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Veronika [31]
3 years ago
7

A circular disk, a ring, and a square (is hollow) have the same mass M and width 2r. For the moment of inertia about their cente

r of mass about an axis perpendicular to the plane of the paper, which statement concerning their moments of inertia is true?
a. Iring > Idisk > Isquare
b. Isquare > Iring > Idisk
c. Idisk > Isquare > Iring
d. Iring > Isquare > Idisk
e. Idisk > Iring > Isquare
f. Isquare > Idisk > Iring
Physics
1 answer:
andre [41]3 years ago
4 0

Answer:

d is your answer

Explanation:

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Two equal resistors are connected in series with a 1.50V battery. In order to keep the current at 0.030 A, the resistors much ea
AlexFokin [52]

Answer: 25 Ohms

Explanation:

From this question, the following parameters are given:

Voltage V = 1.5 v

Current I = 0.03A

From Ohm's law;

V = IR

Where R = resultant resistance of the two resistors.

Substitute V and I into the formula and make resultant R the subject of formula.

1.5 = 0.03 × R

R = 1.5/0.03

R = 50 Ohms

From the question, it is given that Thr two equal resistors are connected in series.

R = R1 + R2

But R1 = R2

50 = 2R1

R1 = 50/2

R1 = 25

R1 = R2 = 25 Ohms

Therefore, the resistors must each have a value of 25 Ohms

4 0
3 years ago
How might a <br> Theory relate to a model
Alexxx [7]
A theory can help create a model
3 0
3 years ago
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = \frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}

  = 4.59\times 10^{-19} \ J

or,

  = \frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}

  = 2.87 \ eV

(b)

As we know,

⇒ Vq=\frac{hc}{\lambda}-\Phi_0

By substituting the values, we get

⇒ 1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0

⇒                       \Phi_0=2.3\times 10^{-19} \ J

or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

5 0
3 years ago
Object a has a mass of 20 g
Anarel [89]
And.. where is the rest of the question?
7 0
3 years ago
Question 4 (1 point)
NNADVOKAT [17]

Answer: 0.42 Amperes

Explanation:

Given that:

Current, I = ?

Electric charge Q = 100 coulomb

Time, T = 4.0 minutes

(The SI unit of time is seconds. so, convert 4.0 minutes to seconds)

If 1 minute = 60 seconds

4.0 minutes = 4.0 x 60 = 240 seconds

Since electric charge, Q = current x time

i.e Q = I x T

100 coulomb = I x 240 seconds

I = 100 coulomb / 240 seconds

I = 0.4167 Amperes (round to the nearest hundredth which is 0.42 amperes)

Thus, 0.42 Amperes of current flows in the circuit.

6 0
3 years ago
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