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2 years ago

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Starting at 48th Street, Dylan rides his bike due east on Meridian Road with the wind at his back. He rides for 20 minmin at 15

mphmph. He then stops for 5 minmin, turns around, and rides back to 48th Street; because of the headwind, his speed is only 10 mph.How long does his trip take?

1 answer:

Ivan2 years ago

8 0

To solve this problem we will apply the linear motion kinematic equations. We will start by finding the quantity shifted through the relationship given by the product between speed and time (Time will be converted to hours since the speed is thus specified).

Later, with the given speed we will find the return time.

The general formula of velocity tells us that it is the distance traveled in a certain time, so from it we will obtain the subsequent derivations.

$v = \frac{x}{t} \Rightarrow x = vt \Rightarrow t = \frac{x}{v}$

Here,

x = Distance

t = Time

Amount of distance traveled for 20 mins

$x = (15\frac{miles}{hour})(20 min \frac{1hour}{60min})$

$x = 5miles$

Amount time taken to return back

$t = \frac{x}{v}$

$t = \frac{5 miles}{10mph}$

$t = \frac{1}{2} h = 30minutes$

Total time taken

$T = 20+5+30$

$T = 55minutes$

Therefore he takes around of 55 minutes to do that trip.

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A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn

Cerrena [4.2K]

Answer:

e. Not enough information to determine

Explanation:

This question is incomplete. Here is the complete question with my solution afterwards;

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turntable (initially at rest) begins to rotate with its rate of rotation constantly increasing.

What is the first event that will occur?(Assume non-zero frictional force and the same coefficients of friction for both bugs.)

a. The ladybug begins to slide

b. The gentleman bug begins to slide

c. Both bugs begin to slide at the same time

d. Nothing ever happens

e. Not enough information to determine

The centripetal force acting on a rotating body or bugs can be written as,

$F=mrw^2$

$m=$ mass of the corresponding bugs

$r=$ corresponding radial distance of each bug

$w=$ angular speed of the turntable

The centripetal force tries to slide the bugs in an outward direction and it is directly proportional to the products of its mass and radial distance from the axis of rotation of the turntable

$F$ ∝ $mr$

Since the radial distance from the axis of rotation of the turntable for each bug is given, but the mass is not given, the given information is therefore not enough to determine which bugs will slide first.

Option "e" is correct.

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2 years ago

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(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500 - μC charge and flies due west at a sp

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(a) $2.64\cdot 10^{-8} N$ north

We can treat the aircraft as a single point charge moving in a magnetic field. In this case, the magnetic force exerted on the plane is

$F=qvB sin \theta$

where

$q=0.500 \mu C = 0.500\cdot 10^{-6} C$ is the charge on the plane

v = 660 m/s is the velocity

$B=8.00\cdot 10^{-5} T$ is the magnitude of the magnetic field

$\theta=90^{\circ}$ is the angle between the direction of motion of the jet and of the magnetic field

Substituting,

$F=(0.5\cdot 10^{-6})(660)(8.0\cdot 10^{-5})=2.64\cdot 10^{-8} N$

The direction can be found by using Fleming's left hand rule. We have:

- index finger: magnetic field direction (straight up)

- middle finger: velocity of the plane (due west)

- force: thumb --> north

(b) Not negligible

As we can see from part (a), the magnitude of the force is not really big, so the effects are negligible.

For instance, we can compare this force with the weight of a plane. If we take a Boeing 737, its mass is about 80,000 kg, so its weight is

$W=mg=(80000)(9.8)=784,000 N$

As we can see, this is several orders of magnitude bigger than the magnetic force calculated at point (a), so the effects of the magnetic force are negligible.

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2 years ago

Consider a father pushing a child on a playground merry-go-round. the system has a moment of inertia of 84.4 kg · m2. the father

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<span>At time t1 = 0 since the body is at rest, the body has an angular velocity, v1, of 0. At time t = X, the body has an angular velocity of 1.43rad/s2. Since Angular acceleration is just the difference in angular speed by time. We have 4.44 = v2 -v1/t2 -t1 where V and t are angular velocity and time. So we have 4.44 = 1.43 -0/X - 0. Hence X = 1.43/4.44 = 0.33s.</span>

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yami pours powdered cocoa mix into milk and stirs it. then she microwaves the mixture for three minutes. when she takes the cup

MrRa [10]

Convection can best be observed as she blows the warm steam air that rises.

As the warm steam rises, she forces displaces it with cool air from her mouth. Because the warm steam is less dense it rises and because the cool air is more dense, it displaces the warm air.

This scenario is an example of convection.

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Combine Newton's 2nd law and Hooke's law for a spring to find the acceleration of the block a(t) as a function of time. Express

Inga [223]

Answer:

$a=-\dfrac{k}{m}x(t)$

Explanation:

From Newton's second law,

$F = ma$

where $F$ is the force, $m$ is the mass and $a$ is the acceleration.

From Hooke's law,

$F = -kx(t)$

where $k$ is the spring constant and $x(t)$ is the displacement function measured from the origin. The negative sign indicates the force acts in opposite direction to the displacement. In fact, it is a restoring force; it acts to return the spring to its original undisturbed position.

Since both forces are the same,

$F = ma= - kx(t)$

$a=-\dfrac{k}{m}x(t)$

The implication of this is that the acceleration is proportional to the displacement but opposite to it. That last statement is the definition of a simple harmonic motion which this is.

The ratio $\dfrac{k}{m}$ is a constant except in situations where the mass is varying (say, the mass on the spring is a decaying material).

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2 years ago