All categories

1 year ago

Pls solve this numerical.

Physics

1 answer:

harkovskaia [24]1 year ago

4 0

Answer:

You should write the vectors in their proper sequence: i , j, k

If the vectors are perpendicular their dot product must be zero:

1) A dot B is obviously zero since A and B have no common terms in i , j , k

2) looks like A = 2 i - 4 j + 3 k and B = i + 2 j + 2 k

A dot B = (2 * 1) + (-4 * 2) + (3 * 2) = 0

Note that the dot product produces a scalar with no direction (vector components)

Only i dot i, j dot j, and k dot k have components i dot j = zero, etc.

You might be interested in

A sprinter starts from rest and reaches a speed of 15 m/s in 4.25 s. Find his acceleration

kakasveta [241]

Answer:

a=3.53 m/s^2

Explanation:

Vo=0 m/s (because he is not moving at the start)

V1=15 m/s

t= 4.25 s

a = (V1-Vo) / t = 15/4.25 = 3.53 m/s^2

5 0

2 years ago

What name is given to the process that occurs when light passes from air to water?

-Dominant- [34]

Answer:

Refraction

Explanation:

5 0

3 years ago

A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1

balandron [24]

Answer:

The answer is " $143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}$ "

Explanation:

For point a:

Energy balance equation:

$\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\$

$W=0\\\\Q=0\\\\m_e=0$

From the above equation:

$\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\$

because the rate of air entering the tank that is $h_i$ constant.

$\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\$

Since the tank was initially empty and the inlet is constant hence, $m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\$

Interpolate the enthalpy between $T = 300 \ K \ and\ T=295\ K$ . The surrounding air

temperature:

$T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}$

Substituting the value from ideal gas:

$\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}$

Follow the ideal gas table.

The $u_2= 298.33\ \frac{kJ}{kg}$ and between temperature $T =410 \ K \ and\ T=240\ K.$

Interpolate

$\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}$

Substitute values from the table.

$\frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\$

For point b:

Consider the ideal gas equation. therefore, p is pressure, V is the volume, m is mass of gas. $\bar{R} \ is\ \frac{R}{M}$ (M is the molar mass of the gas that is $28.97 \ \frac{kg}{mol}$ and R is gas constant), and T is the temperature.

$n=\frac{pV}{TR}\\\\$

$=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\$

For point c:

Entropy is given by the following formula:

$\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}$

5 0

2 years ago

He chart below compares the characteristics of four objects, A, B, C, and D, discovered in the solar system.

Leto [7]

Answer:

Object A and C (Second choice)

Let's observe one by one

#A..

Kupier belt is a belt associated at Neptune outerside similar to asteroid belt

Hence it's correct

#B.

Has enough gravity to keep other objects far away from its orbit

It's any planet or may be sun /star

Is in orbit around the sun

Sure it's correct

Is almost circular in shape

It's any planet or moons

5 0

2 years ago

A silver tea spoon is placed in a cup filled with hot tea. After some time, the exposed end of the spoon becomes hot even withou

EastWind [94]

Answer:

As atoms in the spoon vibrates about their equilibrium positions and transfer energy form one end to other end. This process is called conduction.

4 0

2 years ago

Pls solve this numerical.​

Pls solve this numerical.