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2 years ago

Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?.

1 answer:

3 0

For a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s is mathematically given as

F= 618.9 N

<h3>What is the centripetal force?</h3>

Generally, the equation for the angular speed is mathematically given as

w = v/R

Therefore

w= 4.7/1.8

w= 2.611 rad/s

Where total momentum

Tm= 642.96 + 272.32

Tm= 915.28

and total inertia

Ti= 184 + 246.24

Ti= 430.24

In conclusion, centripetal force

F= mrw^2

F = m*R*w2^2

F = 76*1.8*2.127^2

F= 618.9 N

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If a transformer has 50 turns in the primary winding and 10 turns on the secondary winding, what is the reflected resistance in

Elza [17]

Answer:

The reflected resistance in the primary winding is 6250 Ω

Explanation:

Given;

number of turns in the primary winding, $N_P$ = 50 turns

number of turns in the secondary winding, $N_S$ = 10 turns

the secondary load resistance, $R_S$ = 250 Ω

Determine the turns ratio;

$K = \frac{N_P}{N_S} \\\\K = \frac{50}{10} \\\\K = 5$

Now, determine the reflected resistance in the primary winding;

$\frac{R_P}{R_S} = K^2\\\\R_P = R_SK^2\\\\R_P = 250(5)^2\\\\R_P = 6250 \ Ohms$

Therefore, the reflected resistance in the primary winding is 6250 Ω

6 0

3 years ago

Near the surface of Earth an electric field points radially downward and has a magnitude of approximately 100 N/C. What charge (

pentagon [3]

Answer:

$q=2.997\times 10^{-4}$ C

Sign-Negative

Explanation:

We are given that

Electric field = $E=100NC^{-1}$ (Radially downward)

Acceleration= $0.19 ms^{-2}$ (Upward)

Mass of charge=3 g= $3\times 10^{-3}$ kg

1kg=1000g

We have to find the magnitude and sign of charge would have to be placed on a penny .

By newton's second law

$\sum F_y=ma$

$\sum F_y=qE-mg$

Substitute the values then we get

$qE-mg=ma$

Substitute the values then we get

$q(100)-3\times 10^{-3}(9.8)=3\times 10^{-3}(0.19)$

$100q-29.4\times 10^{-3}=0.57\times 10^{-3}$

$100q=0.57\times 10^{-3}+29.4\times 10^{-3}=29.97\times 10^{-3}$

$q=\frac{29.97\times 10^{-3}}{100}$

$q=2.997\times 10^{-4}$ C

Sign of charge =Negative

Because electric force acting in opposite direction of electric field therefore,charge on penny will be negative.

4 0

3 years ago

When is a secondary source more helpful than a primary source?

UNO [17]

Answer:

I think the answer is C.

Explanation:

A primary source is a first hand account of an event while a secondary source is a retelling or second hand account meaning as many details will be prevalent.

8 0

2 years ago

A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic

miss Akunina [59]

Answer:

0.07756 m

Explanation:

Given mass of object =0.20 kg

spring constant = 120 n/m

maximum speed = 1.9 m/sec

We have to find the amplitude of the motion

We know that maximum speed of the object when it is in harmonic motion is given by $v_{max}=A\omega$ where A is amplitude and $\omega$ is angular velocity

Angular velocity is given by $\omega=\sqrt{\frac{k}{m}}$ where k is spring constant and m is mass

So $v_{max}=A\sqrt{\frac{k}{m}}$

$A=V_{max}\sqrt{\frac{m}{k}}=1.9\times \sqrt{\frac{0.2}{120}}=0.07756 \ m$

3 0

3 years ago

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A 403.0-kg copper bar is melted in a smelter. The initial temperature of the copper is 320.0 K. How much heat must the smelter p

aleksklad [387]

The answer is 2.49 x 10^5 KJ. This was obtained (1) use the formula for specific heat to achieve Q or heat then (2) get the energy to melt the copper lastly (3) Subtract both work and the total energy required to completely melt the copper bar is achieved.

7 0

3 years ago

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