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mario62 [17]
3 years ago
8

A thermometer has its stem marked in millimeters instead of degree celsius. The lower fixed point is 30mm, and the upper fixed p

oint is 180mm. Calculate the temperature im degree celsius when the thermometer reads 45mm. ​
Physics
1 answer:
Masja [62]3 years ago
3 0

Answer:θ = (x/y) x 100.

where x = UFP -LFP -another scale, and y = UFP - LFP for celsius scale.

=> {(45-30)/(180-30)} x 100 = 10.0oC

Explanation:

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A small bead of mass m is constrained to slide without friction inside a circular vertical hoop of radius r which rotates about
Stolb23 [73]

Answer:

Explanation:

The bead is moving on a vertical circular path so it must have a centripetal force towards the centre.

This force is equal to m v² / r

v is velocity of bead and r is radius of the circular path.

The vertical hoop is also rotating about a vertical axis passing through the centre at frequency f so the bead will experience a cetrifugal force due to rotation of the hoop. Its value is

m ω² r . Only at the point o degree and 180 degree , these forces are opposite to each other so at these points , the bead will be in equilibrium .

mv² / r = m ω² r

v² = ω² r²

v = ω r

= 2π f r

= 2 x 3.14 x 2 x 0.22

v = 2.76 m /s

For the bead to rise upto point θ = 90 degree , height achieved is radius R

required velocity = √ 2gR

= √ 2x 9.8x.22

= 2.076 m/s

This velocity is less than the velocity calculated earlier so the bead will be able to ride the required height.

v = 2.76 m/s

5 0
3 years ago
A mass of 148 g stretches a spring 13 cm. The mass is set in motion from its equlibrium position with a downward velocity of 10
Charra [1.4K]

Answer:

u(t)=1.15 \sin (8.68t)cm

0.3619sec

Explanation:

Given that

Mass,m=148 g

Length,L=13 cm

Velocity,u'(0)=10 cm/s

We have to find the position u of the mass at any time t

We know that

\omega_0=\sqrt{\frac{g}{L}}\\\\=\sqrt{\frac{980}{13}}\\\\=8.68 rad/s

Where g=980 cm/s^2

u(t)=Acos8.68 t+Bsin 8.68t

u(0)=0

Substitute the value

A=0\\u'(t)=-8.68Asin8.68t+8.68 Bcos8.86 t

Substitute u'(0)=10

8.68B=10

B=\frac{10}{8.68}=1.15

Substitute the values

u(t)=1.15 \sin (8.68t)cm

Period =T = 2π/8.68

After half period

π/8.68 it returns to equilibruim

π/8.68 = 0.3619sec

6 0
3 years ago
A visitor to the observation deck of a skyscraper manages to drop a penny over the edge. As the penny falls faster, the force du
pentagon [3]
If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
8 0
3 years ago
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kati45 [8]
Im pretty sure its Concave.
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2 years ago
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The loudness of a sound is the waves amplitude <br><br><br><br> True or false
kakasveta [241]

Answer:

true i think

Explanation:

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4 0
2 years ago
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