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2 years ago

6. What is not an example of mechanical energy?

Physics

1 answer:

zhannawk [14.2K]2 years ago

3 0

Answer:

the answer to your question is c

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As the mass of object increased it is density increased

Elan Coil [88]

Answer:

True.

Explanation:

The density of an object is given by its mass divided by its volume. It can be given as follows :

$d=\dfrac{m}{V}$

It can be seen that the density of an object is directly proportional to its mass. It means if the mass of an object increase, its density will also increase. Hence, the given statement is true.

3 0

3 years ago

Determina la velocidad de un avión que recorre 459 km en 67 min

satela [25.4K]

Answer:

122397.7

Explanation:

4 0

3 years ago

A proton (mass m = 1.67 × 10-27 kg) is being accelerated along a straight line at 2.50 × 1012 m/s2 in a machine. If the proton h

Veronika [31]

Answer:

(A) Speed will be $44.18\times 10^4m/sec$

(b) Change in kinetic energy = $1560\times 10^{-19}$

Explanation:

We have given mass of proton $m=1.67\times 10^{-27}kg$

Acceleration of the proton $a=2.50\times 10^{12}m/sec^2$

Initial velocity u = $1.60\times 10^4$ m/sec

Distance traveled by proton s = 3.90 cm = 0.039 m

(a) From third equation of motion we know that

$v^2=u^2+2as$

$v^2=(1.60\times 10^4)^2+2\times 2.5\times 10^{12}\times 0.039$

$v=44.18\times 10^4m/sec$

(b) Initial kinetic energy $KE_I=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (1.6\times 10^4)^2$

Final kinetic energy $KE_F=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (44.18\times 10^4)^2$

So change in kinetic energy $\Delta KE=KE_F-KE_I=\frac{1}{2}\times 1.6\times 10^{-27}\times 10^8\times (44.18^2-1.6^2)=1560\times 10^{-19}J$

6 0

3 years ago

What is 34 + (5) × (1.2465) written with the correct number of significant figures?

bonufazy [111]

Answer: 40

Explanation:

= 34 + 5 * 1.2465

= ‭40.2325‬

= 40

The number of significant figures in the answer should be the same as the number with the least number of significant figures that any of the digits in the equation have.

32 has 2 significant figures so the answer has to be 2 significant figures which is 40.

7 0

3 years ago

Determine the slope of end a of the cantilevered beam. E = 200 gpa and i = 65. 0(106) mm4

DENIUS [597]

For E = 200 gpa and i = 65. 0(106) mm4, the slope of end a of the cantilevered beam is mathematically given as

A=0.0048rads

<h3>What is the slope of end a of the cantilevered beam?</h3>

Generally, the equation for the is mathematically given as

$A=\frac{PL^2}{2EI}+\frac{ML}{EI}$

Therefore

A=\frac{10+10^2+3^2}{2*240*10^9*65*10^6}+\frac{10+10^3*3}{240*10^9*65*10^{-6}}

A=0.00288+0.00192=0.0048rads

A=0.0048rads

In conclusion, the slope is

A=0.0048rads