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4 years ago

How do you calculate torque?

1 answer:

3 0

Torque is equal to the magnitude of the force applied times the distance from the radius of the object. As an equation this looks like t = F x r

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A 13,000 kg helicopter accelerates upward a 0.5 m/s^2 while lifting a 2000 pound car. to the nearest newton, what is the lift fo

Vinil7 [7]

Lift force exerted by the air on the rotors=143244 N

Explanation:

we use Newtons second law

F- (M+m)g=(M+m)a

F= lift force

m= mass of helicopter= 13000 Kg

M= mass of car= 2000 lb=907.2 kg

a= acceleration= 0.5 m/s²

g= acceleration due to gravity

F- (M+m)g=(M+m)a

F=(M+m)(a+g)

F=(13000+907.2)(0.5+9.8)

F=143244 N

8 0

3 years ago

Pllssss answer thiss ... its about the reading of VOLTMETER AND AMMETER

Afina-wow [57]

Answer:

The answer is C 1.8V and 0.38A

4 0

3 years ago

What is the magnitude of the electric force between a proton and an electron when they are at a distance of 4.09 angstrom from e

Zinaida [17]

Answer:

$F=1.38*10^{-9}N$

Explanation:

According to Coulomb's law, the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

$F=\frac{kq_1q_2}{d^2}$

Here k is the Coulomb constant. In this case, we have $q_1=-e$ , $q_2=e$ and $d=4.09*10^-10m$ . Replacing the values:

$F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-1.6*10^{-19}C)(1.6*10^{-19}C)}{(4.09*10^{-10})^2}\\F=-1.38*10^{-9}N$

The negative sign indicates that it is an attractive force. So, the magnitude of the electric force is:

$F=1.38*10^{-9}N$

5 0

3 years ago

A red blood cell contains 4.8 107 free electrons. What is the total charge of these electrons in the red blood cell?

tatuchka [14]

Answer:

Charge, $q=7.68\times 10^{-12}\ C$

Explanation:

It is given that,

The number of electron in a RBCs, $n=4.8\times 10^7$

We need to find the total charge of these electrons in the red blood cell. Let it is q. Using the quantization of charge as follows :

q = ne

e is the change on electron

$q=4.8\times 10^7\times 1.6\times 10^{-19}\\\\q=7.68\times 10^{-12}\ C$

So, the net charge is $7.68\times 10^{-12}\ C$ .

7 0

3 years ago

The force between a pair of charges is 900 newtons. The distance between the charges is 0.01 meters. If one of the charges is 2e

Maksim231197 [3]

Answer:

$\fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}$

Explanation:

Given:

Force between pair of charges= 900 newtons

The distance between the charges = 0.01 meters

Strength of Charge first q1 = 2e-10 Coulomb

To find:

Strength of Charge second q2 = ____ Coulomb?

Solution:

We know that,

Force between two charges separate by distance r is given by the equation,

$|F| = K_e \frac{q1 \cdot \: q2}{ {r}^{2} } \\ 900 =K_e \frac{(2e - 10)\cdot \: q2}{ {0.01}^{2} } \\ 900 \times {10}^{ - 4} = K_e {(2e - 10)\cdot \: q2} \\ q2 = \frac{9 \times {10}^{ - 2} }{(2e - 10) K_e} \\ \\ \fbox{We \: know \: that \: e = 2.71 } \\ substituting \: the \: value \: \\ q2 = \frac{9 \times {10}^{ - 2} }{(2 \times 2.71 - 10)K_e} \\ q2 = \frac{0.09}{ - 4.58 K_e} \\ q2 = \frac{-0.0196}{K_e}\: coulomb$

$\fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}$

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3 0

2 years ago