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3 years ago

9

Carmen is helping load furniture and boxes onto a moving truck. She picks up boxes of her things, places them on a cart, and pus

hes the cart to the moving van. Which term best describes the type of force Carmen uses?
A. contact
B. friction
C. gravity
D. noncontact

2 answers:

AleksandrR [38]3 years ago

6 0

Answer:

B because of the friction from the wheels

Savatey [412]3 years ago

6 0

B . because the friction of the wheels .

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Answer:

The vertical distance is $d = \frac{2}{k} *[mg + f]$

Explanation:

From the question we are told that

The mass of the cylinder is m

The kinetic frictional force is f

Generally from the work energy theorem

$E = P + W_f$

Here E the the energy of the spring which is increasing and this is mathematically represented as

$E = \frac{1}{2} * k * d^2$

Here k is the spring constant

P is the potential energy of the cylinder which is mathematically represented as

$P = mgd$

And

$W_f$ is the workdone by friction which is mathematically represented as

$W_f = f * d$

So

$\frac{1}{2} * k * d^2 = mgd + f * d$

=> $\frac{1}{2} * k * d^2 = d[mg + f ]$

=> $\frac{1}{2} * k * d = [mg + f ]$

=> $d = \frac{2}{k} *[mg + f]$

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4 years ago

Bromine, a liquid at room temperature, has a boiling point

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3 years ago

A car starts from rest and travels for t1 seconds with a uniform acceleration a1. The driver then applies the brakes, causing a

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The answer for this question is b because it says how far it goes before he begins to take brake

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2 years ago

Daily life examples of pressure of solids on solids?

Ipatiy [6.2K]

Explanation:

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3 years ago

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 50.9 c

seraphim [82]

Answer:

Flow rate 2.34 m3/s

Diameter 0.754 m

Explanation:

Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.

The area at the well head is

$A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2$

So the volume flow rate along the pipe is

$\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s$

We can use the similar logic to find the cross-section area at the refinery

$A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2$

The radius of the pipe at the refinery is:

$A_r = \pi r^2$

$r^2 =A_r/\pi = 0.446/\pi = 0.141$

$r = \sqrt{0.141} = 0.377m$

So the diameter is twice the radius = 0.38*2 = 0.754m

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3 years ago