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2 years ago

Which air mass is responsible for bringing warm, moist air to an area?

Physics

2 answers:

babymother [125]2 years ago

6 0

Air mass C for the area

Vika [28.1K]2 years ago

5 0

Answer:

Pretty sure it is air mass C

Explanation:

It looks like yhe air has warm air where as the other 2 options have cold air.

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HELP PLEASE!! An illustration of a pendulum at 3 positions of a pendulum. The equilibrium position is labeled B. The other two p

Sedbober [7]

Answer:

Both A and C

Explanation:

I just got it correct on Edg

7 0

2 years ago

Read 2 more answers

A mass attached to a spring is displaced from its equilibrium position by 5cm and released. The system then oscillates in simple

aliina [53]

Answer:

T= 1 s

Explanation:

Given that

When x= cm ,T= 1

we know that time period of spring mas system given as

$T=2\pi \sqrt{\dfrac{m}{k}}$

T= Time period

m= mass

k=spring constant

So from above equation we can say that time period of system does not depends on the value of x.

So when x= 10 cm ,still time period will be 1 s.

T= 1 s

5 0

3 years ago

What is the temperature of a 3.72 mm cube (e=0.288) that radiates 56.6 W?

blsea [12.9K]

Answer:

The temperature is 2541.799 K

Explanation:

The formula for black body radiation is given by the relation;

Q = eσAT⁴

Where:

Q = Rate of heat transfer 56.6

σ = Stefan-Boltzman constant = 5.67 × 10⁻⁸ W/(m²·k⁴)

A = Surface area of the cube = 6×(3.72 mm)² = 8.3 × 10⁻⁵ m²

e = emissivity = 0.288

T = Temperature

Therefore, we have;

T⁴ = Q/(e×σ×A) = 56.6/(5.67 × 10⁻⁸ × 8.3 × 10⁻⁵ × 0.288) = 4.174 × 10¹⁴ K⁴

T = 2541.799 K

The temperature = 2541.799 K.

7 0

3 years ago

An incompressible fluid flows steadily through a pipe that has a change in diameter. The fluid speed at a location where the pip

OverLord2011 [107]

Answer:

The value is $v_2 = 5.53 \ m /s$

Explanation:

From the question we are told

The pipe diameter at location 1 is $d = 8.8 \ cm = \frac{8.8 }{10} = 0.88 \ m$

The velocity at location 1 is $v_1 = 2.4 \ m /s$

The diameter at location 2 is $d_2 = 5.80 \ cm = 0.58 \ m$

Generally the area at location 1 is

$A_1 = \pi * \frac{d^2}{ 2}$

=> $A_1 = \pi * \frac{0.88^2}{ 2}$

=> $A_1 = 3.142 * \frac{0.88^2}{ 2}$

=> $A_1 = 1.2166 \ m^2$

Generally the area at location 1 is

$A_2 = \pi * \frac{d_1^2}{ 2}$

=> $A_2 = \pi * \frac{0.58^2}{ 2}$

=> $A_2 = 0.528 \ m^2$

Generally from continuity equation we have that

$A_1 * v_1 = A_2 * v_2$

=> $1.2166 * 2.4 = 0.528 * v_2$

=> $v_2 = 5.53 \ m /s$

3 0

3 years ago

21) A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest

zlopas [31]

Answer:

-22,150 N

Explanation:

When the youngster jumps off the platform, during the fall her initial potential energy is converted into kinetic energy, according to the law of conservation of energy. Therefore, we can write:

$mgh=\frac{1}{2}mu^2$

where the term on the left is the potential energy while the term on the right is the kinetic energy, and where

m = 50.0 kg is the mass of the youngster

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 1.00 m is the heigth of the platform

u is the speed of the youngster as she reaches the floor

Solving for u,

$u=\sqrt{2gh}=\sqrt{2(9.8)(1.00)}=4.43 m/s$

Then, when the youngster hits the floor, the force exerted on her during the deceleration is given by:

$F=\frac{\Delta p}{\Delta t}=\frac{m(v-u)}{\Delta t}$

where $\Delta p$ is her change in momentum, and where

m is the mass

v = 0 is the final velocity (she comes to a stop)

u = 4.43 m/s is the initial velocity

$\Delta t=10.0 ms =0.010 s$ is the duration of the collision

Substituting,

$F=\frac{(50.0)(0-4.43)}{0.010}=-22150 N$

And the negative sign means the direction of the force is opposite to the motion (so, upward).

6 0

3 years ago