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2 years ago

7

car driving on a circular test track shows a constant speedometer reading of 100 kph for one lap. a. Describe the car's speed du

ring this time. b.

1 answer:

Annette [7]2 years ago

3 0

Answer:

Speed = 100 km/h

Explanation:

Given:

Speedometer reading = 100 kph for one lap

Assume;

Time taken to complete one lap = 1 hour

Computation:

Speed = Distance / Time

Speed = 100 / 1

Speed = 100 km/h

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A series circuit has a voltage supply of 12 V and a resistor of 2.4 kΩ. How much power is dissipated by the resistor?

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Answer:

(B) 0.06W

Explanation:

power absorbed by the resistor is given by $\frac{V^2}{R}=I^2R$

Where V = voltage

R= resistance

I = current through the circuit

we have given V =12 Volt and resistance =2.4 K $\Omega$

current $I=\frac{V}{R}=\frac{12}{2.4\times 1000}=5\times 10^{-3}A=5mA$

power using voltage and resistance equation

$p=\frac{12^2}{2.4\times 1000}=60mW$ =0.06W

using current equation $P=I^2R=5^2\times 12=60mW$ = 0.06W

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3 years ago

Orchestra instruments are commonly tuned to match an A-note played by the principal oboe. The Baltimore Symphony Orchestra tunes

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Answer:

Δλ = 3*10⁻³ m.

Explanation:

At any wave, there exists a fixed relationship between the speed of the wave, the wavelength, and the frequency, as follows:

$v = \lambda* f (1)$

where v is the speed, λ is the wavelength and f is the frequency.

Rearranging terms, we can get λ from the other two parameters, as follows:

$\lambda = \frac{v}{f} (2)$

Since v is constant for sound at 343 m/s, we can find the different wavelengths at different frequencies, as follows:

$\lambda_{1} =\frac{v}{f_{1}} = \frac{343m/s}{440(1/s)} = 0.779 m (3)$

$\lambda_{2} =\frac{v}{f_{2}} = \frac{343m/s}{442(1/s)} = 0.776 m (4)$

The difference between both wavelengths, is just the difference between (3) and (4):

$\Delta \lambda = \lambda_{1} - \lambda_{2} = 0.779 m - 0.776m = 3e-3 m (5)$

⇒ Δλ = 3*10⁻³ m.

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2 years ago

Kevin goes bowling. Whenever he bowls the ball, he transfers energy from his hand to the bowling ball. The amount of energy befo

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Answer:the answer is C

Explanation:

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2 years ago

A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio

Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

$\sum F_{x}=0$

we can see there are only two forces in x, which are the weight on x and the static friction, so:

$-W_{x}+f_{s}=0$

when solving for the static friction we get:

$f_{s}=W_{x}$

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

$W_{x}=mg sin \alpha$

we can substitute this on our sum of forces equation:

$f_{s}=mg sin \alpha$

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

$f_{s}=N\mu_{s}$

so we can substitute this on our equation:

$N\mu_{s}=mg sin \alpha$

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

$\sum F_{y}=0$

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

$N-W_{y}=0$

when solving for N we get:

$N=W_{y}$

When seeing the free-body diagram we can determine that:

$W_{y}=mg cos \alpha$

so we can substitute that in the sum of y-forces equation, so we get:

$N=mg cos \alpha$

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

$mg cos \alpha \mu_{s}=mg sin \alpha$

we can divide both sides of the equation into mg so we get:

$cos \alpha \mu_{s}=sin \alpha$

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for $\mu_{s}$ we get:

$\mu_{s}=\frac{sin \alpha}{cos \alpha}$

when simplifying this we get that:

$\mu_{s}=tan \alpha$

now we can solve for the angle so we get:

$\alpha= tan^{-1}(\mu_{s})$

and we can substitute the given value so we get:

$\alpha= tan^{-1}(0.350)$

which yields:

α=19.29°

which rounds to:

α=19°

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3 years ago

A car, starting from the origin, travels

sweet [91]

Answer:

The net displacement of the car is 3 km West

Explanation:

Please see the attached drawing to understand the car's trajectory: First in the East direction for 4 km (indicated by the green arrow that starts at the origin (zero), and stops at position 4 on the right (East).

Then from that position, it moves back towards the West going over its initial path, it goes through the origin and continues for 3 more km completing a moving to the West a total of 7 km. This is indicated in the drawing with an orange trace that end in position 3 to the left (West) of zero.

So, its NET displacement considered from the point of departure (origin at zero) to the final point where the trip ended, is 3 km to the west.

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2 years ago

Read 2 more answers