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gayaneshka [121]

2 years ago

12

A parcel of air at 5 km. and an initial temperature of 10°C altitude descends to 3 km altitude at the unsaturated adiabatic laps

e rate. What is the temperature at 3 km?

1 answer:

Alchen [17]2 years ago

7 0

Answer:

30 degrees Celsius

Explanation:

With unsaturated adiabatic lapse rate water vapour does not have to change state from gaseous to liquid.

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A 6.0-kg object moving at 5.0 m/s collides with and sticks to a 2.0-kg object. After the collision the composite object is movin

gogolik [260]

Answer:

a) 23 m/s

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

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

The initial momentum p₀, can be written as follows:

$p_{o} = m_{1} * v_{1o} + m_{2}* v_{2o} = 6.0 kg * 5.0 m/s + 2.0 kg * v_{2o} (2)$

The final momentum pf, can be written as follows:

$p_{f} = (m_{1} + m_{2} )* v_{f} = 8.0 kg* (-2.0 m/s) (3)$

Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:

$v_{2o} = \frac{-6.0kg* 5m/s -8.0 kg*2.0m/s}{2.0kg} = \frac{-46kg*m/s}{2.0kg} = -23.0 m/s (4)$

This means that the 2.0-kg object was moving at 23 m/s in a direction opposite to the 6.0-kg object, so its initial speed, before the collision, was 23.0 m/s.

6 0

3 years ago

In electronics, the formula V = IR is called Ohm's Law. It gives the relationship in a circuit between the voltage V (in volts),

Tomtit [17]

Answer:

3+2i ohm.

Explanation:

Here we have a complex number operation. According to Ohm's law V=I*R, so:

$R=\frac{V}{I}\\\\R=\frac{21+i}{5-3i}$

we need to remove the complex number on the denominator, so we have to multiply by its complement on both side

$R=\frac{21+i}{5-3i}*\frac{5+3i}{5+3i}$

that gives us:

$R=\frac{102+68i}{34}=3+2i$

6 0

3 years ago

An ideal parallel-plate capacitor consists of a set of two parallel plates of area A separated by a very small distance d. When

astraxan [27]

Answer:

option A

Explanation:

given,

area of the plate = A

distance = d

$U = \dfrac{1}{2}CV^2$

where

C is the capacitance

V is the potential difference

The capacitance of the parallel plate capacitor, at the beginning, is given by

$C = \dfrac{\epsilon_0 A}{d}$

where ε₀ is the permittivity of free space,

$U_0 = \dfrac{1}{2}\dfrac{\epsilon_0 A}{d}V^2$

now, the distance is doubled

d' = 2 d

while the potential difference is kept constant. Therefore, we can calculate the new potential energy:

$U = \dfrac{1}{2}\dfrac{\epsilon_0 A}{d'}V^2$

$U = \dfrac{1}{2}\dfrac{\epsilon_0 A}{2d}V^2$

$U = \dfrac{U_0}{2}$

Hence, the correct answer is option A

6 0

3 years ago

Which therapies are associated with ultrasound? Check all that apply.

goldenfox [79]

Answer:

Need a picture

-Malignant tissue

-Over the lower abdomen

-during pregnancy

-Over an infection

Explanation:

IDK if this helps bc there is no picture, but here ya go

3 0

3 years ago

Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.320 m and carries a current

Damm [24]

Answer:

y = 0.108m

Explanation:

The magnitude of the magnetic force is given by:

$B=\frac{\mu_oI}{2\pi r}$ ( 1 )

and the force per unit length is:

$\frac{F}{\Delta L}=\frac{\mu_oI_1I_2}{2\pi d}$ ( 2 )

you first calculate I2 from (2):

$I_2=(\frac{F}{\Delta L})\frac{2\pi d}{\mu_o I_1}=(290*10^{-6}N/m)\frac{2\pi (0.320m)}{(4\pi*10^{-7}m)(30.0A)}=15.46A$

With this values of I2 you can the position in which the magnitude of the magnetic field is zero, by using (1) for both wires:

$\frac{\mu_o}{2\pi}[\frac{I_1}{0.320-r}-\frac{I_2}{r}]=0\\\\I_1r=I_2(0.320-r)\\\\r(I_1+I_2)=0.320I_2\\\\r=\frac{0.320I_2}{I_1+I_2}=\frac{0.320(15.46)}{15.46+30.0}=0.108m$

hence, for y=0.108m the magnitude of B is zero

8 0

3 years ago