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3 years ago

What two units of air pressure are commonly used in weather reports?

Physics

1 answer:

konstantin123 [22]3 years ago

7 0

The two units are Inches of Mercury and Millibars

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Two astronauts are floating together with zero speed in a gravity-free region of space. The mass of astronaut A is 110 kg and th

Kobotan [32]

Answer:

The recoil speed of Astronaut A is 0.26 m/s.

Explanation:

Given that,

Mass of astronaut A, $m_A=110\ kg$

Mass of astronaut B, $m_B=74\ kg$

Astronaut A pushes B away, with B attaining a final speed of 0.4, $v_B=0.4\ m/s$

We need to find the recoil speed of astronaut A. The momentum remains conserved here. Using the law of conservation of linear momentum as :

$m_Av_A=m_Bv_B\\\\v_A=\dfrac{m_Bv_B}{m_A}\\\\v_A=\dfrac{74\times 0.4}{110}\\\\v_A=0.26\ m/s$

So, the recoil speed of Astronaut A is 0.26 m/s.

4 0

3 years ago

Explain why your PE and KE are usually not both high at the same time (If PE is high then usually KE is low)

levacccp [35]

Answer:

An object can have both kinetic and potential energy at the same time. ... As an object falls its potential energy decreases, while its kinetic energy increases. The decrease in potential energy is exactly equal to the increase in kinetic energy. Another important concept is work.

Explanation:

5 0

3 years ago

Read 2 more answers

a circuit contains 2 ohm and 4 ohm resistors connected in parallel, find the total resistance & the amount of current in eac

Lynna [10]

Answer:

Explanation:

Resistance

R = R1 * R2 / (R1 + R2)

R1 = 2

R2 = 4

R = 2*4 / (2 + 4)

R = 8 / 6

R = 1.3333

Current R1

R1 = 2 ohms

V = 12 volts

I = ?

V = I * R

12 = I * 2

12/2 = I

I = 6 amperes.

Current R2

V = 12 volts

R = 4 ohms

I = ?

V = I * R

12 = I * 4

12/4 = I

I = 3 amps

7 0

3 years ago

A 20 kg wagon is pulled along the level ground by a rope inclined at 30 degree above the horizontal. A friction force of 30 N op

Elan Coil [88]

(a) 34.6 N

To solve the problem, we have to analyze the forces acting along the horizontal direction.

We have:

- Forward: the component of the pull parallel to the ground, which is

$F cos \theta$

where

F is the magnitude of the pull

$\theta=30^{\circ}$ is the angle

- Backward: the force of friction, which is

$F_f = 30 N$

So, the equation of motion is

$F cos \theta - F_f = ma$

where

m = 20 kg is the mass of the wagon

a is the acceleration

In this part, the wagon is moving at constant speed, so a =0 and the equation becomes

$F cos \theta - F_f = 0$

Therefore, we can find the pulling force:

$F=\frac{F_f}{cos \theta}=\frac{30}{cos 30}=34.6 N$

(b) 43.9 N

In this case, the acceleration is

$a=0.40 m/s^2$

So, the equation of motion in this case is

$F cos \theta - F_f = ma$

So this time we have to take into account the term (ma).

Using the same data as before:

m = 20 kg

$\theta=30^{\circ}$

$F_f = 30 N$

We find the new magnitude of F:

$F=\frac{ma+F_f}{cos \theta}=\frac{(20)(0.40)+30}{cos 30}=43.9 N$

6 0

3 years ago

The gasoline in a car does 40,000 J of work on a car and generates a constant force of 20 N. How far did the car go?

AnnyKZ [126]

L=F•d=>d=L/F=40,000/20=2,000 m

7 0

3 years ago