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3 years ago

8

If you pull up on a bucket with a tension force of 15 N and the bucket has a weight of 15 N, what is the net force acting on the

bucket?
15 Nup
15 N down
30 N up
O N

1 answer:

fredd [130]3 years ago

3 0

Answer:

F = 0 [N]

Explanation:

To solve this problem we must perform a summation of forces in the direction of the vertical axis. Where the positive force is that of the tension of the upward force, while the force exerted by the weight is directed downward with a negative sign.

ΣF = 0

15 - 15 + F = 0

F = 0 [N]

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Glass has a _____________ index of refraction than air

rewona [7]

Glass has a <u>grater </u>index of refraction than air. The glass's and air's indexes of refraction will be 1.5 and 1, respectively.

<h3>What is an index of refraction?</h3>

The refractive index of a substance is a dimensionless quantity that specifies how quickly light passes through it in optics.

The index of refraction of the glass and air will be 1.5 and 1 respectively.

Hence,glass has a <u>grater </u>index of refraction than air.

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2 years ago

A 1.50 µF capacitor and a 3.50 µF capacitor are connected in series across a 2.50 V battery. How much charge (in µC) is stored o

Nataly_w [17]

Explanation:

The given data is as follows.

$C_{1} = 1.50 \times 10^{-6} F$

$C_{1} = 3.50 \times 10^{-6} F$

Voltage = 2.50 V

Hence, calculate the equivalence capacitor as follows.

$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$

$\frac{1}{C} = \frac{1}{1.50 \times 10^{-6} F} + \frac{1}{3.50 \times 10^{-6} F}$

= $0.945 \times 10^{-6} F$

C = $1.06 \times 10^{-6} F$

Now, we will calculate the charge across each capacitance as follows.

Q = CV

= $1.06 \times 10^{-6} F \times 2.50 V$

= $2.65 \times 10^{-6} C$

= $2.65 \mu C$

Thus, we can conclude that $2.65 \mu C$ is the charge stored on each given capacitor.

5 0

3 years ago

dybincka [34]

Answer:

1. 218.55 N

2. $30.96^{o}$

3. $2.1 m/s^{2}$

Explanation:

Part 1;

Net force $F=mg sin \theta$ where m is mass, g is gravitational force and $\theta$ is the angle of inclination

$F= 46*9.8*sin 29^{o}= 218.55N$

Frictional force, $F_{r}$ is given by

$F_{r} = \mu_{s}mg cos \theta$ where $\mu_{s}$ is the coefficient of static friction

$F_{r} = 0.6*46*9.8 cos 29$

$F_{r}=236.57N$

Since $F_{r}>F$ , therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force $F_{s} = \mu_{s} mg cos \theta$

$mg sin \theta =\mu_{s} mg cos \theta$

$\mu_{s}= \frac {sin \theta}{cos \theta}$

$\mu_{s}= tan \theta$

The maximum angle of inclination $\theta = tan^{-1} \mu_{s}$

$\theta = tan^{-1} (0.6)$

$\theta= 30.96^{o}$

Part 3:

Net force on the object is given by

$ma = mg sin 38 - \mu_{k} mg cos 38$ where $\mu_{k}$ is the coefficient of kinetic friction

$a = g ( sin 38 - \mu_{k} cos 38 )$

= 9.8 ( sin 38 - (0.51) cos 38 )

= $2.1m/s^{2}$

7 0

3 years ago

Considerable scientific work is currently under way to determine whether weak oscillating magnetic fields such as those found ne

VLD [36.1K]

Answer:

The the maximum emf is $2.52\times10^{-11}\ V$

Explanation:

Given that,

Magnetic field $B = 1.40\times10^{-3}\ T$

Frequency = 60 Hz

Diameter = 7.8 μm

We need to calculate the maximum emf

Using formula of emf

$\epsilon=NBA\omega$

Where, N = number of turns

B= magnetic field

A = area

Put the value in to the formula

$\epsilon=1\times1.40\times10^{-3}\times\pi\times(\dfrac{7.8\times10^{-6}}{2})^2\times2\times\pi\times60$

$\epsilon=2.52\times10^{-11}\ V$

Hence, The the maximum emf is $2.52\times10^{-11}\ V$

5 0

3 years ago

2. What energy source produces most of the electrical energy in the United States?

Minchanka [31]

Answer:

Nat gas

Explanation:

Approx:

Natural gas 38% coal 21% oil 19%

6 0

1 year ago