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3 years ago

Which of the following is not a measurement taken by a radiosonde?

Physics

2 answers:

MariettaO [177]3 years ago

5 0

Correct answer choice is :

A) Atmospheric composition

Explanation:

A radiosonde is a battery-powered telemetry device transmitted into the environment normally by a weather balloon that regulates several atmospheric parameters and sends them by radio to a ground receiver. The radiosonde is a small, expandable device unit that is recessed below a six-foot-wide balloon inflated with hydrogen or helium. As the radiosonde stands at about 1,000 feet/minute, sensors on the radiosonde measure forms of pressure, temperature, and related moisture.

AlladinOne [14]3 years ago

4 0

I think the correct answer from the choices listed above is option A. Atmospheric composition is not a measurement taken by a radiosonde. A radiosonde is an instrument used to measure different atmospheric parameters.

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Which diagram best represents the electric field around a negatively charged conducting sphere? (See pic)

dalvyx [7]

The answer is D !!!!!!!

3 0

3 years ago

Oil having a density of 922kg/m^3 floats on water. A rectangular block of wood 3.97 cm high and with a density of 963 kg/m^3 flo

Blizzard [7]

Explanation:

For the equilibrium:

\rho_{wood}gh-\rho_{oil}g(h-x)-\rho_{water}gx=0ρ

wood

gh−ρ

oil

g(h−x)−ρ

water

gx=0

\rho_{wood}h-\rho_{oil}(h-x)-\rho_{water}x=0ρ

wood

h−ρ

oil

(h−x)−ρ

water

x=0

(974)(3.97)-928(3.97-x)-1000x=0(974)(3.97)−928(3.97−x)−1000x=0

x=2.54\ cmx=2.54 cm

3 0

2 years ago

Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f

elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

$W_T+W_F=K_f-K_i$ (1)

Where:

$W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}$

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to $\frac{1}{2}mv^{2}$ , so if the initial velocity $v_i$ is equal to zero, the initial kinetic energy $K_i$ is equal to zero.

Then, replacing the values on the equation and solving for $v_f$ , we get:

$W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}$

$\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f$

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0

3 years ago

Productivity increases when A. inputs increase while outputs remain the same. B. inputs decrease while outputs remain the same.

Blizzard [7]

Answer: Productivity increases when inputs and outputs increase proportionately.

Explanation:

Productivity increases when inputs and outputs increase proportionately. Input has to be directly proportional to output to be productive. This means increase in input to a system must leads to drastic increase in the output. When the output is not balanced with the amount of input, it leads to unproductivity.

Being productive can be business wise or in terms if personal growth and development.

3 0

3 years ago

Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v

wlad13 [49]

1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we call $v_f$ the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
$m_a v_{Ai} + m_b v_{Bi} = (m_A+m_B)v_f$ (1)
where
$m_A=7 kg$ is the mass of ball A
$m_B=2 kg$ is the mass of ball B
$v_{Ai}=6 m/s$ is the initial velocity of ball A
$v_{Bi}=-12 m/s$ is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)

If we solve (1) to find $v_f$ , we find that the final velocity of the balls is
$v_f= \frac{m_Av_{Ai}+m_Bv_{Bi}}{m_A+m_B}= \frac{(7\cdot 6)+(2 \cdot (- 12))}{7+2}= \frac{18}{9}=2 m/s$
and the positive sign means the two balls are going to the right.

2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
$m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}$
$\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2$
where
$v_{fA}$ is the final velocity of ball A
$v_{fB}$ is the final velocity of ball B

If we solve simultaneously the two equations, we find:
$v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s$
$v_{fB}= \frac{v_{Bi}(m_B-m_A)+2m_Av_{Ai}}{m_A+m_B} = \frac{(-12)(2-7)+2(7)(6)}{7+2}= \frac{144}{9}=16 m/s$
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.

3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
$p_i = m_A v_{A} + m_B v_{fB} = (7\cdot 6)+(2 \cdot (-12))=42-24=18 m/s$
and the total momentum after the collision is:
$p_f = m_A v_{A} + m_B v_{fB} = (7\cdot (-2))+(2 \cdot 16)=-14+32=18 m/s$

3 0

3 years ago