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2 years ago

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For the first 10 seconds a squirrel runs 3 m/s to look for an acorn. The next 5 seconds he eats an acorn that he finds. Afterwar

ds the squirrel runs 2 m/s back to where he started. (A) How long does it take for the squirrel to get back to where he started?

1 answer:

Gala2k [10]2 years ago

7 0

Distance covered by the squirrel to look for an acorn :

d = ( 3 m/s ) × 10 s = 30 m.

Time taken to eat an Acron is 5 seconds.

Time taken to cover distance of 30 m with 2 m/s speed is :

$T=\dfrac{30}{2}\ s= 15 \ s$

Therefore, total time take to get back to where he started is ( 10+5+15 ) = 30 s.

Hence, this is the required solution.

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When a particle is a distance r from the origin, its potential energy function is given by the equation U(r)=kr, where k is a co

Reika [66]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

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2 years ago

A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn

EastWind [94]

(a) 12.0 V

In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, $C=5.00 \mu F$ , the charged stored on the capacitor at the end of the process is

$Q=CV=(5.00 \mu F)(12.0 V)=60 \mu C$

When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

$V=\frac{Q}{C}$

and since neither Q nor C change, the voltage V remains the same, 12.0 V.

(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

$C=\frac{\epsilon_0 \epsilon_r A}{d}$

where:

$\epsilon_0$ is the permittivity of free space

$\epsilon_r$ is the relative permittivity of the material inside the capacitor

A is the area of the plates

d is the separation between the plates

As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: $C'=\frac{C}{2}$ . The new voltage across the plate is given by

$V'=\frac{Q}{C'}$

and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is

$V'=\frac{Q}{C'}=\frac{Q}{C/2}=2 \frac{Q}{C}=2V$

So, the new voltage is

$V'=2 (12.0 V)=24.0 V$

(c) (ii) 3.0 V

The area of each plate of the capacitor is given by:

$A=\pi r^2$

where r is the radius of the plate. In this case, the radius is doubled: r'=2r. Therefore, the new area will be

$A'=\pi (2r)^2 = 4 \pi r^2 = 4A$

While the separation between the plate was unchanged (d); so, the new capacitance will be

$C'=\frac{\epsilon_0 \epsilon_r A'}{d}=4\frac{\epsilon_0 \epsilon_r A}{d}=4C$

So, the capacitance has increased by a factor 4; therefore, the new voltage is

$V'=\frac{Q}{C'}=\frac{Q}{4C}=\frac{1}{4} \frac{Q}{C}=\frac{V}{4}$

which means

$V'=\frac{12.0 V}{4}=3.0 V$

3 0

1 year ago

You are driving downhill on a rural road with a 3% grade at a speed of 45 mph. While playing on the side of the road, a child ac

Dennis_Churaev [7]

Answer: a) 95.07m b) 81.88 m

Explanation:

a)

For finding the distance when vehicle is going downhill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1 + $\frac{(0.28*72.4)^{2} }{2*9.81*(0.31-0.03)}$

Stop sight distance= 95.07 m

b)

For finding the distance when vehicle is going uphill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1 + $\frac{(0.28*72.4)^{2} }{2*9.81*(0.31+0.03)}$

Stop sight distance= 81.88 m

5 0

2 years ago

A 2600-m-high mountain is located on the equator. how much faster does a climber on top of the mountain move than a surfer at a

puteri [66]

The climber move 0.19 m/s faster than surfer on the nearby beach.

Since both the person are on the earth, and moves with the constant angular velocity of earth, however there linear velocity is different.

Number of seconds in a day, t=24*60*60=86400 sec

The linear speed on the beach is calculated as

V1= $\frac{2πr}{t}$

Here, t is the time

Plugging the values in the above equation

V1= $\frac{2π*6.4*10^6}{86400}$ =465.421 m/s

Velocity on the mountain is calculated as

V2= $\frac{2π(r+h)}{t}$

Plugging the values in the above equation

V2= $\frac{2π(6.4*10^6+2600}{86400}$ =465.61 m/s

Therefore person on the mountain moves faster than the person on the beach by 465.61-465.421=0.19 m/s

5 0

2 years ago

A laser with wavelength d/8 is shining light on a double slit with slit separation 0.500 mm. This results in an interference pat

ipn [44]

Answer:

0.000109375 m

Explanation:

d = Distance between grating = 0.5 mm

m = Order

$\lambda_1=\dfrac{d}{8}$

Minima relation

$m-\dfrac{1}{2}$

For fourth order minima

$dsin\theta=(4-\dfrac{1}{2})\lambda_1$

For second maxima

$dsin\theta=2\lambda_2$

From the two equations we get

$(4-\dfrac{1}{2})\lambda_1=2\lambda_2\\\Rightarrow \lambda_2=\dfrac{(4-\dfrac{1}{2})\lambda_1}{2}\\\Rightarrow \lambda_2=\dfrac{(4-\dfrac{1}{2})\dfrac{0.5\times 10^{-3}}{8}}{2}\\\Rightarrow \lambda_2=0.000109375\ m$

The wavelength is 0.000109375 m

7 0

2 years ago