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n200080 [17]
3 years ago
14

Use the simulation to compare the masses of the three colored and unlabeled weights of different sizes. To do so, set the spring

constant of both springs to the same value. Hang known weights on the left spring and an unknown weight on the right spring, and compare the two. Use as many known weights as necessary to determine the unknown masses, and then place each into the appropriate mass bins in the ranking task below. A. M<50 g
B. M = 50 g
C. 50 g D. M = 100 g
E. 100 g F. M = 250 g
G. M> 250 g
1. Blue medium sized weight
2. Magenta small sized weight
3. Gold large sized weight
Physics
1 answer:
neonofarm [45]3 years ago
8 0

Answer:

The answer is given as follows,

Explanation:

Gold large-sized weight 100 g < M < 250g

50 g < Magenta small-sized weight < 100g

100g < Blue medium-sized weight < 250g

Hence,

100g < Blue medium-sized weight < 250g

50 g < Magenta small-sized weight < 100g

100 g < Gold large-sized weight < 250g.

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Musya8 [376]

Answer:

1.) U = 39.2 m/s

2.) t = 4s

Explanation: Given that the

height H = 78.4m

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Let's assume that the maximum height = 78.4m. And at maximum height, final velocity V = 0

Velocity of projections can be achieved by using the formula

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0 = U^2 - 2 × 9.8 × 78.4

U^2 = 1536.64

U = sqrt( 1536.64 )

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The time it takes to reach its highest point can be calculated by using the formula;

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7 0
4 years ago
A 2kg block of which material would require 450 joules of thermal energy to increase its temperature by 1 degree Celsius?
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The block is made of A) Tin, as its specific heat capacity is 0.225 J/(g^{\circ}C)

Explanation:

When an amount of energy Q is supplied to a sample of material of mass m, the temperature of the material increases by \Delta T, according to the following equation :

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where  C_s is the specific heat capacity of the material.

In this problem, we have:

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Q = 450 J is the amount of energy supplied to the block

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Solving the equation for C_s, we can find the specific heat capacity of the unknown sample:

C_s = \frac{Q}{m \Delta T}=\frac{450}{(2000)(1)}=0.225 J/(g^{\circ}C)

And by comparing with tabular values, we can find that this value is approximately the specific heat capacity of tin.

Learn more about specific heat capacity:

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