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3 years ago

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A truck moving at 13.3 m/s hits a concrete wall. As a result of the collision, a 6-kg wrench moves forwards and strikes the wall

of the tool compartment. If the wrench stops after being in contact with the wall for 0.07 s, what is the average force exerted on the wrench by the wall

1 answer:

Naddik [55]3 years ago

8 0

Answer:

Explanation:

The velocity of the wrench must be equal to the velocity of the truck . So momentum of the wrench before it hits the wall

= mv = 6 x 13.3 = 79.8 kg m /s

If resisting force of wall be F , impulse on the wrench = F x time

= F x .07

Impulse = change in momentum of the wrench = mv - 0 = mv = 79.8 kgm/s

So F x .07 = 79.8

F = 1140 N .

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Consult Multiple-Concept Example 15 to review the concepts on which this problem depends. Water flowing out of a horizontal pipe

kiruha [24]

Answer:

The pressure of the water in the pipe is 129554 Pa.

Explanation:

<em>There are wrongly written values on the proposal, the atmospheric pressure must be 101105 Pa, and the density of water 1001.03 kg/m3, those values are the ones that make sense with the known ones.</em>

We start usign the continuity equation, and always considering point 1 a point inside the pipe and point 2 a point in the nozzle:

$A_1v_1=A_2v_2$

We want $v_2$ , and take into account that the areas are circular:

$v_2=\frac{A_1v_1}{A_2}=\frac{\pi r_1^2 v_1}{\pi r_2^2}=\frac{r_1^2 v_1}{r_2^2}$

Substituting values we have (we don't need to convert the cm because they cancel out between them anyway):

$v_2=\frac{r_1^2 v_1}{r_2^2}=\frac{(1.8cm)^2 (0.56m/s)}{(0.49cm)^2}=7.56m/s$

For determining the absolute pressure of the water in the pipe we use the Bernoulli equation:

$P_1+\frac{\rho v_1^2}{2}+\rho gh_1=P_2+\frac{\rho v_2^2}{2}+\rho gh_2$

Since the tube is horizontal $h_1=h_2$ and those terms cancel out, so the pressure of the water in the pipe will be:

$P_1=P_2+\frac{\rho v_2^2}{2}-\frac{\rho v_1^2}{2}=P_2+\frac{\rho (v_2^2-v_1^2)}{2}$

And substituting for the values we have, considering the pressure in the nozzle is the atmosphere pressure since it is exposed to it we obtain:

$P_1=101105 Pa+\frac{1001.03Kg/m^3 ((7.56m/s)^2-(0.56m/2)^2)}{2}=129554 Pa$

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3 years ago

You are comparing a reaction that produces a chemical change and one that produces a physical change. What evidence could you us

Lynna [10]

Answer: A chemical change results from a chemical reaction, while a physical change is when matter changes forms but not chemical identity. Examples of chemical changes are burning, cooking, rusting, and rotting. Examples of physical changes are boiling, melting, freezing, and shredding. Often, physical changes can be undone, if energy is input.

Explanation: hope this helps have a good day

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4 years ago

Which one A, B, C, D or the last one?

Finger [1]

Answer:

C

Explanation:

Answer is C.

8 0

3 years ago

Calculate the flux of the vector field F⃗ =−6i⃗ +5x2j⃗ −5k⃗ , through the square of side 8 in the plane y=1, centered on the y-a

Tasya [4]

Answer:

The flux is 682.6 Wb.

Explanation:

Given that,

Vector field $F=-6i+5x^2j-5k$

We need to calculate the flux

Using formula of flux

$\phi=\int_{-4}^{4}\int_{-4}^{4}(F\cdot j\ dxdz)$

Put the value into the formula

$\phi=\int_{-4}^{4}\int_{-4}^{4}(-6i+5x^2j-5k)1\ dxdz$

$\phi=\int_{-4}^{4}\int_{-4}^{4}(5x^2)dxdz$

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Hence, The flux is 682.6 Wb.

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3 years ago

A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or

Svetlanka [38]

<u>Answer:</u>

Mass of base ball $m_b=3.992*10^{14}kg$

<u>Explanation:</u>

Circumference of baseball = 2πr = 23 cm

So radius of baseball = 3.66 cm = $3.66*10^{-2}$ m

Mass per unit volume of baseball = Mass per unit volume of neutron or proton.

Mass of proton = $10^{-27}$ kg

Diameter of proton = $10^{-15}$ m

Radius of proton = $5*10^{-16}$ m

Volume of ball = $\frac{4}{3} \pi r^3$

Now substituting all values in Mass per unit volume of baseball = Mass per unit volume of neutron or proton.

$\frac{m_b}{\frac{4}{3}\pi *(3.66*10^{-2})^3} =\frac{10^{-27}}{\frac{4}{3}\pi *(5*10^{-16})^3}$

$\frac{m_b}{(3.66*10^{-2})^3} =\frac{10^{-27}}{(5*10^{-16})^3}$

$m_b=3.992*10^{14}kg$

So mass of base ball $m_b=3.992*10^{14}kg$

5 0

4 years ago

Read 2 more answers