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1 year ago

Explain how the forces need to change so the aeroplane can land

1 answer:

7 0

When an airplane is flying straight and level at a constant speed, the lift it produces balances its weight, and the thrust it produces balances its drag. However, this balance of forces changes as the airplane rises and descends, as it speeds up and slows down, and as it turns.

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A pendulum is used in a large clock. The pendulum has a mass of 2kg. If the pendulum is moving at a speed of 2.9 m/s when it rea

Vanyuwa [196]

You first us 1/2(mv^2) to solve for the potential energy and then put that in to PE=m*g*h and solve for hight

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1 year ago

Read 2 more answers

A 5.0 kg cannonball is dropped from the top of a tower. It falls for 1.6 seconds before slamming into a sand pile at the base of

stepan [7]

Answer:

15.7 m/s

Explanation:

The motion of the cannonball is a accelerated motion with constant acceleration g = 9.8 m/s^2 towards the ground (gravitational acceleration). Therefore, the velocity of the ball at time t is given by:

$v(t)=u + gt$

where

u = 0 is the initial velocity

g = 9.8 m/s^2 is the acceleration

t is the time

If we substitute t=1.6 s into the equation, we find the final velocity of the cannonball:

$v(1.6 s)=0+(9.8 m/s^2)(1.6 s)=15.7 m/s$

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2 years ago

Block A with a mass of 10 kg rests on a 30 degree incline. the coefficient of kinetic friction is 0.20. theattatched string is p

Tomtit [17]

Answer:

Please find attached

Explanation:

6 0

1 year ago

A particular planet has a moment of inertia of 9.74 × 1037 kg ⋅ m2 and a mass of 5.98 × 1024 kg. Based on these values, what is

malfutka [58]

Answer: A) $6.38(10)^{6} m$

Explanation:

The equation for the moment of inertia $I$ of a sphere is:

$I=\frac{2}{5}mr^{2}$ (1)

Where:

$I=9.74(10)^{37}kg m^{2}$ is the moment of inertia of the planet (assumed with the shape of a sphere)

$m=5.98(10)^{24}kg$ is the mass of the planet

$r$ is the radius of the planet

Isolating $r$ from (1):

$r=\sqrt{\frac{5I}{2m}}$ (2)

Solving:

$r=\sqrt{\frac{5(9.74(10)^{37}kg m^{2})}{2(5.98(10)^{24}kg)}}$ (3)

Finally:

$r=6381149.077m \approx 6.38(10)^{6} m$

Therefore, the correct option is A.

4 0

1 year ago

A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It rolls down the ramp without slipping. The

Oksi-84 [34.3K]

Answer:

The horizontal distance d does the ball travel before landing is 1.72 m.

Explanation:

Given that,

Height of ramp $h_{1}=2.30\ m$

Height of bottom of ramp $h_{2}=1.69\ m$

Diameter = 0.17 m

Suppose we need to calculate the horizontal distance d does the ball travel before landing?

We need to calculate the time

Using equation of motion

$h_{2}=ut+\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2h_{2}}{g}}$

$t=\sqrt{\dfrac{2\times1.69}{9.8}}$

$t=0.587\ sec$

We need to calculate the velocity of the ball

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2$

$K.E=\dfrac{7}{10}mv^2$

Using conservation of energy

$K.E=mg(h_{1}-h_{2})$

$\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})$

$v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})$

Put the value into the formula

$v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}$

$v=2.922\ m/s$

We need to calculate the horizontal distance d does the ball travel before landing

Using formula of distance

$d =vt$

Where. d = distance

t = time

v = velocity

Put the value into the formula

$d=2.922\times 0.587$

$d=1.72\ m$

Hence, The horizontal distance d does the ball travel before landing is 1.72 m.

8 0

1 year ago