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balu736 [363]
3 years ago
12

A sphere has a radius of 3.9cm and a density of 7.58 g/cm cubed. What is the mass?

Physics
1 answer:
snow_lady [41]3 years ago
3 0

Answer:

m=ρV

V=4/3 * pi * r3

V=1.3 * 3.14 * 3.9^3

V=242.14 cm^3

m=7.58 * 242.14

m=1.8 kG

Explanation:

1. We calculate volume for sphere.

2. Then we calculate mass of sphere.

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Which organ system do you think is the most interesting? Why?
klasskru [66]

The reproductive system because without that we would not exist , but the same thing could be said about any other systems but the reproductive system is the reason why were here today .

3 0
3 years ago
What is the force felt by the electrons and the nuclei in the rod when the external field described in the problem introduction
fomenos

Complete Question

The complete question is shown on the uploaded image

Answer:

The nuclei experience a force that will move it to the right of the conductor rod while the electrons experience a force that will move it to the left side of the conductor rod.

Explanation:

The force that act on the charges(both the positive and the negative charge ) is Mathematically expressed as

                      F = qE   => E = F/q

where F is the force, q is the charge and E is the electric field

we can see that the force is directly proportional to the electric field,so an increase in the electric field would increase the force.

we can also see that the electric field is given as force per unit charge and generally the direction of this field is taken to be the direction of the force it would exert on a positive test charge

Now from the question we are being told that the external electric field is the direction of the positive x axis

Hence this field would drive the positive charge i.e the nuclei to the right.

In order to further explain let consider this

Generally the electric field is always radially outward  originating from a positive point charge and radially in toward a negative point charge.

what  this means for this question, is that the positive point charge is on the left side of the electric field while the negative point charge is at the right side of the field.

According to Coulomb's law which states that unlike term attract while like terms repel, it means that  the electron would move to the left of the conductor rod   while the nuclei would move to the right side of the conductor rod.

7 0
3 years ago
The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum
Tanya [424]

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case  h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W;  we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this  acts to slow down the ejected electrons from the catode.

5 0
3 years ago
A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav
Vinvika [58]
<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}    (1)

V=V_{o}-g.t    (2)

Where:

y  is the rock's final position

y_{o}=0  is the rock's initial position

V_{o}=30\frac{m}{s} is the rock's initial velocity

V is the final velocity

t is the time the parabolic movement lasts

g=1.62\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of the moon

As we know y_{o}=0 , equation (2) is rewritten as:

y=V_{o}.t+\frac{1}{2}g.t^{2}    (3)

On the other hand, the maximum height  is accomplished when V=0:

V=V_{o}-g.t=0    (4)

V_{o}-g.t=0    

V_{o}=g.t    (5)

Finding t:

t=\frac{V_{o}}{g}    (6)

Substituting (6) in (3):

y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}    (7)

y_{max}=\frac{{V_{o}}^{2}}{2g}    (8)  Now we can calculate the maximum height of the rock

y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}   (9)

Finally:

y_{max}=277.777m  

4 0
3 years ago
Given Vout = 17.33 vpp and R1 = 3 kΩ, find the value of RF required to provide Av = 4.33. (Round your answer to 2 decimal places
Olenka [21]

Answer:

The magnitude of V_{2} is 4 V and phase of input voltage is zero

Explanation:

Given:

Output voltage V_{out} = 17.33

Resistance R_{1} = 3 kΩ

Voltage gain A_{v} = 4.33

For finding feedback resistance we use gain equation

Gain equation for non inverting op-amp is given by,

     A_{v} = 1+\frac{R_{f} }{R_{1} }

   4.33 = 1+ \frac{R_{f} }{3 k }

     R_{f} ≅ 10 kΩ

For finding input voltage we use,

   A_{v} = \frac{V_{out} }{V_{2} }

    V_{2} = \frac{17.33}{4.33}

    V_{2} = 4 V

The Phase of V_{2} is zero because output voltage phase is 360°

Therefore, the magnitude of V_{2} is 4 V and phase of input voltage is zero

7 0
3 years ago
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