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1 year ago

15

Bjorn is holding a tennis ball outside a second floor window (3.5 meters from the ground) and billie jean is holding one outside

a third floor window (6.25 meters from the ground) how much more GPE does billie jean's tennis ball have? (each tennis ball has a mass of 0.06 kg)

1 answer:

MArishka [77]1 year ago

4 0

The answer is 1.01 x 10^(-11) N. I arrived to this answer through calculating the GPEs of both balls. Bjorn's ball has a GPE of 1.402 x 10^(-11) N. Billie Jean's ball has a GPE of <span>2.503 x 10^(-11) N. I subtracted the two and I found that Billie Jean's tennis ball has a GPE of 1.01 x 10^(-11) more than Bjorn's tennis ball.</span>

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When driving in heavy rain, or on a flooded road, your tires can ride on a thin film of water like skis;

Simora [160]

The answer is letter a. It is best to slow down in situations of heavy rain or flooded road as skid could be the result if you lose out of control because the driver isn't slowing down. That is why it is being said that tires can ride on a thin film of water skis as it could skid if it has lost control if the driver hadn't slowed down.

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1 year ago

A toroidal solenoid has an inner radius of 12.0 cm and an outer radius of 15.0 cm . It carries a current of 1.50 A . Part A How

tensa zangetsu [6.8K]

Answer:

The number of turns is $N = 1750 \ turns$

Explanation:

From the question we are told that

The inner radius is $r_i = 12.0 \ cm = 0.12 \ m$

The outer radius is $r_o = 15.0 \ cm = 0.15 \ m$

The current it carries is $I = 1.50 \ A$

The magnetic field is $B = 3.75 mT = 3.75 *10^{-3} \ T$

The distance from the center is $d = 14.0 \ cm = 0.14 \ m$

Generally the number of turns is mathematically represented as

$N = \frac{2 * \pi * d * B}{ \mu_o * r_o }$

Generally $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} \ N/A^2$

So

$N = \frac{2 * 3.142 * 0.14 * 3.75 *10^{-3} }{ 4\pi * 10^{-7} * 0.15 }$

$N = 1750 \ turns$

5 0

1 year ago

You are piloting a small airplane in which you want to reach a destination that is 750 km due north of your starting location. O

alexira [117]

Answer:

v_wind = 101.46 km / h , θ = 61.8

Explanation:

This is a velocity composition exercise.

Let's do the problem in parts. Let's start by knowing the speed of the plane without air.

v = d / t

v = 750 / 3.14

v = 238.85 km / h

This is the speed of the plane relative to the Earth and it does not change.

In the second part, when there is wind, the travel time is greater than when there is no wind, therefore the wind delays the plane. To be more general, suppose that the wind has two components vₓ and $v_{y}$

Let's use trigonometry to find the components of the plane's speed

cos θ = v_N / v

sin θ = v_W / v

v_N = v cos θ

v_W = v sin θ

let's calculate

V _N = 238.85 cos 22 = 221.46 km / h

v_W = -238.85 sin 22 = -89.47

the negative sign is because the plane is going west and the positive sign is the east direction.

As it indicates that the destination of the avine is towards the north, the x component of the wind must be

vₓ - v_W = 0

vₓ = v-w

vₓ = 89.47 km / h

in the direction to the East.

Now let's analyze the component of the wind in the Nort-South direction,

Indicate the travel time, let's calculate the speed that the component must have the speed of the plane

v_total = d / t

v_total = 750 / 4.32

v_total = 173.61 km / h

This is the final speed of the plane, which can be written

v_total = v_n - vy

vy = v_n - v_total

vy = 221.46 - 173.61

vy = 47.85 km

this component is directed towards the south

Let's use the Pythagorean Theorem, to find the magnitude

v_wind² = vₓ² + vy²

v_wind = √ (89.47² + 47.85²)

v_wind = 101.46 km / h

the address will then be found using trigonometry

θ = Vy / vx

θ = tan⁻¹ (vy / vx)

θ = tan⁻¹1 (47.85 / 89.47)

θ = 28.14

Therefore, the magnitude of the wind speed is 101.5 km / h and its direction is 28º south of the East, to give this value

90- θtea = 90- 28.2

θ = 61.8

East of South

5 0

1 year ago

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit

inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses $x_1$ distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1$

When speed doubles

$\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2$

divide 1 and 2

$\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}$

$x_2=2x_1$

Therefore spring compresses twice the initial value

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1 year ago

A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 6 inches below the

hoa [83]

Answer:

Since the spring mass system will execute simple harmonic motion the position as a function of time can be written as $x(t)=Asin(\omega t+\phi)$

'A' is the amplitude = 6 inches (given)

$\omega =\sqrt{\frac{k}{m}}$ is the natural frequency of the system

At equilibrium we have

$mg=kx\\\\k=\frac{mg}{x}$

Applying values we get

$k=40 lb/ft$

thus natural frequency equals

$\omega =\sqrt{\frac{40}{\frac{20}{32}}}\\\\\omega =8s^{-1}$

Thus the equation of motion becomes

$x(t)=6sin(8t+\phi)$

At time t=0 since mass is at it's maximum position thus we have

$A=Asin(\omega t+\phi)\\\\\therefore sin(\omega\times 0+\phi)=1\\\\\phi=\frac{\pi}{2}\\\\\therefore x(t)=Asin(\omega t+\frac{\pi}{2})$

Thus the position of mass at the given times is as follows

1) at $\frac{\pi}{12}$ $x(t)=5.99inches$

2) at $\frac{\pi}{8}$ $x(t)=5.9909inches$

3) at $\frac{\pi}{6}$ $x(t)=5.98397inches$

4) at $\frac{\pi}{4}$ $x(t)=5.9639inches$

5) at $\frac{9\pi}{32}$ $x(t)=5.954inches$

4 0

1 year ago