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3 years ago

The missile s engine exerts a 260-kN force F.

Physics

1 answer:

alexdok [17]3 years ago

6 0

Answer:

a)F=(208 i+156 j) kN

b)F+W=219 kN

Explanation:

Solution:

To determine the components of F

F=(260 kN).( $\frac{4}{\sqrt{4^{2}+3^{2}}}i+\frac{3}{\sqrt{4^{2}+3^{2}}}j$ )

=(208 i+156 j) kN

F=(208 i+156 j) kN

b)weight W expressed in component and add into the

Force F.

W=-(8800 kg).(9.81 m/s^2)j

F+W=(208 i+[156-86.3]j) kN

F+W= $\sqrt{(208kN)^{2}+(-69.7kN)^{2} }$

=219 kN

F+W=219 kN

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A vector is parallel to the y axis, what is it x component?

slavikrds [6]

Answer:

Explanation:

A vector is parallel to the y axis .

Let its magnitude be A . So the vector can be represented as A j .

where i and j are unit vectors in x and y axis direction .

The x component of A j will be dot product of A j with i

The x component of A j = A j . i

= A x 0 [ Since j . i = 0 ]

= 0

4 0

3 years ago

A porter can carry 40 bricks of 10 N load of each. He can carry up to 75m in 40 sec, calculate his power.

alexira [117]

Answer:

750W

Explanation:

40×10= 400N

work done= force × distance

=400 × 75

=30000 J

Power= work done/ time

= 30000 ÷ 40

= 750 W

4 0

1 year ago

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

a) 6738.27 J

b) 61.908 J

c) $\frac{4492.18}{v_{car} ^{2} }$

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = 6738.27 J

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

$(I_{1} +I_{2} )w$

where the subscripts 1 and 2 indicates the values first and second flywheels

$(I_{1} +I_{2} )w$ = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = $Iw^{2}$ = 6.655 x $3.05^{2}$ = 61.908 J

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = $\frac{1}{2}mv_{car} ^{2}$

where m is the mass of the car

$v_{car}$ is the velocity of the car

Equating the energy

2246.09 = $\frac{1}{2}mv_{car} ^{2}$

making m the subject of the formula

mass of the car m = $\frac{4492.18}{v_{car} ^{2} }$

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3 years ago

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katen-ka-za [31]

so the 1st on is the one on the left, middle is right and the 3rd one is the right one

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3 years ago

Read 2 more answers

Science please helppp

Daniel [21]

The red box must way more. Gravitational potential energy is the product of a an objects mass times the acceleration due to gravity (which is constant on earth) times its height. Since the objects are on the same shelf they are at the same height, and since gravitational acceleration is constant as long as we stay on planet earth, then the mass is the only possible thing that could have changed. This means that the red box must weigh more than the blue box.

3 0

3 years ago