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yuradex [85]
3 years ago
6

I WILLL MARK YOUUUUUUU HELPPPP

Physics
2 answers:
Ksivusya [100]3 years ago
3 0

Answer:

10.75 kg

Explanation:

moon = (weight on earth/9.81m/s^2)*1.622m/s^2

Firlakuza [10]3 years ago
3 0

Answer:

weight of the moon = mass × gravity

= 65kg × 1•6

=104N/kg

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An adiabatic nozzle has an inlet area of 1 m^2 and an outlet area of 0.25 m^2. Water enters the nozzle at a rate of 5 m^3/s and
svlad2 [7]

Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed

Q=VA

for he exitt

Q=flow=5m^3/s

A=area=0.25m^2

V=Speed

solving for V

V=\frac{Q}{A} \\V=\frac{5}{0.25} =20m/s

velocity at the exit=20m/s

for entry

V=\frac{5}{1} =5m/s

2.

To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

\frac{P1}{\alpha } +\frac{v1^2}{2g} =\frac{P2} {\alpha } +\frac{v2^2}{2g}

where

P=presure

α=9.810KN/m^3 specific weight for water

V=speed

g=gravity

solving for P1

(\frac{p1}{\alpha } +\frac{V1^2-V2^2}{2g})\alpha  =p2\\(\frac{150}{9.81 } +\frac{5^2-20^2}{2(9.81)})9.81  =p2\\P2=-37.5kPa

the pressure at exit is -37.5kPa

7 0
3 years ago
A student is preparing to bake a cake at home and places several ingredients on the countertop to use. First he preheats the ove
Evgen [1.6K]

Answer:

baking the cake batter

Explanation:

Baking the cake batter will indicate that chemical change has occurred here. What is a chemical change?

  • A chemical change is one in which a new kind of matter is formed.
  • It is usually accompanied by energy either evolution or absorption of energy in form of heat or light or both.
  • The process is irreversible.
  • When the batter bakes, a new substance different from the cake mix is obtainable.
  • We cannot get back the ingredient from this baked cake. It is impossible.
  • This is good indicator of chemical change.
7 0
3 years ago
While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)
ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration=2.40m/s^{2}, c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = 2.40m/s^{2}

time = 12.0 s

initial position = 0

c)

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}

after cancelling the pertinent units, I get that  my answer will be given in meters. So I get:

x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}

which solves to:

x=172.8m

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

a=\frac{V_{f}-V_{0} }{t}

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

at=V_{f}-V_{0}

and finally I can add V_{0} to both sides so I get:

V_{f}=at+V_{0}

and now I can proceed and substitute the known values:

V_{f}=at+V_{0}

V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0

which solves to:

V_{f}=28.8m/s

8 0
3 years ago
Read 2 more answers
7. Calculations.
kotykmax [81]

Answer:

5 ms-2

Explanation:

F = ma

F = 100N

m = 20kg ( you should make sure the unit is kg before you answer the question)

100 = 20a

a = 100÷ 20

a = 5 ms-2

4 0
3 years ago
A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it
natka813 [3]

Answer:

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Explanation:

The rotated angle is given by:

\theta=\omega0*t+1/2*\alpha*t^2

Since this is a quadratic equation it can be solved using:

x=\frac{-b \± \sqrt{b^2-4*a*c}  }{2*a}

Rewriting our equation:

1/2*\alpha*t^2+\omega0*t-\theta=0

t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Since \sqrt{\omega0^2+2*\theta*\alpha} >\omega0 we discard the negative solution.

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

8 0
3 years ago
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