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2 years ago

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WallyGPX accelerates from 0 m/s to 8 m/s in 3 seconds. What is his acceleration? Is this acceleration higher than that of a car

which accelerates from 0 to 30 m/s in 8 seconds?

1 answer:

olga nikolaevna [1]2 years ago

4 0

My Phone is +2348181686682

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The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine th

Luba_88 [7]

Here is the complete question

The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine the horizontal shear stress at point H, which is located L = 20 mm below the centriod

The missing image which is the remaining part of this question is attached in the image below.

Answer:

The horizontal shear stress at point H is $\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

Explanation:

Given that :

The internal shear force V = 80 kN = 80 × 10³ N

The moment of inertia = 64,900,000

The length = 20 mm below the centriod

The horizontal shear stress $\tau$ can be calculated by using the equation:

$\tau = \dfrac{VQ}{Ib}$

where;

Q = moment of area above or below the point H

b = thickness of the beam = 10 mm

From the centroid ;

Q = $Q_1 + Q_{2}$

Q = $A_1y_1 + A_{2}y_{2}$

Q = ( ( 70 × 10) × (55) + ( 210 × 15) (90 + 15/2) ) mm³

Q = ( ( 700) × (55) + ( 3150 ) ( 97.5) ) mm³

Q = ( 38500 + 307125 ) mm³

Q = 345625 mm³

$\tau_H = \dfrac{VQ}{Ib}$

$\tau_H = \dfrac{80*10^3 * 345625}{64900000*10 }$

$\tau_H = \dfrac{2.765*10^{10}}{649000000 }$

$\tau_H = 42.60400616 \ N/mm^2$

$\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

The horizontal shear stress at point H is $\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

7 0

2 years ago

Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of

schepotkina [342]

If speed = distance/time , then time = speed/distance.

So...

Speed of light = 3*10^8(m/s)
Average distance from Earth to Sun = 149.6*10^9(m)

Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))

I hope this was a helpful explanation, please reply if you have further questions about the problem.

Good luck!

5 0

1 year ago

An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary

Assoli18 [71]

Answer:

The tension in the cable when the craft was being lowered to the seafloor is 4700 N.

Explanation:

Given that,

When the craft was stationary, the tension in the cable was 6500 N.

When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N.

The drag force of 1800 N will act in the upward direction. As it was lowered or raised at a steady rate, so its acceleration is 0. As a result, net force is 0. So,

T + F = W

Here, T is tension

F = 1800 N

W = 6500 N

Tension becomes :

$T=W-F\\\\T=6500-1800\\\\T=4700\ N$

So, the tension in the cable when the craft was being lowered to the seafloor is 4700 N.

7 0

2 years ago

If the cold temperature reservoir of a Carnot engine is held at a constant 306 K, what temperature should the hot reservoir be k

Paraphin [41]

Efficiency η of a Carnot engine is defined to be:
<span>η = 1 - Tc / Th = (Th - Tc) / Th </span>
<span>where </span>
<span>Tc is the absolute temperature of the cold reservoir, and </span>
<span>Th is the absolute temperature of the hot reservoir. </span>

<span>In this case, given is η=22% and Th - Tc = 75K </span>
<span>Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). </span>

<span>Th = (Th - Tc) / η </span>
<span>Th = 75 / 0.22 = 341 K (rounded to closest number) </span>
<span>Tc = Th - 75 = 266 K </span>

<span>Lower temperature is Tc = 266 K </span>
<span>Higher temperature is Th = 341 K</span>

6 0

2 years ago

Two conducting spheres, one having twice the diameter of the other, are separated by a distance that is large compared to their

Snezhnost [94]

Answer: Option (a) is the correct answer.

Explanation:

When these two conducting spheres are connected together through a thin wire then charge from the smaller sphere will travel through the wire. And, this charge will continue to travel towards the neutral sphere until the charge on both the spheres will become equal to each other.

For example, charge on small sphere is 5 C then this charge will continue to travel towards the neutral sphere until its charge also becomes equal to 5 C.

Hence, then their potential will also become equal.

Thus, we can conclude that the spheres are connected by a long, thin wire, then after a sufficiently long time the two spheres are at the same potential.

6 0

2 years ago