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Natasha_Volkova [10]
3 years ago
12

Answer question and I will give brainliest

Physics
1 answer:
8090 [49]3 years ago
8 0
D. mechanical energy is transformed into sound energy.
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At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81m/s2 . A watermelon has a weight of 58.0N at the sur
kvasek [131]

Answer: A) mass on earth surface = 5.91kg

B) mass on surface of jupiter = 5.91kg

C) weight on surface of jupiter = 10.697N

Explanation:

The relationship between weight (W), mass (m) and acceleration due gravity (g) is given below

W=mg

From the question, g= 9.8m/s² and weight on the surface on the earth is 58N

A) The mass of watermelon on earth is

m = 58/ 9.8 = 5.91kg

B) the mass of the watermelon on jupiter is 5.91kg.

You will notice this is the same as the mass of watermelon on earth and that is so because mass is a scalar quantity that does not depends on the distance away from the center of the earth (unlike weight which is a vector) thus making it constant all through any location.

C) mass of watermelon is 5.91kg, g=9.8m/s² weight of watermelon on jupiter is given below as

W = mg

W = 5.91 x 9.8

= 10.697N.

6 0
2 years ago
Please help me guys never mind the calculations ​
vlada-n [284]

The shape is connected in parallel so;

5.1) Ans;

\frac{1}{R}  =  \frac{1}{R1} +  \frac{1}{R2}   \\  \frac{1}{R}  =  \frac{1}{2}  +  \frac{1}{3}  \\  \frac{1}{R}  =  \frac{3 + 2}{6}  =  \frac{5}{6}  \\ R =  \frac{6}{5}  = 1.2 \:  \: ohm

5.2) Ans;

\frac{1}{R}  =  \frac{1}{R1} +  \frac{1}{R2}   \\  \frac{1}{R}  =  \frac{1}{8}  +  \frac{1}{10}  \\  \frac{1}{R}  =  \frac{5 + 4}{40}  =  \frac{9}{40}  \\ R =  \frac{40}{9}  = 4.4 \:  \: ohm

I hope I helped you^_^

7 0
2 years ago
Look at the graph
viva [34]
Here, Carefully look at the graph.
When it is on x=10, it is approximately 10, (slightly less than 10)
Closest value would be 90, so y/x = 90/10 = 9 

So, the density of the graph would be 9 g/cm³

In short, Your Answer would be Option D

Hope this helps!
4 0
2 years ago
Read 2 more answers
Amy throws a softball through the air. What are the different forces acting on the ball while it’s in the air?
aniked [119]
An applied force<span> is a </span>force<span> that is </span>applied<span> to an object by a person or another object.
An attractive force is a force of an attraction (where object are attracted by each other). Gravitation is an example of attractive force.
</span>Normal force<span> is the component, perpendicular to the surface (surface being a plane) of contact.
</span><span>The softball experiences an applied force as a result of Amy’s throw. As the ball moves, it experiences attractive force from the air it passes through. It also experiences a downward pull because of the normal force.
Solution A.</span>
4 0
3 years ago
Read 2 more answers
Suppose that you observe light emitted from a distant star to be at a wavelength of 525 nm. The wavelength of light to an observ
Tomtit [17]

Answer:

The velocity of the star is 0.532 c.

Explanation:

Given that,

Wavelength of observer = 525 nm

Wave length of source = 950 nm

We need to calculate the velocity

If the direction is from observer to star.

From Doppler effect

\lambda_{0}=\sqrt{\dfrac{c+v}{c-v}}\times\lambda_{s}

Put the value into the formula

525=\sqrt{\dfrac{c+v}{c-v}}\times950

\dfrac{c+v}{c-v}=(\dfrac{525}{950})^2

\dfrac{c+v}{c-v}=0.305

c+v=0.305\times(c-v)

v(1+0.305)=c(0.305-1)

v=\dfrac{0.305-1}{1+0.305}c

v=−0.532c

Negative sign shows the star is moving toward the observer.

Hence, The velocity of the star is 0.532 c.

7 0
3 years ago
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