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Kay [80]
2 years ago
9

A weightlifter lifts a 13.0-kg barbel from the ground an moves it a distance of 1.3 meters. What is the work se does on the barb

ell?
Physics
1 answer:
marta [7]2 years ago
6 0

The work done on the barbell is -165.62 Nm.

Explanation:

Work done on any object is the measure of force required to move that object from one position to another. So it is determined by the product of force acting on the object with the displacement of the object.

In the present problem, the displacement of the object on acting of force is given as 1.3 m. And the weight of the object which is a barbel is given as 13 kg. As the work is to lift the object from the ground, so the acceleration due to gravity will be acting on the object. In other words, the force applied on the object to lift it should be in opposite direction to the acting of acceleration due to gravity.

Thus, Force = - Mass * Acceleration due to gravity = - 13 * 9.8 =-127.4 N

Now, the force is -127.4 N and the displacement is 1.3 m.

So, Work done = F*d

Work done = -127.4* 1.3 = -165.62 Nm

So, the work done on the barbell is -165.62 Nm.

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Wire A has the same length and twice the radius of wire B. Both wires are made of the same material and carry the same current.
WARRIOR [948]

Answer:

V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}

V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}

And as we can see we have that:

V_A = \frac{1}{4} V_B

So then the best answer would be:

a. vA = vB/4

Explanation:

For this case we know the following conditions:

L_A = L_B =L same length

I_A = I_B =I both wires with the same current

Both wires are made of he same material, so then the number of electrons per cubic meter (n) are the same for both wires n_A = n_B =n

We also know that r_A = 2 r_B where r represent the radius.

Since we know that a wire have a cylindrical form we can find the area for each case:

A_A= \pi r^2_A = \pi (2r_B)^2 = 4 \pi r^2_B= 4 A_B

A_B = \pi r^2_B

So then we have that A_A = 4 A_B

Now we know that from the definition the drift velocity of electron in a wire is given by:

v_d = \frac{I}{neA}

Where I is the current, n the number of electrons per cubic meter, e is the charge for the electron and A the area.

If we replace we have this:

V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}

V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}

And as we can see we have that:

V_A = \frac{1}{4} V_B

So then the best answer would be:

a. vA = vB/4

6 0
2 years ago
A lab technician uses laser light with a wavelength of 670 nm to test a diffraction grating. When the grating is 40.0 cm from th
Juliette [100K]

Answer:

N = 221.4 lines / mm

Explanation:

Given:

- The wavelength of the source λ = 670 nm

- Distance of the grating from screen B = 40.0 cm

- The distance of first bright fringe from central order P = 6.0 cm

Find:

How many lines per millimeter does this grating have?

Solution:

- The derived results from Young's experiment that relates the order of bright fringes about the central order is given by:

                                          sin (Q) = n*λ*N

Where,

n is the order number 0, 1 , 2, 3 , ....

λ  is the wavelength of the light source

Q is the angle of sweep respective fringe from central order

N is the number of lines/mm the grating has

- We will first compute the length along which the light travels for the first bright fringe:

                                            L^2 = P^2 + B^2

                                            L^2 = 40^2 + 6^2

                                            L^2 = 1636

                                            L = 40.45 cm  

- Now calculate the sin(Q) that the fringe makes with the central order:

                                            sin (Q) = P / L

                                            sin (Q) = 6 / 40.45

- Now we will use the derived results:

                                           N = sin(Q) / n*λ

  Where, n = 1 - First order

  Plug values in                N = (6 / 40.45) / (670 *10^-6)

                                          N = 221.4 lines / mm

8 0
2 years ago
A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it
natka813 [3]

Answer:

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Explanation:

The rotated angle is given by:

\theta=\omega0*t+1/2*\alpha*t^2

Since this is a quadratic equation it can be solved using:

x=\frac{-b \± \sqrt{b^2-4*a*c}  }{2*a}

Rewriting our equation:

1/2*\alpha*t^2+\omega0*t-\theta=0

t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Since \sqrt{\omega0^2+2*\theta*\alpha} >\omega0 we discard the negative solution.

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

8 0
2 years ago
01 – (Valor – 2,0) O maior campo de testes de veículos da América Latina, localizado na cidade de Indaiatuba (SP), tem forma cir
Scilla [17]

Answer:

a) Calcule a frequência em RPM

= 0.6 RPM

b) a velocidade escalar do carro em m/s.

= 20m/s

Explanation:

a) Calcule a frequência em RPM

A fórmula para calcular a frequência é: 1/T

onde T= Tempo (seconds)

T = 100s

A frequência = 1/100s

A frequência = 0.01Hz

em RPM

A fórmula para calcular a frequência em RPM =

1 Hz = 60RPM

0.01Hz =

A frequência em RPM = 0.01Hz × 60

= 0.6 RPM

b) a velocidade escalar do carro em m/s.

A fórmula para calcular a velocidade escalar = diâmetro ou distância (m) ÷ tempo (s)

Diâmetro ou Distância = 2.0km

Converter 2.0km para m

1 km = 1000m

2km =

2 km × 1000m

= 2000m

A velocidade escalar = 2000m ÷ 100s

A velocidade escalar = 20m/s

Answer:

a) Frequency in RPM

= 0.6 RPM

b) Scalar Velocity

= 20m/s

Explanation:

a) Frequently in RPM

Formula : 1/T

Where T= Time (seconds)

T = 100s

= 1/100s

= 0.01Hz

Frequency in RPM =

1 Hz = 60RPM

0.01Hz = 0.01Hz × 60

= 0.6 RPM

b) Scalar velocity

The formula = Diameter or Distance ÷ Time

Diameter or Distance = 2.0km

Convert 2.0km to m

1 km = 1000m

2km =

2 km × 1000m

= 2000m

Scalar Velocity = 2000m ÷ 100s

Scalar Velocity = 20m/s

8 0
2 years ago
Let’s consider tunneling of an electron outside of a potential well. The formula for the transmission coefficient is T \simeq e^
ioda

Answer:

L' = 1.231L

Explanation:

The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:

T \approx e^{-2CL}

L: width of the barrier

C: constant that includes particle energy and barrier height

You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.

To find the new value of the L' you can write down both situation for T and T', as in the following:

0.050=e^{-2CL}\ \ \ \ (1)\\\\0.025=e^{-2CL'}\ \ \ \ (2)

Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):

ln(0.050)=ln(e^{-2CL})=-2CL\ \ \ \ (3)\\\\ln(0.025)=ln(e^{-2CL'})=-2CL'\ \ \ \ (4)

Next, you divide the equation (3) into (4), and finally, you solve for L':

\frac{ln(0.050)}{ln(0.025)}=\frac{-2CL}{-2CL'}=\frac{L}{L'}\\\\0.812=\frac{L}{L'}\\\\L'=\frac{L}{0.812}=1.231L

hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L

8 0
2 years ago
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