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Inessa05 [86]
4 years ago
10

An ant sits on a cd at a distance of 17 cm from the center. If it sits there for 42

Physics
1 answer:
nydimaria [60]4 years ago
4 0

Answer:

a) N = 8.54[rev]; b) v = 21.73 [m/s]; c) T = 4.918[s]; d) 12.2[rev/min]

Explanation:

We know that the arc length is calculated by the following expression:

L = α * r

where:

r = radius = 17 [cm] or 0.17 [m]

α = 360° or 2*pi [rad] = 6.283 [rad]

L = (6.283*0.17) = 1.068 [m] = 106.81 [cm] is the distance travelled in one revolution of the CD

a) N = number of revolutions

N = 913 / 106.81

N = 8.54 [rev]

b ) The speed can be calculated by the following expression:

v = d/t

Where:

d = distance = 913[cm]

t = time = 42[sec]

v = 913/42 = 21.73 [cm/s]

c)

We have the number of revolutions and the time therefore we can calculate the number of revolutions per second

w = 8.54 / 42 =0.2033 [rev/s]

And we know that the period is the reciprocal of the time

T = 1 / 0.2033

T = 4.918 [s]

d )

We need to convert from [rev/s] into [rev/min]

0.2033[\frac{rev}{s}]*60[\frac{s}{1min}]\\12.198[rev/min]

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Colt1911 [192]

Answer:

50 degrees

Explanation:

Usually in physics in solving such questions refractive index of air is taken as 1.00 (n=1.00) by assuming that air is a vacuum. Hence, the light ray passes from water to air at an angle of incident of 40 degrees, it means that it has angle of 50 degrees to the normal line.

5 0
4 years ago
A jet airliner moving initially at 503 mph
Varvara68 [4.7K]

Answer:

The new speed is 1230.28 m/h

Explanation:

The jet airliner moving initially at 503 mph to the east

The wind is blowing at  855 mph in a direction 52° north of east

At first let us distribute the velocity of the wind into east component

and north component

→ The east component is 855 cos(52) m/h

→ The north component is 855 sin(52) m/h

Now we have two components of velocity in the east direction

and one component of velocity in the north direction

The new speed is the resultant of the east and north components

→ The east components are 503 m/h and 855 cos(52) m/h

→ The north component is 855 sin(52) m/h

Add the components of the speeds in direction of east

→ The east component = 503 + 855 cos(52) = 1029.39 m/h

→ The north component = 855 sin(52) = 673.75 m/h

Now we can find the new speed as a resultant speed of the east and

north components

→ The new speed = \sqrt{(1029.39)^{2}+(673.75)^{2}}

→ The new speed = 1230.28 m/h

<em>The new speed is 1230.28 m/h</em>

3 0
4 years ago
Why would a dewar flask not need a lid?
snow_tiger [21]

Answer:

because it does ent

Explanation:

5 0
3 years ago
1. What would be the final temperature of a mixture of 50g of water at 200C temperature and
ad-work [718]
It would be the average I guess..300c temperature.Heat flows from higher temp. to lower temp. till the temp. of both the liquids are same.
8 0
3 years ago
Ignoring the mass of the spring, a 5 kg mass hanging from a coiled spring having a constant k= 50 N/m will have a period of osci
Marrrta [24]

Answer:

Period of oscillation, T = 2 sec

Explanation:

It is given that,

Mass of the object, m = 5 kg

Spring constant of the spring, k = 50 N/m

This object is hanging from a coiled spring. We need to find the period of oscillation of the spring. The time period of oscillation of the spring is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

T=2\pi\sqrt{\dfrac{5\ kg}{50\ N/m}}

T = 1.98 sec

or

T = 2 sec

So, the period of oscillation is about 2 seconds. Hence, this is the required solution.

4 0
3 years ago
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