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Nastasia [14]
3 years ago
9

a disk of a radius 50 cm rotates at a constant rate of 100 rpm. what distance in meters will a point on the outside rim travel d

uring 30 seconds of rotation?​
Physics
1 answer:
vlabodo [156]3 years ago
8 0

Answer: 50π m ≈ 157 m

Explanation:

100 rev/min (2π rad/rev) / (60 sec/min) = 3⅓π rad/s

d = ωrt = 3⅓π(0.50)(30) = 50π m ≈ 157 m

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If vector C is added to vector D, the result is a third vector that is perpendicular to D and has a magnitude equal to 3D. What
Jet001 [13]

Answer:

(e) 3.2

Explanation:

We are given that vector C and D.

Let R be the magnitude of C+D.

According to question

R=3D

We have to find the ratio of the magnitude of C to that of D.

By using right triangle property

C^2=R^2+D^2

C^2=(3D)^2+D^2

C^2=9D^2+D^2

C^2=10D^2

C=\sqrt{10D^2}=3.2D

\frac{C}{D}=3.2

Hence, the ratio of the magnitude of C to that of D=3.2

(e) 3.2

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3 years ago
What is the average speed in mph for a car that travels for 5 hours and 20 minutes?
Ymorist [56]

Answer: If you meant 5 miles in 20 minutes than it’s 1 mile in 5 minutes

Explanation:

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3 years ago
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4 years ago
How do the amplitudes of a 120 decibel sound and a 100-decibel sound compare?​
NemiM [27]
The 120 decibel sound has more amplitude than the 100 decibel sound.

In Physics, the relation between amplitude and intensity is that the intensity of the wave is directly proportional to the square of its amplitude.
4 0
2 years ago
A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A
Sphinxa [80]

Answer:

Density of 127 I = \rm 1.79\times 10^{14}\ g/cm^3.

Also, \rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

Explanation:

Given, the radius of a nucleus is given as

\rm r=kA^{1/3}.

where,

  • \rm k = 1.3\times 10^{-13} cm.
  • A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.

For the nucleus 127 I,

Mass, M = \rm 2.1\times 10^{-22}\ g.

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.

On comparing with the density of the solid iodine,

\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

7 0
3 years ago
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