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3 years ago

The mass of _______ makes up 70 percent of all planetary mass in the solar system. A. Uranus B. Neptune C. Jupiter D. Saturn

Physics

2 answers:

yawa3891 [41]3 years ago

4 0

C. Jupiter's mass is the largest

mr_godi [17]3 years ago

3 0

the mass of Jupiter makes up 70 percent of the planetary mass in the solar system because it is the largest plant in the solar system

You might be interested in

State and briefly describe the applications of expansion<br>

zlopas [31]

When solid material expands in response to an increase in temperature (thermal expansion), it can increase in length in a process known as linear expansion. for an example application of expansion and contraction.

examples =
(1) Changing of shape and dimensions of objects such as doors.
(2) Wall collapsing due to bulging.
(3) Cracking of glass tumbler due to heating.
(4) Bursting of metal pipes carrying hot water or steam are some of the disadvantages of thermal expansion of matter.

6 0

2 years ago

A rock is thrown from the top of a 20-m building at an angle of 53Â° above the horizontal. If the horizontal range of the throw

NemiM [27]

Answer:

28.5 m/s

18.22 m/s

Explanation:

h = 20 m, R = 20 m, theta = 53 degree

Let the speed of throwing is u and the speed with which it strikes the ground is v.

Horizontal distance, R = horizontal velocity x time

Let t be the time taken

20 = u Cos 53 x t

u t = 20/0.6 = 33.33 ..... (1)

Now use second equation of motion in vertical direction

h = u Sin 53 t - 1/2 g t^2

20 = 33.33 x 0.8 - 4.9 t^2 (ut = 33.33 from equation 1)

t = 1.17 s

Put in equation (1)

u = 33.33 / 1.17 = 28.5 m/s

Let v be the velocity just before striking the ground

vx = u Cos 53 = 28.5 x 0.6 = 17.15 m/s

vy = uSin 53 - 9.8 x 1.17

vy = 28.5 x 0.8 - 16.66

vy = 6.14 m/s

v^2 = vx^2 + vy^2 = 17.15^2 + 6.14^2

v = 18.22 m/s

6 0

3 years ago

4. I drop a pufferfish of mass 5 kg from a height of 5.5 m onto an upright spring of total length 0.5 m and spring constant 3000

KatRina [158]

Answer:

a) 0.28 m or 28 cm is the minimum height above ground the fish reaches.

b) at the height of 0.484 m height , the pufferfish will eventually come to rest.

c) There exists two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy = 23.72J

Spring potential energy = 0.384 J

Explanation:

Given that :

Mass of the pufferfish m =5kg

initial height of the fish h =5.5m

length of the spring l =0.5m

Spring constant K =3000N/m

Assuming no energy loss to friction, what is the minimum height above the ground that the pufferfish reaches?

Lets assume that the minimum height the fish reaches is = x meters

Now by using the conservation of energy; we realize that :

Initial total energy = final total energy

Gravitational potential energy =

Gravitational potential energy' + Spring potential energy (kinetic energy is zero in both cases)

$mgh = mgx + \frac{1}{2}K(l-x)^2$

Replacing our given values into the above equation; we have :

$(5)(9.8)(5.5) = (5)(9.5)(x) + \frac{1}{2}(3000)(0.5-x)^2$

269.5 = 47.5 x + 1500(0.5 -x )²

269.5 = 47.5 x + 1500(0.25 - x²)

269.5 = 47.5 x + 375 - 1500 x²

269.5 - 375 = 47.5 x - 1500 x²

-105.5 = 47.5 x - 1500 x²

-105.5 + 1500 x² - 47.5 x = 0

1500 x² - 47.5 x - 105.5 = 0

By using quadratic equation and taking the positive value;

x = 0.28 m or 28 cm is the minimum height above ground the fish reaches.

At the equilibrium position the weight of fish will be equal to the force applied by the spring thus

mg = kx

substituting our given values ; we have:

(5)(9.8) = 3000x

x = 61.22

x = 0.016m : so this is the compression in the spring

Now; to determine the height the pufferfish gets to before it eventually come to rest; we have

(0.5-0.016) m = 0.484m

therefore, at the height of 0.484 m height , the pufferfish will eventually come to rest.

There exists two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy = mgh' = (5)(9.8)(0.484)

= 23.72J

and spring potential energy

$=\frac{1}{2}Kx^2\\ = \frac{1}{2}(3000)(0.016)^2\\= 0.384J$

8 0

3 years ago

A wave has a frequency of 35 Hz and a wavelength of 15 meters,<br> what is the speed of the wave?

Mila [183]

Answer:

f lamda = c

Explanation:

525 m/s is the speed

5 0

3 years ago

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

6 0

3 years ago