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LuckyWell [14K]

3 years ago

14

With the piston head locked in place, will the volume of the gas increase, decrease, or stay the same whenthe piston is placed a

bove the flameThe volume increases.The volume decreases.The volume stays the same

1 answer:

Nuetrik [128]3 years ago

6 0

Answer:

For real gas the volume of a given mass of gas will increase with increase in temperature.

Explanation:

With the piston head locked in place and place above the fire,the volume of the gas will increase,because the volume of a given mass of gas increases with increase temperature.

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Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.

Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and $g=9.8m/s^2$ the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

$\displaystyle d={\frac {v_o^{2}\sin(2\theta )}{g}}$

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

$\displaystyle d={\frac {20^{2}\sin(2\cdot 30^\circ )}{9.8}}$

$\displaystyle d={\frac {400\sin(60^\circ )}{9.8}}$

$d=35.35\ m$

The range is 35.35 m

7 0

3 years ago

A wire with mass 50 g is stretched so that it’s ends are tied down at points 100 cm apart. The wire vibrates in its fundamental

ivolga24 [154]

I think the answer is 2500 N

6 0

3 years ago

Read 2 more answers

A concave mirror has a focal length of 13.5 cm. This mirror forms an image located 37.5 cm in front of the mirror. Find the magn

77julia77 [94]

Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

$\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}$

$\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}$

u = -21.09 cm

The magnification of the mirror is given by :

$m=\dfrac{-v}{u}$

$m=\dfrac{-(-37.5)}{(-21.09)}$

m = -1.77

So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

7 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

An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the origination o

Ann [662]

Answer:

θ = 4.78º

with respect to the vertical or 4.78 to the east - north

Explanation:

This is a velocity compound exercise since it is a vector quantity.

The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed

v_fly² = v_nort² + v_air²

v_nort² = v_fly² + - v_air²

Let's use trigonometry to find the direction of the plane

sin θ = v_air / v_fly

θ = sin⁻¹ (v_air / v_fly)

let's calculate

θ = sin⁻¹ (10/120)

θ = 4.78º

with respect to the vertical or 4.78 to the north-east

5 0

3 years ago