has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed <em>v</em> when it emerges from the block.

If the block exerts a resistive force <em>F</em> on the bullet and the thickness of the block is <em>x</em> then, the work done by the resistive force is given by,

W=Fx

This is equal to the change in the bullet's kinetic energy.

W=Fx=\frac{1}{2} m(u^2-v^2)......(1)

If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v<em>₁.</em>

Assuming the same resistive forces to act on the bullet,

F(\frac{x}{2} )=\frac{1}{2} m(u^2-v_1^2)......(2)

Divide equation (2) by equation (1) and simplify for v<em>₁.</em>

\frac{\frac{Fx}{2} }{Fx} =\frac{(u^2-v_1^2)}{(u^2-v^2)} \\\frac{100^2-v_1^2}{100^2-10^2} =\frac{1}{2} \\v_1^2=5050\\v_1=71.06 m/s

Thus the speed of the bullet is 71 m/s


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Professional Application: A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 11
Nataliya [291]

Answer:

0.182 m/s

Explanation:

m1 = 30,000 kg, m2 = 110,000 kg, u1 = 0.85 m/s

let the velocity of loaded freight car is v

Use the conservation of momentum

m1 x u1 + m2 x 0 = (m1 + m2) x v

30,000 x 0.85 = (30,000 + 110,000) x v

v = 0.182 m/s

5 0
3 years ago
Calculate the current in a circuit if 500 C of charge passes through it in 10 minutes. *
Sloan [31]

Answer:

Choice a. Approximately 0.83\; \rm A on average.

Explanation:

The electric current through a wire is the rate at which electric charge flows through a cross-section of this wire.

Assume that electric charge of size Q flowed through a wire cross-section over a period of time t. The average current in that wire would be:

\displaystyle I = \frac{Q}{t}.

For this question:

  • Q = 500\; \rm C, whereas
  • t = 10\; \rm \text{minutes}.

Therefore, the average current in this circuit would be:

\displaystyle I = \frac{Q}{t} = \frac{500\; \rm C}{10\; \text{minutes}} = 50\; \rm C /\text{minute}.

However, the units in the choices are all in \rm A (for Amperes.) One Ampere is equal to one \rm C / \text{second}. It will take some unit conversations to change the unit of I = 50\; \rm C/ \text{minute} (coulombs-per-minute) to coulombs-per-second.

\begin{aligned}I &= 50\; \rm C/ \text{minute} \\ &= \frac{50\; \rm C}{1\; \rm \text{minute}} \times \frac{1 \; \text{minute}}{60\; \rm \text{seconds}} \approx 0.83\; \rm C/ \text{second} = 0.83 \; \rm A\end{aligned}.

Hence, the most accurate choice here would be choice a.

7 0
3 years ago
A charge of 63.0 nC is located at a distance of 3.40 cm from a charge of -47.0 nC. What are the x- and y-components of the elect
Serjik [45]

Answer:

Ep_x = 288.97*10^3\frac{N}{C}

Ep_y = 2770.6*10^3\frac{N}{C}

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/r²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

r: distance from charge q to point P in meters (m)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Equivalences

1nC= 10⁻⁹ C

1cm= 10⁻² m

Graphic attached

The attached graph shows the field due to the charges:

Ep₁: Total field at point P due to charge q₁. As the charge is positive ,the field leaves the charge.

Ep₂: Total field at point P due to charge q₂. As the charge is negative, the field enters the charge.

Known data

q₁ = 63 nC = 63×10⁻⁹ C

q₂ = -47 nC = -47×10⁻⁹ C

k = 8.99*10⁹ N×m²/C²

d₁ = 1.4cm = 1.4×10⁻² m

d₂ = 3.4cm = 3.4×10⁻² m

Calculation of r and β

r=\sqrt{d_1^2 + d_2^2} = \sqrt{(1.4*10^{-2})^2 + (3.4*10^{-2})^2} = 3.677*10^{-2}m

\beta = tan^{-1}(\frac{d_1}{d_2}) = tan^{-1}(\frac{1.4}{3.4}) = 22.38^o

Problem development

Ep: Total field at point P due to charges q₁ and q₂.

Ep = Ep_x i + Ep_y j

Ep₁ₓ = 0

Ep_{2x}=\frac{-k*q_2*Cos\beta}{r^2}=\frac{8.99*10^9*47*10^{-9}*Cos(22.38)}{(3.677*10^{-2})^2}=288.97*10^3\frac{N}{C}

Ep_{1y}=\frac{-k*q_1}{d_1^2}=\frac{8.99*10^9*63*10^{-9}}{(1.4*10^{-2})^2}=2889.6*10^3\frac{N}{C}

Ep_{2y}=\frac{-k*q_2*Sen\beta}{r^2}=\frac{-8.99*10^9*47*10^{-9}*Sen(22.38)}{(3.677*10^{-2})^2}=-119*10^3\frac{N}{C}

Calculation of the electric field components at point P

Ep_x = Ep_{1x} + Ep_{2x} = 0 + 288.97*10^3 = 288.97*10^3\frac{N}{C}

Ep_y = Ep_{1y} + Ep_{2y} = 2889.6*10^3 - 119*10^3 = 2770.6*10^3\frac{N}{C}

6 0
3 years ago
What is the power generation in a machine that produces 76 j in 3.7s
ICE Princess25 [194]
Power is the ratio between energy and time:
P= \frac{E}{t}
In our problem we have E=76 J and t=3.7 s. Therefore, the power is
P= \frac{76 J}{3.7 s} =20.5 W
3 0
3 years ago
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pishuonlain [190]

yup..........................................                                                

7 0
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