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weqwewe [10]
1 year ago
14

How to calculate final speed when the mass is 40,000kg height id 2.5km and 500,000N of force

Physics
1 answer:
Tju [1.3M]1 year ago
8 0

The final speed when the mass is 40,000kg height is 2.5km and 500,000N of force is 176.8 m / s

According to Newton's second law of motion,

F = m a

F = Force

m = Mass

a = Acceleration

m = 40000 kg

F = 500000 N

a = F / m

a = 500000 / 40000

a = 12.5 m / s²

a = v / t

v = d / t

v = Velocity

t = Time

d = Distance

d = 2.5 km = 2500 m

a = d / t²

12.5 = 2500 / t²

t² = 200

t = 14.14 s

v = 2500 / 14.14

v = 176.8 m / s

Therefore, the final speed is 176.8 m / s

To know more about Newton's second law of motion

brainly.com/question/13447525

#SPJ9

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Many counties in Florida missed many school days in the fall of 2004 due to hurricanes that year. A solution for how to make up
Ivenika [448]

Answer:

108 extended days

Explanation:

Regular school hours a day = 6 hr

No. of school days to make up by extending the  regular hours = 3 days

Amount of time added to the regular hours of school = 10 min

No. of extended school days to make up the 3 school days by following the above mentioned criteria be x.

Time of school hours in 3 days = 3\times 6= 18\,hr\,\,\,\, ;or\,\,\, (18\times 60)\,minutes

\therefore x=\frac{18\times 60}{10}

x=108\,\,days are required to make up 3 days of school having 6 hours of regular timing with 10 minutes of add-on time each day.

4 0
3 years ago
Technician A says some compressor service procedures can be performed on the vehicle if space permits. Technician B says modern
stiv31 [10]

Answer:

Technicians A.

Explanation:

Since air compressor uses series of processes that turn incoming ambient air into a power source for tools and machinery. This means that air compressor has many different parts, and each of these parts must be maintained to ensure they function properly and optimally.

These are the basis when it comes to servicing a compressor

You need to change its oil

And clean its filters.

Inspected it's filters every three months, and have its filters replaced and connections tightened at least once every year.

To do all these can be performed on the vehicle if there is enough space just as Technician A said for the question context.

5 0
3 years ago
A 2.0-kg laptop sits on the horizontal surface of the seat of a car moving at 8.0 m/s. The driver starts slowing down to stop. F
ivanzaharov [21]

Answer: 32.65\ m

Explanation:

Given

mass of laptop m=2 kg

The velocity of car u=8 m/s

The coefficient of static friction is \mu_s=0.4

The coefficient of kinetic friction is \mu_k=0.2

As the car is moving, so the coefficient of kinetic friction comes into play

deceleration offered by friction \mu_kg=0.2\times 9.8\ m/s^2

Using the equation of motion v^2-u^2=2as\\

insert the values

0^2-8^2=2(-0.2\times 9.8)s\\\\s=\dfrac{64}{1.96}\\\\s=32.65\ m

4 0
3 years ago
A projectile is launched at an angle above the
gtnhenbr [62]
The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.

The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.

v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s

Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.

The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.

Combining this together we get:
(1) vx=40m/s and vy=10m/s

7 0
3 years ago
A block of mass 12.2 kg is sliding at an initial velocity of 3.9 m/s in the positive x-direction. The surface has a coefficient
Studentka2010 [4]

Answer:

Explanation:

a) Force of friction = μ R where μ is coefficient of kinetic friction and R is reaction force

R = mg where m is mass of the block

Force of friction F = μ x mg

= .173 x 12.2 x 9.8

= 20.68 N

b ) Only force of friction is acting on the body so

deceleration = force / mass = 20.68 / 12.2 = 1.7 m /s²

acceleration = - 1.7 m /s²

c )

v² = u² - 2 a s

v = 0 , u = 3.9 m /s

a = 1.7 m /s

0 = 3.9² - 2 x 1.7 x s

s = 4.47  m

5 0
3 years ago
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