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3 years ago

1. Impulse is the product of force and distance. True or false?

Physics

1 answer:

daser333 [38]3 years ago

8 0

Answer:

Explanation:

1. Impulse, I = F.t

The statement impulse is the product of Force and distance is false.

2. F = m g

Force necessary to lift the object depends on the mass of the object.

statement 2 is false.

3. Joule is equal to Newton times meter.

Statement 3 is false.

4. Work done to lift an object is correct statement.

Statement 4 is true.

5. Kinetic energy of an object is due to motion.

Statement 5 is false.

6. Stopping distance is directly proportional to the square of velocity.

If velocity is doubled, stopping distance is quadrupled.

Statement 6 is false.

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Two forces Upper FSubscript Upper A Baseline Overscript right-arrow EndScripts and Upper FSubscript Upper B Baseline Overscript

Pavel [41]

Answer:

Part a)

$F_A = 4.59 N$

Part B)

$F_B = 1.28 N$

Explanation:

As we know that when both the forces are acting on the object in same direction then we will have

$F_A + F_B = ma$

as we know that

$a = 0.554 m/s^2$

m = 10.6 kg

now we will have

$F_A + F_B = 10.6(0.554)$

$F_A + F_B = 5.87 N$

Now two forces are in opposite direction then we have

$F_A - F_B = 10.6(0.313)$

$F_A - F_B = 3.32 N$

Part A)

Now we will have from above two equation

$F_A = 4.59 N$

Part B)

Similarly for other force we have

$F_B = 1.28 N$

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3 years ago

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The tallest building in the world, according to some architectural standards, is the Taipei 101 in Taiwan, at a height of 1671 f

Andrej [43]

Answer:

35.14°C

Explanation:

The equation for linear thermal expansion is $\Delta L = \alpha L_0\Delta T$ , which means that a bar of length $L_0$ with a thermal expansion coefficient $\alpha$ under a temperature variation $\Delta T$ will experiment a length variation $\Delta L$ .

We have then $\Delta L$ = 0.481 foot, $L_0$ = 1671 feet and $\alpha$ = 0.000013 per centigrade degree (this is just the linear thermal expansion of steel that you must find in a table), which means from the equation for linear thermal expansion that we have a $\Delta T =\frac{\Delta L }{\alpha L_0}$ = 22.14°. As said before, these degrees are centigrades (Celsius or Kelvin, it does not matter since it is only a variation), and the foot units cancel on the equation, showing no further conversion was needed.

Since our temperature on a cool spring day was 13.0°C, our new temperature must be $T_f=T_0+\Delta T$ = 35.14°C

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3 years ago

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Answer:

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