A 64.0-kg ice skater is moving at 4.04 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.890 m around the pole. (a) Determine the magnitude of the force exerted by the horizontal rope on her arms. kN (b) Compare this force with her weight. Frope W =

Answers

Answer 1

Answer:

We have that for the Question it can be said that the magnitude of the force exerted by the horizontal rope on her arms and the ratio of the Force to the weight is

F=1150.561N

F/W=1.8325

From the question we are told

A 64.0-kg ice skater is moving at 4.04 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.890 m around the pole. (a) Determine the magnitude of the force exerted by the horizontal rope on her arms. kN (b) Compare this force with her weight. F-rope W =

Generally the equation for the force applied is mathematically given as

$F=(( mv^2))/(R)\n\nTherefore\n\nF=(( mv^2))/(R)\n\nF=(( (64)(4.0)^2))/(0.890)\n\n$

F=1150.561N

Generally the equation for the Weight is mathematically given as

W=mg

Therefore

W=64*9.81

W=627.84N

Therefore

The Force to weight ratio is

$F/W=1150.561N/627.84N$

F/W=1.8325

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Answer 2

Answer:

Final answer:

The force exerted by the rope on the skater's arms as she moves in a circular path is 1.167 kN. This force is about 1.860 times her weight, which is 627.2 N.

Explanation:

The skater is experiencing centripetal force exerted by the rope, which causes her to move in a circular path. The magnitude F of this force can be calculated using the formula F = mv²/r, where m is the skater's mass (64.0 kg), v is her velocity (4.04 m/s), and r is the radius of her circular path (0.890 m).

By substituting the given numbers into this formula, we get: F = (64.0 kg)(4.04 m/s)² / 0.890 m = 1166.67 N. In kilonewtons, this force is 1.167 kN.

To compare this force with her weight, we can calculate the weight (W) using the formula W = mg, where g is the acceleration due to gravity (around 9.8 m/s²). Substituting the given mass into this formula gives us: W = (64.0 kg)(9.8 m/s²) = 627.2 N.

Comparing these two forces shows that the force exerted by the rope on her arms is about 1.860 times her weight.

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A hemispherical surface (half of a spherical surface) of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere. What is the magnitude of the electric flux through the hemisphere surface?

Answers

Answer:

π*R²*E

Explanation:

According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.

As the electric field E is uniform and parallel to the hemisphere axis, and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.

The flux across the open surface can be expressed as follows:

$\int\ {E} \, dA = E*A = E*\pi *R^(2)$

As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².

⇒Flux = E*π*R²

Each mass in the figure is 3 kg. Find the magnitude and direction of the net gravitational force on mass A due to the other masses.A. 2.45 × 10–7 N toward B
B. 3.75 × 10–7 N toward C
C. 2.00 × 10–7 N toward D
D. 1.15 × 10–7 N toward D

Answers

The magnitude and direction of the net gravitationalforce on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.

The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In this case, all of the masses are equal to 3 kg, and the distance between mass A and mass D is 3 m.

The gravitational force between mass A and mass D is therefore:

F = G * m_A * m_D / r²

= 6.674 × 10⁻¹¹ N m² / kg² * 3 kg * 3 kg / 3 m²

= 1.15 × 10⁻⁷ N

The direction of the gravitational force is towards mass D.

Therefore, the net gravitational force on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.

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Answer:

THE ANSER IS B

Explanation:

A toy car that is 0.12 m long is used to model the actions of an actual car thatis 6 m long. Which ratio shows the relationship between the sizes of the
model and the actual car?
A: 5:1
B: 1:50
C: 50:1
D: 1:5

Answers

Answer:

it is B 1:50

Explanation:

just did it on apex

Final answer:

The ratio that shows the relationship between the sizes of the model car and the actual car is 1:50. This is because the actual car is 50 times longer than the model car.

Explanation:

The relationship between the sizes of the model car and the actual car is represented by a ratio. To find this ratio, we can divide the length of the actual car by the length of the model car. So, 6 m / 0.12 m = 50. This means that the actual car is 50 times longer than the model car, or in other words, the model car is 1/50th the size of the actual car. Therefore, the correct ratio is 1:50.

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When a 20.0-ohm resistor is connected across the terminals of a 12.0-V battery, the voltage across the terminals of the battery falls by 0.300 V. What is the internal resistance of this battery

Answers

The internal resistance of the battery is 0.5 ohms.

To calculate the internal resistance of the battery, we use the formula below

Formula:

(V/R)r = V'............. Equation 1

Where:

V = Voltage across the terminal of the battery
R = Resistance connected across the battery
r = internal resistance of the battery
V' = voltage drop of the battery.

Make r the subject of the equation

r = V'R/V............ Equation 2

From the question,

Given:

V = 12 V
R = 20 ohms
V' = 0.3 V

Substitute these values into equation 2

r = (0.3×20)/12
r = 6/12
r = 0.5 ohms.

Hence, The internal resistance of the battery is 0.5 ohms.

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Answer:

The internal resistance is $r = 0.5 \ \Omega$

Explanation:

From the question we are told that the resistance of

The resistance of the resistor is $R = 20.0\ \Omega$

The voltage is $V = 12.0 \ V$

The magnitude of the voltage fall is $e = 0.300\ V$

Generally the current flowing through the terminal due to the voltage of the battery is mathematically represented as

$I = (V)/(R)$

substituting values

$I = (12.0 )/(20 )$

$I = 0.6 \ A$

The internal resistance of the battery is mathematically represented as

$r = (e)/(I)$

substituting values

$r = (0.300)/( 0.6 )$

$r = 0.5 \ \Omega$

What is the frequency of a photon that has the same momentum as a neutron moving with a speed of 1.90 × 103 m/s?

Answers

The mass of a neutron is:
$m=1.67 \cdot 10^(-27)kg$
Since we know its speed, we can calculate the neutron's momentum:
$p=mv=(1.67 \cdot 10^(-27)kg)(1.90 \cdot 10^3 m/s)=3.17 \cdot 10^(-24) kg m/s$

The problem says the photon has the same momentum of the neutron, p. The photon momentum is given by
$p= (h)/(\lambda)$
where h is the Planck constant, and $\lambda$ is the photon wavelength. If we re-arrange the equation and we use the momentum we found before, we can calculate the photon's wavelength:
$\lambda= (h)/(p)= (6.6 \cdot 10^(-34)Js)/(3.17 \cdot 10^(-24) kg m/s)=2.08 \cdot 10^(-10) m$

And since we know the photon travels at speed of light c, we can now calculate the photon frequency:
$f= (c)/(\lambda)= (3 \cdot 10^8 m/s)/(2.08 \cdot 10^(-10) m)= 1.44 \cdot 10^(18) Hz$

The New York Wheel is the world's largest Ferris wheel. It's 183 meters in diameter and rotates once every 37.3 min.1. Find the magnitude of the average velocity at the wheel's rim, over a 7.40-
min interval.

2.Find the magnitude of the average acceleration at the wheel's rim, over a 7.40-
min interval.

Answers

Answer:

Velocity =0.241 m/s

Acceleration = 7.21e-4 m/s²

Explanation:

The wheel travels through

Θ = (7.40/37.3)*360º = 71.42º

and so the length of the line segment connecting the initial and final position is

L = 2*L*sin(Θ/2) = 2 * (183m/2) * sin(71.42º/2) = 107 m

so the average velocity is

v = L / t = 107m / 7.40*60s = 0.241 m/s

Initially, let's say the velocity is along the +x axis:

Vi = π * 183m / (37.3*60s) i = 0.257 m/s i

Later, it's rotated through 71.42º, so

Vf = 0.257m/s * (cos71.42º i + sin71.42º j) = [0.0819 i + 0.244 j] m/s

ΔV = Vf - Vi = [(0.0819 - 0.257) i + 0.244 j] m/s = [-0.175 i + 0.244 j] m/s

which has magnitude

|ΔV| = √(0.175² + 0.244²) m/s = 0.300 m/s

Then the average acceleration is

a_avg = |ΔV| / t = 0.300m/s / (7.40*60s) = 6.76e-4 m/s²

The instantaneous acceleration is centripetal: a = ω²r

a = (2π rads / (37.3*60s)² * 183m/2 = 7.21e-4 m/s²

Answer:

$v = 0.24 m/s$

$a = 6.75 * 10^(-4) m/s^2$

Explanation:

Given that wheel completes one round in total time T = 37.3 min

so angular speed of the wheel is given as

$\omega = (2\pi)/(T)$

$\omega = (2\pi)/(37.3) rad/min$

now the angle turned by the wheel in time interval of t = 7.40 min

$\theta = \omega t$

$\theta = ((2\pi)/(37.3))(7.40) = 0.4\pi$

PART 1)

Now the average velocity is defined as the ratio of displacement and time

here displacement in given time interval is

$d = 2Rsin(\theta)/(2)$

R = radius = 91.5 m

$d = 183sin(0.2\pi) = 106.8 m$

Now time to turn the wheel is given as

$t = 7.40 min = 444 s$

now we have

$v = (d)/(t) = (106.8)/(444)$

$v = 0.24 m/s$

PART 2)

Now average acceleration is defined as ratio of change in velocity in given time interval

here velocity of a point on its rim is given as

$v = R\omega$

$v = (91.5)((2\pi)/(37.3* 60))$

$v = 0.257 m/s$

now change in velocity when wheel turned by the above mentioned angle is given as

$\Delta v = 2vsin(\theta)/(2)$

$\Delta v = 2(0.257)sin(0.2\pi)$

$\Delta v = 0.3 m/s$

time interval is given as

$t = 7.40 min = 444 s$

now average acceleration is given as

$a = (0.3)/(444)$

$a = 6.75 * 10^(-4) m/s^2$

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