Ask question

Ask question

All categories

3 years ago

13

What is the length of the shadow cast on the vertical screen by your 10.0 cm hand if it is held at an angle of θ=30.0∘ above hor

izontal? Express your answer in centimeters to three significant figures. View Available Hint(s)

1 answer:

Alchen [17]3 years ago

5 0

Answer:

The length is $D = 5 \ cm$

Explanation:

From the question we are told that

The length of the hand is $l = 10.0 \ cm$

The angle at the hand is held is $\theta = 30 ^o$

Generally resolving the length the length of the hand to it vertical component we obtain that the length of the shadow on the vertical wall is mathematically evaluated as

$D = l * sin(\theta )$

substituting values

$D = 10 * sin (30)$

$D = 5 \ cm$

You might be interested in

We want to construct a solenoid with a resistance of 4.30 Ω and generate a magnetic field of 3.70 × 10−2 T at its center when ap

marshall27 [118]

Answer with Explanation:

We are given that

Resistance of solenoid,R=4.3 ohm

Magnetic field,B= $3.7\times 10^{-2} T$

Current,I=4.6 A

Diameter of wire,d=0.5 mm= $0.5\times 10^{-3} m$

Radius of wire,r= $\frac{d}{2}=\frac{0.5\times 10^{-3}}{2}=0.25\times 10^{-3} m$

$1mm=10^{-3} m$

Radius of solenoid,r'=1 cm= $1\times 10^{-2} m$

$1 cm=10^{-2} m$

Resistivity of copper, $\rho=1.68\times 10^{-8}\Omega m$

We know that

$R=\frac{\rho l}{A}$

Where $A=\pi r^2$

Using the formula

$4.3=\frac{1.68\times 10^{-8}\times l}{\pi(0.25\times 10^{-3})^2}$

$l=\frac{4.3\times \pi(0.25\times 10^{-3})^2}{1.68\times 10^{-8}}=50.23 m$

Number of turns of wire= $\frac{l}{2\pi r'}$

Number of turns of wire= $\frac{50.26}{2\pi(1\times 10^{-2}}=800$

Hence, the number of turns of the solenoid,N=799

Magnetic field in solenoid,B= $\mu_0 nI$

$3.7\times 10^{-2}=4\pi\times 10^{-7} n\times 4.6$

$n=\frac{3.7\times 10^{-2}}{4\times 3.14\times 10^{-7}\times 4.6}$

$n=6404 turns/m$

$n=\frac{N}{L}$

$L=\frac{N}{n}$

$L=\frac{799}{6404}$

$L=0.125 m=0.125\times 100=12.5 cm$

Length of solenoid=12.5 cm

1m=100 cm

8 0

3 years ago

Why would an egg break immediately when it hits the ground?

Elza [17]

As an egg falls towards the floor, it begins to travel faster and faster. When it slams into the floor, the egg is stopped almost immediately. This force of the floor against the eggshell is too large, so it breaks.

6 0

2 years ago

Read 2 more answers

Suppose the force of gravity on the electron was comparable to the electric force created by the deflection plates. how would th

Alexandra [31]

"F=Vector Sum Of The Two Forces" Is the answer.

3 0

3 years ago

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0

3 years ago

Infer why the output force exerted by a rake must be less than input force?

adell [148]

<h3><u>Answer and Explanation</u>;</h3>

input force refers to the force exerted on a machine, also known as the effort, while the output force is the force machines produce or the Load. The ratio of output force to input force gives the mechanical advantage of a simple machine
<em><u>The output force exerted by the rake must be less than the input force because one has to use force while raking. The force used to move the rake is the input force. </u></em>
<em><u>The rake is not going to be able to convert all of the input force into output force, the force the rake applies to move the leaves, because of friction.</u></em>

5 0

3 years ago