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3 years ago

7

Pls give me an example of newton's 1 ,2 ,3 law of motion pls this is due tomorrow

1 answer:

kompoz [17]3 years ago

6 0

"He who lays up treasures on earth spends his life backing away from his treasures. To him, death is loss. He who lays up treasures in heaven looks forward to eternity; he’s moving daily toward his treasures. To him, death is gain." - Randy Alcorn

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A maintenance worker wants to torque an engine bolt to 65.0 N m. If the torque wrench is 35cm in length, what is force applied t

harina [27]

Answer:

Force = 186 N

Explanation:

Torque is the rotational equivalent of linear force. It can be easely calculated using the formula :

$Torque = \vec{r} \times \vec{F}$

Where $\vec{r}$ is a vector that from the origin of the coordinate system to the point at which the force is applied (the position vector), $\vec{F}$ is the applied force.

The easiest way of computing the force is by setting the origin of the coordinate system to the lowest point of the torque wrench. By doing this we have that $r$ (the magnitud of the position vector) is 35cm.

Before computing the force we need to set all our values to the international system of units (SI). The torque is already in SI. The one missing is the length of the torque wrench (it is in centimeters and we need it in meters). So :

$35cm * \dfrac{1m}{100cm} = 0.35m$

Now using the torque formula:

$Torque = \vec{r} \times \vec{F}$

$Torque = rFsin(\theta)}$

Where $\theta$ is the smaller angle between the force and the position vector. Because the force is applied perpendiculary to the position vector $\theta = 90°$ , thus :

$Torque = rFsin(\theta)}$

$65 N m = (0.35m)Fsin(90°)}$

$F = \dfrac{65N m} {(0.35m)sin(90°)}$

$F = \dfrac{65N m} {(0.35m)sin(90°)}$

$F = \dfrac{1300} {7}N$

so the force is approximately 186 N.

4 0

3 years ago

. A student claims that if lighting strikes a metal flagpole, the force exerted by the Earth’s magnetic field on the current in

Ratling [72]

Answer:

Explanation:

Given that, current generated from lightning range from

10⁴ A < I < 10^5 A

We know that,

The magnetic force is given as

F = iLB

The magnetic field on the earth surface is

B = 10^-5 T

So, let assume the worst case of a 15m flag pole

L = 15m

Then,

F = iLB

F = 10^5 × 10 × 10^-5

F = 15 N

Therefore, 15N is fairly strong so it will come to the material that was use for the material of the flag pole.

Therefore, it is possible that the student is right depending on the material of the flag pole.

7 0

3 years ago

(30 points) Air enters a compressor at 1 bar, 310 K, and is compressed adiabatically to 12 bar, 630 K. The air exiting the compr

SashulF [63]

Answer:

‍♀️‍♀️

Explanation:

6 0

3 years ago

Jackson wants to fit into last year's Halloween costume, but he has been building muscle and has gained a few healthy pounds sin

katrin [286]

It would be A. Because think of the explanations Jasons friend could say to them that would be a negative 'statement'.

7 0

3 years ago

A car battery with a 12 V emf and an internal resistance of 0.11 Ω is being charged with a current of 56 A. What are (a) the pot

denpristay [2]

Answer:

Part a)

V = 18.16 V

Part b)

$P_r = 345 Watt$

Part c)

P = 672 Watt

Part d)

V = 5.84 V

Part e)

$P_r = 345 Watt$

Explanation:

Part a)

When battery is in charging mode

then the potential difference at the terminal of the cell is more than its EMF and it is given as

$\Delta V = E + i r$

here we have

$E = 12 V$

$i = 56 A$

$r = 0.11$

now we have

$\Delta V = 12 + (0.11)(56) = 18.16 V$

Part b)

Rate of energy dissipation inside the battery is the energy across internal resistance

so it is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

Part c)

Rate of energy conversion into EMF is given as

$P_{emf} = i E$

$P_{emf} = (56)(12)$

$P_{emf} = 672 Watt$

Now battery is giving current to other circuit so now it is discharging

now we have

Part d)

$V = E - i r$

$V = 12 - (56)(0.11)$

$V = 12 - 6.16 = 5.84 V$

Part e)

now the rate of energy dissipation is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

7 0

3 years ago