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pashok25 [27]
3 years ago
5

A spring scale hung from the ceiling stretches by 6.1cm when a 2.0kg mass is hung from it. The 2.0kg mass is removed and replace

d with a 2.8kg mass.What is the stretch of the spring?
Physics
1 answer:
Anarel [89]3 years ago
8 0
2.0kg and the mass of sometimes makes it hard 2.8kg
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A 2.03 kg book is placed on a flat desk. Suppose the coefficient of static friction between the book and the desk is 0.602 and t
diamong [38]

Answer:

11.98 N

Explanation:

Normal force =   mg =  2.03 * 9.81

coeff of static friction must be overcome for the book to begin moving

       .602 = F / (2.03 * 9.81)   = 11.98  N

5 0
2 years ago
A runner of mass 80 kg is moving at 8.0 m/s. Calculate her kinetic energy. ​
lana [24]

Answer:

2560J

Explanation:

By definition the kinetic energy can be calculated in the following way:

K = (mv²)/2 = 80kg·(8.0m/s)²/2 = 2560 J

7 0
2 years ago
If Justin races his Chevy S-10 down highway 37 north for 2,560 meters in 60 seconds, what is
Harrizon [31]

Answer:

A.) 42.7 m/s

B.) 0.33 m/s^2

C.) 90 kg

Explanation:

A.) If Justin races his Chevy S-10 down highway 37 north for 2,560 meters in 60 seconds, what is his velocity? 

Velocity = displacement/time

Velocity = 2560/60

Velocity = 42.67 m/s

B.) The Chevy S-10 started rounding at 10 meters per hour. What is the acceleration at 30 seconds on the highway?

Acceleration = velocity/time

Acceleration = 10/30

Acceleration = 0.33 m/s^2

C.) The S-10 has a force of 30 N. What is the mass of the car?

Force = mass × acceleration

30 = mass × 0.33

Mass = 30/ 0.33

Mass = 90 kg

8 0
3 years ago
Where can radiation be found in nature and how is it affected
Murljashka [212]
The sun is a clear example of objects releasing radiation in nature
8 0
2 years ago
A beam of light, with a wavelength of 4170 Å, is shined on sodium, which has a work function (binding energy) of 4.41 × 10 –19 J
Lyrx [107]

Explanation:

Given that,

Wavelength of the light, \lambda=4170\ A=4170\times 10^{-10}\ m

Work function of sodium, W_o=4.41\times 10^{-19}\ J

The kinetic energy of the ejected electron in terms of work function is given by :

KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J

The formula of kinetic energy is given by :

KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s

Hence, this is the required solution.

7 0
2 years ago
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