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3 years ago

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A 15g bullet is shot into a 5085g wooden block standing on a frictionless surface. The block, with the bullet in it, acquires a

velocity of .94m/s. Calculate the velocity of the bullet before striking the block.

1 answer:

kaheart [24]3 years ago

5 0

To develop this problem we will apply the concepts related to the conservation of momentum. For this purpose, the initial momentum will be equivalent to the initial momentum of the two objects when they have the same speed. Mathematically this can be expressed as,

$m_bv_b + m_wv_w = (m_b+m_w)v$

Where,

$m_b$ = Mass of bullet

$v_b$ = Velocity of bullet

$m_w$ = Mass of wooden block

$v_w =$ Velocity of Wooden block

Inititally $v_w = 0$ then we have that the expression can be rearrange to find the velocity of the bullet,

$v_b = \frac{(m_b+m_w)v}{m_b}$

Replacing with our values

$v_b = \frac{(15+5085)(0.94)}{15}$

$v_b = 319.6m/s$

Therefore the velocity of the bullet before striking the block is 319.6m/s

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Two engineering students, John with a weight of 92 kg and Mary with a weight of 46 kg, are 30 m apart. Suppose each has a 0.04%

Hoochie [10]

Answer:

F = 6.27 x 10 ¹⁹ N

Explanation:

Given

m₁ = 92 kg, m₂ = 46 kg, % = 0.04% N = 6.022 x 10²³ Z = 18, e = 1.6 x 10 ⁻¹⁹ C, M = 0.018 kg/mol

q₁ = % * [m * N * A * e / M ]

q₁ = 0.0004 * [ ( 92 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]

q₁ = 3.54 x 10⁶ C

q₂ = 0.0004 * [ ( 46 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]

q₂ = 1.773 x 10⁶ C

Now to determine the electrostatic force con use the equation

F = K * q₁ * q₂ / d²

K = 8.99 x 10 ⁹

F = 8.99 x 10 ⁹ * 3.54 x 10⁶ C * 1.773 x 10⁶ C / (30m)²

F = 6.27 x 10 ¹⁹ N

3 0

3 years ago

Which quantity below is a derived quantity?

Reil [10]

By definition, the speed of an object is given by:

$v = \frac{dr}{dt}$

Where,

dr/dt: derived from the position with respect to time

Therefore, speed has units of length over units of time.

Thus, speed is a derived quantity, since it depends on the value of two other quantities.

Answer:

a derived quantity is:

C. Speed

4 0

3 years ago

Read 2 more answers

If f(x) = -3x +4 and g(x) = 2, solve for the value of x for<br>which (fx) = g(x) is true.

maksim [4K]

Answer:

$x=-\frac{2}{3}$

Explanation:

Let:

$f(x)=-3x+4\\\\And\\\\g(x)=2$

We need to know for which value of $x$ the function $f(x)$ is equal to $g(x)$ :

$f(x)=g(x)$

Therefore, we need to solve for the previous equation for $x$ :

Replacing the values of $f(x)$ and $g(x)$ :

$-3x+4=2$

Subtract 4 from both sides:

$-3x+4-4=2-4\\\\-3x=-2$

Multiply both sides by -1

$-3x(-1)=-2(-1)\\\\3x=2$

Divide both sides by 3:

$\frac{3x}{3} =\frac{2}{3} \\\\x=\frac{2}{3}$

Therefore the value of $x$ for which $f(x)=g(x)$ is $x=\frac{2}{3}$ .

Verify the result:

$-3(\frac{2}{3} )+4=2\\\\-2+4=2\\\\2=2$

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3 years ago

A good sign that cardiovascular exercise is occurring is that you _____.

grigory [225]

Are breathing hard. This is because cardiovascular exercise makes the heart beat faster which in turn creates a need for more air.

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2 years ago

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How much work is required to accelerate a proton from rest up to a speed of 0.993 c? Express your answer with the appropriate un

Keith_Richards [23]

Answer:

(a). The work done is 7001 MeV.

(b). The momentum of this proton is $4.20\times10^{8}\ kg-m/s$ .

Explanation:

Given that,

Speed = 0.993 c

We need to calculate the work done

Using work energy theorem

The work done is equal to the kinetic energy relative to the proton

$W=K.E$

$W=\dfrac{m_{p}c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}-m_{p}c^2$

Put the value into the formula

$W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(\dfrac{0.993c}{c})^2}}-1.67\times10^{-27}\times(3\times10^{8})^2$

$W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(0.993)^2}}-1.67\times10^{-27}\times(3\times10^{8})^2$

$W=1.122\times10^{-9}\ J$

$W=7001\ MeV$

(b). We need to calculate the momentum of this proton

Using formula of momentum

$p=\dfrac{m_{0}v}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Put the value into the formula

$p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(\dfrac{0.993c}{c})^2}}$

$p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(0.993)^2}}$

$p=1.404\times10^{-26}c$

$p=4.20\times10^{8}\ kg-m/s$

Hence, (a). The work done is 7001 MeV.

(b). The momentum of this proton is $4.20\times10^{8}\ kg-m/s$ .

4 0

3 years ago