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Damm [24]
3 years ago
7

A glowing electric light bulb placed 15 cm from a concave spherical mirror produces a real image 8.9 from the mirror. the light

bulb is moved to a position 20cm from the mirror
a) what is the image position? (answer with -1000 if the image does not exist)
Physics
1 answer:
marta [7]3 years ago
5 0

Answer:

The image will form at distance d = 7.75 cm from the mirror

Explanation:

As we know that the bulb is placed at 15 cm distance from the concave mirror

Then the image will form at distance d = 8.9 cm

now we have

\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}

so we have

\frac{1}{8.9} + \frac{1}{15} = \frac{1}{f}

f = 5.58 cm

Now when bulb is placed at distance 20 cm from the mirror

then again we have

\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}

\frac{1}{d_i} + \frac{1}{20} = \frac{1}{5.58}

now we have

d_i = 7.75 cm

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While participating in a blood drive at school, Keona learns that blood has a density of 1.06 g/mL. She donates one pint of bloo
Sergio [31]

Answer:

m≈501.57 g

Explanation:

The density formula is:

d=m/v

Let’s rearrange the formula for m. m is being divided by v. The inverse of division is multiplication, so multiply both aides by v.

d*v= m/v*v

d*v=m

The mass can be found by multiply the density and the volume.

m=d*v

The density is 1.06 grams per milliliter and the volume is 473.176 milliliters.

d= 1.06 g/mL

v= 473.176 mL

Substitute the values into the formula.

m= 1.06 g/mL * 473.176 mL

Multiply. When multiplying, the mL will cancel out.

m= 501.56656 g

Let’s round to the nearest hundredth. The 6 in the thousandth place tells us to round the 6 to a 7 in the hundredth place.

m ≈501.57 g

The mass is about 501.57 grams.

7 0
3 years ago
Read 2 more answers
The force of the added water produces a torque on the dam. In a simple model, if the torque due to the water were enough to caus
Fittoniya [83]

Answer:

The appropriate response is "\tau=\frac{1}{6} PgLh^3". A further explanation is described below.

Explanation:

The torque (\tau) produced by the force on the dam will be:

⇒  d \tau=XdF

On applying integration both sides, we get

⇒  \tau = \int_{0}^{a}x pgL(h-x)dx

⇒     = pgL\int_{0}^{h}(h-x)dx

⇒     =pgL[\frac{h^3}{2} -\frac{h^3}{3} ]

⇒     =\frac{1}{6} PgLh^3

8 0
2 years ago
Physics B 2020 Unit 3 Test
weqwewe [10]

Answer:

1)

When a charge is in motion in a magnetic field, the charge experiences a force of magnitude

F=qvB sin \theta

where here:

For the proton in this problem:

q=1.602\cdot 10^{-19}C is the charge of the proton

v = 300 m/s is the speed of the proton

B = 19 T is the magnetic field

\theta=65^{\circ} is the angle between the directions of v and B

So the force is

F=(1.602\cdot 10^{-19})(300)(19)(sin 65^{\circ})=8.28\cdot 10^{-16} N

2)

The magnetic field produced by a bar magnet has field lines going from the North pole towards the South Pole.

The density of the field lines at any point tells how strong is the magnetic field at that point.

If we observe the field lines around a magnet, we observe that:

- The density of field lines is higher near the Poles

- The density of field lines is lower far from the Poles

Therefore, this means that the magnetic field of a magnet is stronger near the North and South Pole.

3)

The right hand rule gives the direction of the  force experienced by a charged particle moving in a magnetic field.

It can be applied as follows:

- Direction of index finger = direction of motion of the charge

- Direction of middle finger = direction of magnetic field

- Direction of thumb = direction of the force (for a negative charge, the direction must be reversed)

In this problem:

- Direction of motion = to the right (index finger)

- Direction of field = downward (middle finger)

- Direction of force = into the screen (thumb)

4)

The radius of a particle moving in a magnetic field is given by:

r=\frac{mv}{qB}

where here we have:

m=6.64\cdot 10^{-22} kg is the mass of the alpha particle

v=2155 m/s is the speed of the alpha particle

q=2\cdot 1.602\cdot 10^{-19}=3.204\cdot 10^{-19}C is the charge of the alpha particle

B = 12.2 T is the strength of the magnetic field

Substituting, we find:

r=\frac{(6.64\cdot 10^{-22})(2155)}{(3.204\cdot 10^{-19})(12.2)}=0.366 m

5)

The cyclotron frequency of a charged particle in circular motion in a magnetic field is:

f=\frac{qB}{2\pi m}

where here:

q=1.602\cdot 10^{-19}C is the charge of the electron

B = 0.0045 T is the strength of the magnetic field

m=9.31\cdot 10^{-31} kg is the mass of the electron

Substituting, we find:

f=\frac{(1.602\cdot 10^{-19})(0.0045)}{2\pi (9.31\cdot 10^{-31})}=1.23\cdot 10^8 Hz

6)

When a charged particle moves in a magnetic field, its path has a helical shape, because it is the composition of two motions:

1- A uniform motion in a certain direction

2- A circular motion in the direction perpendicular to the magnetic field

The second motion is due to the presence of the magnetic force. However, we know that the direction of the magnetic force depends on the sign of the charge: when the sign of the charge is changed, the direction of the force is reversed.

Therefore in this case, when the particle gains the opposite charge, the circular motion 2) changes sign, so the path will remains helical, but it reverses direction.

7)

The electromotive force induced in a conducting loop due to electromagnetic induction is given by Faraday-Newmann-Lenz:

\epsilon=-\frac{N\Delta \Phi}{\Delta t}

where

N is the number of turns in the loop

\Delta \Phi is the change in magnetic flux through the loop

\Delta t is the time elapsed

From the formula, we see that the emf is induced in the loop (and so, a current is also induced) only if \Delta \Phi \neq 0, which means only if there is a change in magnetic flux through the loop: this occurs if the magnetic field is changing, or if the area of the loop is changing, or if the angle between the loop and the field is changing.

8)

The flux is calculated as

\Phi = BA sin \theta

where

B = 5.5 T is the strength of the magnetic field

A is the area of the coil

\theta=18^{\circ} is the angle between the  direction of the field and the plane of the loop

Here the loop is rectangular with lenght 15 cm and width 8 cm, so the area is

A=(0.15 m)(0.08 m)=0.012 m^2

So the flux is

\Phi = (5.5)(0.012)(sin 18^{\circ})=0.021 Wb

See the last 7 answers in the attached document.

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5 0
3 years ago
Mohs scale is used in describing which characteristic? luster toughness hardness crystal form
Lady_Fox [76]
The answer is hardness not luster 

3 0
3 years ago
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Why is acceleration of an object moving at a constant velocity always zero?
k0ka [10]

Answer:

If an object is moving with a constant velocity, then by definition it has zero acceleration. So there is no net force acting on the object. The total work done on the object is thus 0 (that's not to say that there isn't work done by individual forces on the object, but the sum is 0 ).

Explanation:

In the middle, when the object was changing position at a constant velocity, the acceleration was 0. This is because the object is no longer changing its velocity and is moving at a constant rate.

8 0
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