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Brrunno [24]
4 years ago
6

If the same amount of force were applied to all four balls in the picture, which would experience the greatest change in motion?

Engineering
2 answers:
nydimaria [60]4 years ago
6 0
There is no image connected to this question. Next time attach something to your question to get a more successful answer.
RUDIKE [14]4 years ago
6 0

Answer:

Answer 40kg

Explanation:

That ball weighs more than the other which means it has more mass so it wont move as far

Hope I Helped!

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Which happens when a wave passes through an opening
Nookie1986 [14]
Reflecting

Explaining
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3 years ago
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An incompressible viscous fluid flows through a pipe with a flow rate of 1 mL/s. The pipe has a uniform diameter D0 and a length
Nataly [62]

Answer:

Q_2 = 32 mL/s

Explanation:

Given :

The flow is incompressible viscous flow.

The initial flow rate, Q_1 = 1 mL/s

Initial diameter, D_1= D_0

Initial length, L_1=L_0

The initial pressure difference to maintain the flow, P_1=P_0

We know for a viscous flow,

$\Delta P = \frac{32 \mu V L}{D^2}$

$\Delta P = \frac{32 \mu Q L}{\frac{\pi}{4}D^4}$

$Q \propto \Delta P \times D^4$

$\frac{Q_1}{Q_2}= \frac{P_1}{P_2} \times \left( \frac{D_1}{D_2} \right)^4$

$\frac{1}{Q_2}= \frac{P_0}{2P_0} \times \left( \frac{D_0}{2D_0} \right)^4$

$\frac{1}{Q_2}= \frac{1}{2} \times \left( \frac{1}{2} \right)^4$

$\frac{1}{Q_2}= \frac{1}{32}$

∴ Q_2 = 32 mL/s

6 0
3 years ago
# 45 You are driving on a two-lane highway and the vehicle behind you wants to pass. You should
hram777 [196]

Answer:

Signal the driver behind you when it is safe to pass by turning on your four-way emergency flashers.

Avoiding casualties is the top priority when driving, the other choices given do not put whether there are cars in the other lane into consideration, therefore making them incorrect. Signaling the driver when it is safe gives you the time and them the gateway to pass, making a nice interaction keep you both alive.

4 0
2 years ago
A square steel bar has a length of 7.2 ft and a 2.5 in by 2.5 in cross section and is subjected to axial tension. The final leng
Nataly_w [17]

Answer:

A) ν = 0.292

B) ν = 0.381

Explanation:

Poisson's ratio = - (Strain in the direction of the load)/(strain in the direction at right angle to the load)

In axial tension, the direction of the load is in the length's direction and the direction at right angle to the load is the side length

Strain = change in length/original length = (Δy)/y or (Δx)/x or (ΔL/L)

A) Strain in the direction of the load = (2.49946 - 2.5)/2.5 = - 0.000216

Strain in the direction at right angle to the load = (7.20532 - 7.2)/7.2 = 0.0007389

Poisson's ratio = - (-0.000216)/(0.0007389) = 0.292

B) Strain in the direction of the load = (2.09929 - 2.1)/2.1 = - 0.0003381

Strain in the direction at right angle to the load = (5.30470 - 5.3)/5.3 = 0.0008868

Poisson's ratio = - (-0.0003381)/(0.0008868) = 0.381

7 0
3 years ago
The hypotenuse of a 45° right triangle is
wlad13 [49]
75+69 and divide by 54
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