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baherus [9]
4 years ago
12

As Aubrey watches this merry-go-round for a total of 2 minutes, she notices the black horse pass by 15 times. What is the period

of the black horse?
Physics
2 answers:
marishachu [46]4 years ago
8 0
2 min = 120 sec

120/15 = 8

The black horse represents 8 seconds.
Wewaii [24]4 years ago
5 0

Answer:

8 seconds

Explanation:

The time taken by the merry go round to complete one round is called time period.

Number of rounds = 15

Time taken = 2 minutes = 2 x 60 = 120 second

15 rounds are completetd in 120 seconds

1 round completed in 120 / 15 = 8 seconds

So, the time period is 8 seconds.

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Define fundamental unit ? write its two example. ​
stealth61 [152]

the International System of Units, the fundamental units are: The meter (symbol: m), used to measure length. The kilogram (symbol: kg), used to measure mass. The second (symbol: s), used to measure time. The ampere (symbol: A), used to measure electric current.

8 0
3 years ago
You would like to store 8.1 J of energy in the magnetic field of a solenoid. The solenoid has 620 circular turns of diameter 6.6
marta [7]

Answer:

(a) The current needed is 56.92 A

(b) The magnitude of the magnetic field inside the solenoid is 0.134 T

(c) The energy density inside the solenoid is 7.144 kJ/m³

Explanation:

Given;

energy stored in the magnetic field of solenoid, E = 8.1 J

number of turns of the solenoid, N = 620 turns

diameter of the solenoid, D = 6.6 cm = 0.066 m

radius of the solenoid, r = D/2 = 0.033 m

length of the solenoid, L = 33 cm = 0.33 m

Inductance of the solenoid is given as;

L= \frac{\mu_o N^2 A}{l}

where;

A is the area of the solenoid = πr² = π (0.033)² = 0.00342 m²

μ₀ is permeability of free space = 4π x 10⁻⁷ H/m

L= \frac{4\pi*10^{-7} *620^2 *0.00342}{0.33} \\\\L = 0.005 \ H

(A). How much current needed

Energy stored in magnetic field of solenoid is given as;

E = \frac{1}{2} LI^2\\\\

Where;

I is the current in the solenoid

E = \frac{1}{2} LI^2\\\\I^2 = \frac{2E}{L}\\\\I = \sqrt{\frac{2*8.1}{0.005}}\\\\ I = 56.92 \ A

(B) The magnitude of the magnetic field inside the solenoid

B = μ₀nI

where;

n is number of turns per unit length

B = μ₀(N/L)I

B = (4π x 10⁻⁷)(620/0.33)(56.92)

B = 0.134 T

(C) The energy density (energy/volume) inside the solenoid

U_B = \frac{B^2}{2\mu_0} \\\\U_B = \frac{(0.134)^2}{2*4\pi*10^{-7}} \\\\U_B = 7143.54 \ J/m^3\\\\U_B = 7.144 \ kJ/m^3

3 0
3 years ago
To analyze the experiment used to determine the properties of an electron. In 1909, Robert Millikan performed an experiment invo
sweet-ann [11.9K]

Answer:

C has 5 electrons

Explanation:

Given:

The data acquired from the experiment performed by Millikan:

Q_a = 3.20 x10^{-19}  C

Q_b = 4.80 x10^{-19}  C

Q_c = 8.00 x 10^{-19}  C

Q_d = 9.60 x 10^{-19}  C

Find:

How many Electrons were present in drop C

Solution:

It is known that the charge of an electron e = 1.602 *10^-19 C / electron.

Hence the number of electrons n in drop C will be:

      n = Q_c / e

      n = 8.00 x 10^{-19}  / 1.602*10^-19

      n = 4.99 = 5 electrons  

Answer: The drop C contains 5 electrons.

5 0
3 years ago
A 12.70 g bullet has a muzzle velocity (at the moment it leaves the end of a firearm) of 430 m/s when rifle with a weight of 25.
Norma-Jean [14]

Answer:

2.1844 m/s

Explanation:

The principle of conservation of momentum can be applied here.

when two objects interact, the total momentum remains the same  provided no external forces are acting.

Consider the whole system , gun and bullet. as an isolated system, so the net momentum is constant. In particular before firing the gun, the net momentum is zero. The conservation of momentum,

0=m_{bullet}*v_{bullet}  + m_{rifle}*v_{rifle}  \\

assume the bullet goes to right side and the gravitational acceleration =10 ms^{-2}

so now the weight of the rifle=\frac{25}{10}  

0=m_{bullet}*v_{bullet}  + m_{rifle}*v_{rifle}  \\\\0=(12.70*10^{-3}) *430ms^{-1} +(\frac{25}{10} )*v_{rifle} \\v_{rifle} =-2.1844ms^{-1}

this is a negative velocity to the right side. that means the rifle recoils to the left side

3 0
3 years ago
A small aircraft, on a heading of 225°, is cruising at 150 km/h. It is encountering a wind blowing from a bearing of 315° at 35
ki77a [65]

As we can see in the figure attached here

The velocity of aircraft is given as

v_p = 150 km/h at 225 degree

velocity of wind is given as

v_a = 35 km/h at 315 degree

so here the net velocity of aircraft will be the vector sum of aircraft speed and wind speed

since in the figure we can see that the two speeds are perpendicular to each other

so we can say that the resultant of two speed can be given by using Pythagoras theorem

so we will have

v_{net} = \sqrt{v_a^2 + v_p^2}

v_{net} = \sqrt{150^2 + 35^2}

v_{net} = 154 km/h

so net speed of aircraft will be 154 km/h

6 0
3 years ago
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