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3 years ago

which of the following instruments is used to measure air pressure? a. thermometer b. barometer c. anemometer d. hygrometer

2 answers:

8 0

The correct answer is B. barometer

Thermometer is used to measure temperature, anemometer is used to measure wind speed, while a hygrometer is used to measure humidity.

Aleks04 [339]3 years ago

5 0

Barometer is used to measure air pressure.

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Why do you think it might be useful to know the distance between a gun fired and it's target?

inysia [295]

Distance plays an incredible factor, especially when taking into consideration especially long distances in terms of shooting a gun. You may be familiar with the Coriolis Effect, and how a bullet's trajectory will change in relation to the rotation of the Earth. The farther the distance, the greater the change in trajectory.

4 0

3 years ago

Read 2 more answers

Name the characteristics of electro magnetic waves that i) increases ii) remains constant in the electromagneticspectrum as one

Paul [167]

As we move from Radiowaves to UV waves in electromagnetic spectrum we know that the wavelength is maximum for radio waves while its frequency is minimum

PART (i)

So as we move from radiowaves to UV region

the Frequency will increase

Part (ii)

As we move from Radiowaves to UV region all the parts must have same speed in same medium

All parts will move with speed of light

so Speed will remain the same

4 0

3 years ago

There is a filing cabinent that is 52 in. tall and 15 in. wide. The center of gravity of the cabinet is right at the center. To

erastova [34]

Answer:

16 degrees

Explanation:

The tipping point of the cabinet is sketched below.

5 0

1 year ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago

A car in an amusement park ride rolls without friction around a track (Fig. P7.42). The hB car starts from rest at point A at a

MrRissso [65]

Answer:

$h>\dfrac{5}{2}R$

Explanation:

Given that

Height = h

Radius = R

From energy conservation

$KE_A+U_A=KE_B+U_B$

At point B

The minimum speed to complete the the circle

$V_B=\sqrt{gR}\ m/s$

So the kinetic energy at point B

$KE_B=\dfrac{1}{2}mV^2$

$KE_B=\dfrac{1}{2}mgR$

$KE_A+U_A=KE_B+U_B$

$0+mgh=\dfrac{1}{2}mgR+2mgR$

Without falling off at the top (point B)

$0+mgh>\dfrac{1}{2}mgR+2mgR$

$mg(h-2R)>\dfrac{1}{2}mgR$

$g(h-2R)>\dfrac{1}{2}gR$

$h>\dfrac{5}{2}R$

6 0

3 years ago