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PilotLPTM [1.2K]

3 years ago

13

It takes you 5 min to walk with an average velocity of .75 m/s to the north from the parking lot to the entrance of the amusemen

t park. What is your displacement?

1 answer:

skad [1K]3 years ago

3 0

A displacement is a vector quantity that takes into account the shortest distance from the starting point to the endpoint.

The given above gave a time interval in minutes which needs to be converted to seconds. Given that each minute is 60 seconds, 5 minutes equal 300 seconds. To determine the distance, multiply time with speed. The product is 225 m.

Thus, the displacement is 225 m.

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A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a complete circle of radius R. The centra

Dmitriy789 [7]

Answer:

(a). If z = 0, The electric field due to the rod is zero.

(b). If z = ∞, The electric field due to the rod is $E\propto\dfrac{1}{z^2}$ .

(c). The positive distance is $\dfrac{R}{\sqrt{2}}$

(d). The maximum magnitude of electric field is $1.54\times10^{4}\ N/C$

Explanation:

Given that,

Radius = 2.00 cm

Charge = 4.00 mC

(a). If the radius and charge are R and Q.

We need to calculate the electric field due to the rod

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}$

Where, Q = charge

z = distance

If z = 0,

Then, The electric field is

$E=0$

(b). If z = ∞, z>>R

So, R = 0

Then, the electric field is

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q}{z^2}$

$E\propto\dfrac{1}{z^2}$

(c). In terms of R,

We need to calculate the positive distance

If $E\rightarrow E_{max}$

Then, $\dfrac{dE}{dz}=0$

$\dfrac{Q}{4\pi\epsilon_{0}}(\dfrac{(z^2+R^2)^\frac{3}{2}-\dfrac{3z}{2}(z^2+R^2)^\dfrac{1}{2}}{(z^2+R^2)^2})=0$

Taking only positive distance

$z=\dfrac{R}{\sqrt{2}}$

(d). If R = 2.00 and Q = 4.00 mC

We need to calculate the maximum magnitude of electric field

Using formula of electric field

$E_{max}=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}$

$E_{max}=9\times10^{9}\times\dfrac{4.0\times10^{-6}\times\dfrac{2.00}{\sqrt{2}}}{((\dfrac{2.00}{\sqrt{2}})^2+(2.00)^2)^{\frac{2}{3}}}$

$E_{max}=15418.7\ N/C$

$E_{max}=1.54\times10^{4}\ N/C$

Hence, (a). If z = 0, The electric field due to the rod is zero.

(b). If z = ∞, The electric field due to the rod is $E\propto\dfrac{1}{z^2}$ .

(c). The positive distance is $\dfrac{R}{\sqrt{2}}$

(d). The maximum magnitude of electric field is $1.54\times10^{4}\ N/C$

6 0

3 years ago

regrine falcons frequently grab prey birds from the air. Sometimes they strike at high enough speeds that the force of the impac

solmaris [256]

Answers:

a) 30 m/s

b) 480 N

Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

<h3>a) Final speed</h3>

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum $p_{i}$ before the collision must be equal to the final momentum $p_{f}$ after the collision:

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=MV_{i}+mU_{i}$

$p_{f}=(M+m) V$

Where:

$M=480 g \frac{1 kg}{1000 g}=0.48 kg$ the mas of the peregrine falcon

$V_{i}=45 m/s$ the initial speed of the falcon

$m=240 g \frac{1 kg}{1000 g}=0.24 kg$ is the mass of the pigeon

$U_{i}=0 m/s$ the initial speed of the pigeon (at rest)

$V$ the final speed of the system falcon-pigeon

Then:

$MV_{i}+mU_{i}=(M+m) V$ (2)

Finding $V$ :

$V=\frac{MV_{i}}{M+m}$ (3)

$V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg}$ (4)

$V=30 m/s$ (5) This is the final speed

<h3>b) Force on the pigeon</h3>

In this part we will use the following equation:

$F=\frac{\Delta p}{\Delta t}$ (6)

Where:

$F$ is the force exerted on the pigeon

$\Delta t=0.015 s$ is the time

$\Delta p$ is the pigeon's change in momentum

Then:

$\Delta p=p_{f}-p_{i}=mV-mU_{i}$ (7)

$\Delta p=mV$ (8) Since $U_{i}=0$

Substituting (8) in (6):

$F=\frac{mV}{\Delta t}$ (9)

$F=\frac{(0.24 kg)(30 m/s)}{0.015 s}$ (10)

Finally:

$F=480 N$

7 0

3 years ago

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0

3 years ago

Which of the following is an example or an orginasm mantaining homeostasis

Kryger [21]

a cat grooming itself

8 0

3 years ago

What does the model below represent?

Viefleur [7K]

Answer:

D

Explanation:

Because we have two reactants and product

5 0

2 years ago