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2 years ago

Match the measuring instrument with the appropriate statement.

Physics

1 answer:

yaroslaw [1]2 years ago

7 0

<u>Answer:</u>

1A, 2E, 3C, 4F, 5B, 6D

<u>Explanation:</u>

Tilt meter is a an equipment used to measure very small changes from vertical level. So it is a correct match for blank space in statement A.

Richter Scale is a scale of measurement of earthquake strength, So it is a correct match for blank space in statement E.

Mercalli Intensity Scale measures effect of an earthquake, like damage caused.So it is a correct match for blank space in statement C.

Seismograph is earthquake wave measuring instrument. So it is a correct match for blank space in statement F.

Correlation Spectrometer (C.O.S.P.E.C.) is used to measure sulfur dioxide content in smoke. So it is a correct match for blank space in statement B.

Moment Magnitude Scale is an earthquake measuring scale for great earthquakes. So it is a correct match for blank space in statement D.

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A car of mass 1100kg moves at 24 m/s. What is the braking force needed to bring the car to a halt in 2.0 seconds? N

LenaWriter [7]

13200N

Explanation:

Given parameters:

Mass = 1100kg

Velocity = 24m/s

time = 2s

unknown:

Braking force = ?

Solution:

The braking force is the force needed to stop the car from moving.

Force = ma = $\frac{mv}{t}$

m is the mass of the car

v is the velocity

t is the time taken

Force = $\frac{1100 x 24}{2}$ = 13200N

Learn more:

Force brainly.com/question/4033012

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8 0

2 years ago

A careful photographic survey of Jupiter’s moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to

Y_Kistochka [10]

Answer:

529.15 m/s

Explanation:

h = Maximum height = 70000 m

g = Acceleration due to gravity = 2 m/s²

m = Mass of sulfur

As the potential and kinetic energies are conserved

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 2\times 70000}\\\Rightarrow v=529.15\ m/s$

The speed with which the liquid sulfur left the volcano is 529.15 m/s

7 0

2 years ago

A rectangular coil of dimensions 5.40cm x 8.50cm consists of25 turns of wire. The coil carries a current of 15.0 mA.

Kazeer [188]

Answer:

(a) Magnetic moment will be $17.212\times 10^{-4}A-m^2$

(b) Torque will be $6.024\times 10^{-4}N-m$

Explanation:

We have given dimension of the rectangular 5.4 cm × 8.5 cm

So area of the rectangular coil $A=5.4\times 8.5=45.9cm^2=45.9\times 10^{-4}m^2$

Current is given as $i=15mA=15\times 10^{-3}A$

Number of turns N = 25

(A) We know that magnetic moment is given by $magnetic\ moment=NiA=25\times 45.9\times 10^{-4}\times 15\times 10^{-3}=17.212\times 10^{-4}A-m^2$

(b) Magnetic field is given as B = 0.350 T

We know that torque is given by $\tau =BINA=0.350\times 15\times 10^{-3}\times 25\times 45.9\times 10^{-4}=6.024\times 10^{-4}N-m$

4 0

1 year ago

calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame

kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

$\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2$

0 - p × Area for trachea =

$\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

⇒ $-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}$

$\frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s$

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

3 0

2 years ago

Read 2 more answers

An airplane flying parallel to the ground undergoes two consecutive dis- placements. The first is 75 km 30.0° west of north, and

torisob [31]

Answer: displacement of airplane is 172 km in direction 34.2 degrees East of North

Explanation:

In constructing the two displacements it is noticed that the angle between the 75 km vector and the 155 km vector is a right angle (90 degrees).

Hence if the plane starts out at A, it travels to B, 75 km away, then turns 90 degrees to the right (clockwise) and travels to C, 155 km away from B. Angle ABC is 90 degrees, hence we can use Pythagoras theorem to solve for AC

AC2 = AB2 + BC2 ; AC^2 = 752 + 1552 ; from this we get AC = 172 km (3 significant figures)

Angle BAC = Tan-1(155/75) ; giving angle BAC = 64.2 degrees

Hence AC is in a direction (64.2 - 30) = 34.2 degrees East of North

Therefore the displacement of the airplane is 172 km in a direction 34.2 degrees East of North

5 0

2 years ago