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kakasveta [241]
2 years ago
9

If a helicopter's mass is 4,500kg and the net force on it is 18,000 N upward, what is its acceleration?

Physics
2 answers:
melamori03 [73]2 years ago
5 0

Answer:

Acceleration, a=4\ m/s^2

Explanation:

It is given that,

Mass of the helicopter, m = 4500 kg

Net force acting on the helicopter, F = 18000 N

Let a is the acceleration of the helicopter. The force acting on an object is equal to the product of mass and acceleration. It is calculated as :

F=ma

a=\dfrac{F}{m}

a=\dfrac{18000\ N}{4500\ kg}

a=4\ m/s^2

So, the acceleration of the helicopter is 4\ m/s^2. Hence, this is the required solution.

Vinvika [58]2 years ago
4 0
The acceleration is0.25m/s^2
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Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long. How long is the ot
s344n2d4d5 [400]

Answer:

The length of the longer pipe is L = 2.30 m

Explanation:

Given that:

Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long.

How long is the other pipe?

From above;

The formula for the frequency of open ended pipes can be expressed as:

f = \dfrac{nv}{2L}

where n = 1 ( since half wavelength exist between those two pipes)

v = 343 m/s  and L = 2.08 m

Thus, the shorter pipe produces a frequency of :

f = \dfrac{1*343}{2*2.08}

f = \dfrac{343}{4.16}

f =82.45 \ Hz

Also; we know that the beat frequency was given as 8.0 Hz

Then,

The lower frequency of the longer pipe = ( 82.45 - 8.0 )Hz

The lower frequency of the longer pipe = 74.45 Hz

Finally;

From the above equation; make Length L the subject of the formula. Then,

The length of the longer pipe is L = \dfrac{nv}{2f}

The length of the longer pipe is L = \dfrac{1*343}{2*74.45}

The length of the longer pipe is L = \dfrac{343}{148.9}

The length of the longer pipe is L = 2.30 m

6 0
2 years ago
Suppose we replace both hover pucks with pucks that are the same size as the originals but twice as massive. otherwise, we keep
kvv77 [185]

This pair of pucks will rotate at the same rate.

Further Explanation:

Hover Puck:  

The Hover Puck coasts effectively on a self-produced pad of air until followed up on by an outside power.  

Employments of hover puck:  

The Hover Puck coasts effectively on a self-produced pad of air until followed up on by an outside power.  

• Use one Hover Puck to show themes from Newton's First Law to impacts to reflection.  

• Use at least two pucks to examine the protection laws in two measurements.  

• Set up a bowling alley utilizing plastic beverage bottles.  

Innovation of hover puck:  

The First Rubber Hockey Pucks Were Made From Sliced-Up Lacrosse Balls. At the point when the game moved inside, entire balls were initially utilized, yet arena proprietors before long thought that it was desirable over cut them into thirds and keep the center segment. This fundamental plan was the standard by 1885.  

hockey puck was developed:  

The hockey puck appeared in 1875. It's hazy who really imagined it. Specialists accept the main hockey puck was likely only an elastic ball cut down the middle. This gave players an article with a level side that would slide over the ice.

Subject: physics

Level: High School

Keywords: Hover Puck, Employments of hover puck, Innovation of hover puck, hockey puck was developed.  

Related links:  

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brainly.com/question/727976

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5 0
3 years ago
Read 2 more answers
Assuming the starting height is 0.0 m, calculate the potential energy of the cart after it has been elevated to a height of 0.5
Bogdan [553]
The potential energy is most often referred to as the "energy at rest" and is dependent on the elevation of an object. This can be calculated through the equation,

     E = mgh

where E is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. In this item, we are not given with the mass of the cart so we assume it to be m. The force is therefore,

   E = m(9.8 m/s²)(0.5 m) = 4.9m

Hence, the potential energy is equal to 4.9m.
8 0
2 years ago
A slender rod is 80.0 cm long and has mass 0.370 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05
nataly862011 [7]

Answer:

1.10 m/s

Explanation:

Linear speed is given by

v=r\omega

Kinetic energy is given by

KE=0.5I\omega^{2}

Potential energy

PE= mgh

From the law of conservation of energy, KE=PE hence

0.5I\omega^{2}=mgh where m is mass, I is moment of inertia, \omega is angular velocity, g is acceleration due to gravity and h is height

Substituting m2-m1 for m and 0.5l for h, \frac {2v}{L} for \omega we obtain

0.5I(\frac {2v}{L})^{2}=0.5Lg(m2-m1)

(\frac {2v}{L})^{2}=\frac {gl(m2-m1)}{I} and making v the subject

v^{2}=\frac {gl^{3}(m2-m1)}{4I}

v=\sqrt {\frac {gl^{3}(m2-m1)}{4I}}

For the rod, moment of inertia I=\frac {ML^{2}}{12} and for sphere I=MR^{2} hence substituting 0.5L for R then I=M(0.5L)^{2}

For the sphere on the left hand side, moment of inertia I

I=m1(0.5L)^{2} while for the sphere on right hand side, I=m2(0.5L)^{2}

The total moment of inertia is therefore given by adding

I=\frac {ML^{2}}{12}+ m1(0.5L)^{2}+ m2(0.5L)^{2}=\frac {L^{2}(M+3m1+3m2)}{12}

Substituting \frac {L^{2}(M+3m1+3m2)}{12} for I in the equation v=\sqrt {\frac {gL^{3}(m2-m1)}{4I}}

Then we obtain

v=\sqrt {\frac {gL^{3}(m2-m1)}{4(\frac {L^{2}(M+3m1+3m2)}{12})}}=\sqrt {\frac {3gL^{3}(m2-m1)}{L^{2}(M+3m1+3m2)}}

This is the expression of linear speed. Substituting values given we get

v=\sqrt {\frac {3*9.81*0.8^{3}(0.05-0.02)}{0.8^{2}(0.39+3(0.02)+3(0.05))}} \approx 1.08 m/s

8 0
2 years ago
Stu wanted to calculate the resistance of a light bulb connected to a 4.0 V battery, with a resulting current of 0.5 A. He used
Dimas [21]
His answer was incorrect because according to ohm's law the formula used should have been R=V/I instead of multiplying and the answer should be 8ohms
4 0
2 years ago
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