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2 years ago

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A coffee cup on the horizontal dashboard of a car slides forward when the driver decelerates from 45 kmh to rest in 3.5 s or les

s, but not if she decelerates in a longer time. What is the coefficient of static friction between the cup and the dash? Assume the road and the dashboard is level (horizontal).

1 answer:

Svetach [21]2 years ago

6 0

Answer:

μs = 0.36

Explanation:

Assuming no other forces acting on the cup while the car is decelerating, the friction force is responsible for any horizontal movement of the cup.

If the cup is on the verge of starting to slide, the friction force can be expressed as follows:

Ff = -μs*N = -μs*m*g

This force produces a deceleration from 45 Kmh to rest, in 3.5 s or more.

Converting 45 kmh to m/s, we have:

$45 kmh *\frac{1000m}{1km} * \frac{1h}{3600 s} = 12.5 m/s$

We can find the acceleration, just applying the definition, with vf =0, as follows:

$a = \frac{-vf}{t} =\frac{-12.5m/s}{3.5s} = -3.57 m/s2$

According to Newton's 2nd law, we can write the following expression:

F = m*a = -μs*m*g

Simplifying common terms, we can solve for μs, as follows:

μs = $\frac{a}{-g} =\frac{-3.57 m/s2}{-9.80m/s2} = 0.36$

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The intensity at a distance of 6.0 m from a source that is radiating equally in all directions is 6.0 × 10-10 w/m2 . what is the

satela [25.4K]

The intensity is defined as the ratio between the power emitted by the source and the area through which the power is calculated:
$I= \frac{P}{A}$ (1)
where
P is the power
A is the area

In our problem, the intensity is $I=6.0 \cdot 10^{-10} W/m^2$ . At a distance of r=6.0 m from the source, the area intercepted by the radiation (which propagates in all directions) is equal to the area of a sphere of radius r, so:
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And so if we re-arrange (1) we find the power emitted by the source:
$P=IA = (6.0 \cdot 10^{-10}W/m^2)(452.2 m^2)=2.7 \cdot 10^{-7} W$

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2 years ago

A spaceship of frontal area 10 m2 moves through a large dust cloud with a speed of 1 x 106 m/s. The mass density of the dust is

Step2247 [10]

Answer:

The decelerating force is $3\times 10^{- 11}\ N$

Solution:

As per the question:

Frontal Area, A = $10\ m^{2}$

Speed of the spaceship, v = $1\times 10^{6}\ m/s$

Mass density of dust, $\rho_{d} = 3\times 10^{- 18}\ kg/m^{3}$

Now, to calculate the average decelerating force exerted by the particle:

$Mass,\ m = \rho_{d}V$ (1)

Volume, $V = A\times v\times t$

Thus substituting the value of volume, V in eqn (1):

$m = \rho_{d}(Avt)$

where

A = Area

v = velocity

t = time

$m = \rho_{d}(A\times v\times t)$ (2)

$Momentum,\ p = \rho_{d}(Avt)v = \rho_{d}Av^{2}t$

From Newton's second law of motion:

$F = \frac{dp}{dt}$

Thus differentiating w.r.t time 't':

$F_{avg} = \frac{d}{dt}(\rho_{d}Av^{2}t) = \rho_{d}Av^{2}$

where

$F_{avg}$ = average decelerating force of the particle

Now, substituting suitable values in the above eqn:

$F_{avg} = 3\times 10^{- 18}\times 10\times 1\times 10^{6} = 3\times 10^{- 11}\ N$

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1 year ago

Why does the closed top of a convertible bulge when the car is riding along a highway? Why does the closed top of a convertible

iren2701 [21]

Answer:

Option D

The air pressure inside the car is greater than the pressure outside.

Explanation:

When considering airflow over and around a surface, from Bernoulli's equation, air flow regions with higher velocity have a lower pressure, and regions with lower velocity have a higher pressure.

The air outside the convertible is moving faster than the air inside the convertible. This leads to a higher pressure zone just below the surface of the roof (inside the car) causing the roof of the convertible to bulge upwards

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2 years ago

When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa

MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

$\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2$

In parallel

$\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}$

$\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0$

Solving the above quadratic equation

$\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}$

$\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega$

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0

2 years ago

Determine the sign (+ or −) of the torque about the elbow caused by the biceps, τbiceps, the sign of the weight of the forearm,

Alex Ar [27]

Ans:
1. τbiceps = +(Positive)
2. τforearm = -(Negative)
3. τball = -(Negative)

Explanation:

The figure is attached down below.

1. T<span>orque about the elbow caused by the biceps, τbiceps:
Since Torque = r x F (where r and F are the vectors)
</span>Where r is the vector from elbow to the biceps.
<span>
We can see in the figure that F(biceps) is in upward direction, and by applying the right hand rule from r to F, we get the counterclockwise direction. The torque in counterclockwise direction is positive(+). Therefore, the sign would be +.

2. </span>Torque about the the weight of the forearm, τforearm:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the forearm.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(forearm) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

3. Torque about the the weight of the ball, τball:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the ball.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(ball) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

8 0

2 years ago