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Aneli [31]
3 years ago
11

Science help answer all

Physics
1 answer:
wlad13 [49]3 years ago
4 0
Idk but i hope you figure it out :)
You might be interested in
A spring on a horizontal surface can be stretched and held 0.2 m from its equilibrium position with a force of 16 N. a. How much
nekit [7.7K]

Answer:

a   W_{3.5} = 490 \  J

b  W_{2.5} =  250 \  J

Explanation:

Generally the force constant is mathematically represented as

       k  = \frac{F}{x}

substituting values given in the question

=>   k  = \frac{16}{0.2}

=>   k  =  80 \ N /m

Generally the workdone  in stretching the spring 3.5 m is mathematically represented as

       W_{3.5} =  \frac{1}{2}  *  k  *  (3.5)^2

=>     W_{3.5} =  \frac{1}{2}  *  80  *  (3.5)^2

=>    W_{3.5} = 490 \  J

Generally the workdone  in compressing the spring 2.5 m is mathematically represented as

        W_{2.5} =  \frac{1}{2}  *  k  *  (2.5)^2

=>      W_{2.5} =  \frac{1}{2}  *  80 *  (2.5)^2

=>       W_{2.5} =  250 \  J

5 0
2 years ago
The density of atmosphere (measured in kilograms/meter3) on a certain planet is found to decrease as altitude increases (as meas
alexgriva [62]

Answer:

B.  inverse plot, 0.51 kilograms/meter3

Explanation:

First of all, we note that the relationship between the altitude and the atmospheric density is an inverse relationship. In fact, an inverse relationship is a relationship between the x-variable and the y-variable of the form

y \propto \frac{1}{x}

Therefore, as the x increases, the y decreases, and as the x decreases, they increases. This is exactly what occurs with the altitude and the atmospheric density in this plot: as the altitude increases, the density decreases, and vice-versa.

Moreover, we can infer the value of the atmospheric density at an altitude of 1,291 km. This point is located between point A (2550 km) and point B(1000 km), so the density must have a value between 0.30 kg/m^3 and 0.54 kg/m^3, so the correct choice is

B.  inverse plot, 0.51 kilograms/meter3


5 0
2 years ago
Superman is flying 54.5 m/s when he sees
Nady [450]

348.34 m/s. When Superman reaches the train, his final velocity will be 348.34 m/s.

To solve this problem, we are going to use the kinematics equations for constant aceleration. The key for this problem are the equations d=v_{0} t+\frac{at^{2} }{2} and v_{f} =v_{0} +at where d is distance, v_{0} is the initial velocity, v_{f} is the final velocity, t is time, and a is aceleration.

Superman's initial velocity is v_{0}=54.5\frac{m}{s}, and he will have to cover a distance d = 850m in a time t = 4.22s. Since we know d, v_{0} and t, we have to find the aceleration a in order to find v_{f}.

From the equation d=v_{0} t+\frac{at^{2} }{2} we have to clear a, getting the equation as follows: a=\frac{2(d-v_{0}t) }{t^{2} }.

Substituting the values:

a=\frac{2(850m-54.5\frac{m}{s}.4.22s) }{(4.22s)^{2}}=69.63\frac{m}{s^{2}}

To find v_{f} we use the equation v_{f} =v_{0} +at.

Substituting the values:

v_{f} =54.5\frac{m}{s} +(69.63\frac{m}{s^{2}}.4.22s)=348.34\frac{m}{s}

5 0
3 years ago
Impact of electricity in the society
Travka [436]

Answer:

<h2>Electricity has many uses in our day to day life. It is used for lighting rooms, working fans and domestic appliances like using electric stoves, A/C and more. All these provide comfort to people. In factories, large machines are worked with the help of electricity.</h2>

5 0
2 years ago
What is the gravitational force on a 70kg that is 6.38x10^6m above the earths surface
NARA [144]

Answer:

171.5 N

Explanation:

The gravitational force on an object due to the Earth is given by

F=mg

where

m is the mass of the object

g is the acceleration due to gravity

The acceleration due to gravity at a certain height h above the Earth is given by

g=\frac{GM}{(R+h)^2}

where:

G is the gravitational constant

M=5.98\cdot 10^{24} kg is the Earth's mass

R=6.37\cdot 10^6 m is the Earth's radius

Here,

h=6.38\cdot 10^6  m

So the acceleration due to gravity is

g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})}{(6.37\cdot 10^6 + 6.38\cdot 10^6)^2}=2.45 m/s^2

We know that the mass of the object is

m = 70 kg

So, the gravitational force on it is

F=mg=(70)(2.45)=171.5 N

5 0
3 years ago
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