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vodka [1.7K]
3 years ago
14

One end of a 7-cm-long spring is attached to the ceiling. When a 5.4 kg mass is hung from the other end, the spring is stretched

by a length of 4.3 cm. How long is the spring when a 3.3 kg mass is suspended from it in cm?
Physics
1 answer:
mash [69]3 years ago
8 0

Answer:

2.63 cm

Explanation:

Hooke's law gives that the force F is equal to cy where c is spring constant and x is extension

Making c the subject of the formula then

c=\frac {F}{y}

Since F is gm but taking the given mass to be F

c=\frac {5.4 kg}{4.3 cm}=1.2558139534883720930232558139534883720930

By substitution now considering F to be 3.3 kg

y=\frac {3.3 kg}{1.2558139534883720930232558139534883720930}=2.6277777777777 cm\approx 2.63 cm

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For each pair of terms below, write a sentence that summarizes how they are different.
Maksim231197 [3]

Answer:

Thermal expansion ,thermal contraction

Explanation:

They are different because it has different cause and chemical reactions

thermal contraction is a chemical reaction

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2 years ago
If an experiment involves a large volume of liquid a _______ would most likely be used to hold it.
Grace [21]
The answer is d a beacker
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3 years ago
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An object is thrown straight up into the air with an initial speed of 8 m/s, and reaches a greatest height of 15 m before it fal
Ugo [173]

Answer:

The value is T_t =  2.5659 \  s    

Explanation:

From the we are told that  

         The initial speed of the object is  u =  8 \  m/s

         The greatest height it reached is  h  =  15 \  m

Generally from kinematic equation we have that

      v^2 =  u^2 + 2gH

At maximum height v  =  0 m/s

So

      0^2 =  8^2 + 2 *  - 9.8 *  H

=>    H  =  3.27 \  m

Here H is the height from the initial height to the maximum height

So the initial height is mathematically represented as  

      s =  h - H

=>    s =  15 - 3.27

=>    s =  11.73 \  m

Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

            v  =  u + (-g) t

At maximum height v  =  0 m/s

           0 = 8 - 9.8t

=>         t = 0.8163 \  s

Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

       h  =  ut_1 + \frac{1}{2}  gt_1^2

Here the initial velocity is  0 m/s given that its the velocity at maximum height

Also  g is positive because we are moving in the direction of gravity  

So

       15  =  0* t  +  4.9 t^2

=>      t_1  =  1.7496

Generally the total time taken is mathematically represented as

          T_t =  t + t_1

=>        T_t =  0.8163  +   1.7496

=>        T_t =  2.5659 \  s            

 

6 0
3 years ago
For a given Prandtl-Meyer expansion, the upstream Mach number is 3 and the pressure ratio across the wave is P2/P1 = 0.4. Calcul
loris [4]

Answer:

The angle for the forward Mach line is 19.47°

The angle for the rearward Mach line is 5.21°

Explanation:

From table A-1 (Modern Compressible Flow: with historical perspective):

\frac{P_{o} }{P_{1} } =36.73 (M₁ = 3)

If Po₁ = Po₂

\frac{P_{o2} }{P_{2} } =\frac{P_{o1} }{P_{1} } *\frac{P_{1} }{P_{2} } =36.73*\frac{1}{4} =91.825

Table A-1:

\frac{P_{o2} }{P_{2} } =91.825,M_{2} =3.63

Table A-5:

v₁ = 49.76°

μ₁ = 19.47°

v₂ = 60.55°

μ₂ = 16°

θ = 60.55 - 49.76 = 10.79°

The angle for the forward Mach line is:

μ₁ = 19.47°

The angle for the rearward Mach line is:

θr = μ₂ - θ = 16 - 10.79 = 5.21°

3 0
3 years ago
What is the term used when a ball is hit and the batter reaches the following bases safely (without being called out)?
alina1380 [7]

Answer:

Shoukd be a base hit

Explanation:

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6 0
3 years ago
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