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3 years ago

Science Will give brainless and 22 points

Physics

2 answers:

fenix001 [56]3 years ago

4 0

Answer:

Mercury 0.39 AU, 36 million miles

57.9 million km

Venus 0.723 AU

67.2 million miles

108.2 million km

Earth 1 AU

93 million miles

149.6 million km

Mars 1.524 AU

141.6 million miles

227.9 million km

Jupiter 5.203 AU

483.6 million miles

778.3 million km

Saturn 9.539 AU

886.7 million miles

1,427.0 million km

Uranus 19.18 AU

1,784.0 million miles

2,871.0 million km

Neptune 30.06 AU

2,794.4 million miles

4,497.1 million km

Pluto (a dwarf planet) 39.53 AU

3,674.5 million miles

5,913 million km

otez555 [7]3 years ago

3 0

Answer:

Mercury 0.39 AU, 36 million miles 57.9 million km

Venus 0.723 AU 67.2 million miles 108.2 million km

Earth 1 AU 93 million miles 149.6 million km

Mars 1.524 AU 141.6 million miles 227.9 million km

Jupiter 5.203 AU 483.6 million miles 778.3 million km

Saturn 9.539 AU 886.7 million miles 1,427.0 million km

Uranus 19.18 AU 1,784.0 million miles 2,871.0 million km

Neptune 30.06 AU 2,794.4 million miles 4,497.1 million km

Pluto (a dwarf planet) 39.53 AU 3,674.5 million miles 5,913 million km

Explanation:

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A rocket is fired vertically upwards with initial velocity 92 m/s at the ground level. Its engines then fire and it is accelerat

Readme [11.4K]

Answer:

The rocket above the ground is in 44 sec.

Explanation:

Given that,

Initial velocity = 92 m/s

Acceleration = 4 m/s²

Altitude = 1200 m

Suppose, How long was the rocket above the ground?

We need to calculate the time

Using equation of motion

$s=ut-\dfrac{1}{2}at^2$

Put the value into the formula

$1200=92t+\dfrac{1}{2}\times4t^2$

$2t^2+92-1200=0$

$t=10\ sec$

We need to calculate the velocity

Using equation of motion

$v=u+at$

Put the value into the formula

$v=92+4\times10$

$v=132\ m/s$

When the rocket hits the ground,

Then, h'=0

We need to calculate the time

Using equation of motion

$h'=h+ut-\dfrac{1}{2}at^2$

Put the value into the formula

$0=1200+132t-\dfrac{1}{2}\times9.8t^2$

$4.9t^2-132t-1200=0$

$t=34\ sec$

When the rocket is in the air it is the sum of the time when it reaches 1000 m and the time when it hits the ground

So, the total time will be

$t'=34+10$

$t'=44\ sec$

Hence, The rocket above the ground is in 44 sec.

7 0

4 years ago

Air friction damps a tuning fork so the amplitude decreases by 1/10 per second. Resonances] 10 Hz. what % chnge in frequency res

aleksandr82 [10.1K]

Answer:

The percentage change in frequency is 10%.

Explanation:

Given that,

Amplitude decreases by 1/10 per second.

Resonance = 10 Hz

We need to calculate the change in frequency

Using formula of change in frequency

$\Delta f=\Delta A\times R$

Put the value into the formula

$\Delta f=\dfrac{1}{10}\times10$

$\Delta f=1\ Hz$

We need to calculate the resonant frequency

Using formula of resonant frequency

$f_{r}=R-\Delta f$

Put the value into the formula

$f_{r}=10-1$

$f_{r}=9\ Hz$

We need to calculate the percentage change in frequency

Using formula of percentage change in frequency

$f=\dfrac{10-9}{10}$

$f=10\%$

Hence, The percentage change in frequency is 10%.

7 0

4 years ago

HELP ME!!PLEASE ANSWER!!!!!!

Phoenix [80]

There are 3 states of matter. Solid, Liquid, Gas. An example of each would be ice, water, steam,

5 0

2 years ago

A gymnast of mass 62.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t

MrRissso [65]

Answer:

a) T = 608.22 N

b) T = 608.22 N

c) T = 682.62 N

d) T = 533.82 N

Explanation:

Given that the mass of gymnast is m = 62.0 kg

Acceleration due to gravity is g = 9.81 m/s²

Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.

So;

To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;

T = mg

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs the rope at a constant rate tension in the string is

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs up the rope with an upward acceleration of magnitude

a = 1.2 m/s²

the tension in the string is T - mg = ma (Since acceleration a is upwards)

T = ma + mg

= m (a + g )

= (62.0 kg)(9.81 m/s² + 1.2 m/s²)

= (62.0 kg) (11.01 m/s²)

= 682.62 N

When the gymnast climbs up the rope with an downward acceleration of magnitude

a = 1.2 m/s² the tension in the string is mg - T = ma (Since acceleration a is downwards)

T = mg - ma

= m (g - a )

= (62.0 kg)(9.81 m/s² - 1.2 m/s²)

= (62.0 kg)(8.61 m/s²)

= 533.82 N

5 0

3 years ago

A 296-kg motorcycle is accelerating up along a ramp that is inclined 26.2° above the horizontal. The propulsion force pushing th

Alexeev081 [22]

Answer:

The magnitude of the motorcycle's acceleration is 5.20 m/s².

Explanation:

Given that,

Mass of motorcycle = 296 kg

Angle = 26.2°

Force on motorcycle= 3106 N

Force = 286 N

We need to calculate the magnitude of the motorcycle's acceleration

The net force acting on the motorcycle

Using newton's second law

$ma=F_{p}-F_{air}-F_{g}$

$ma=F_{p}-F_{air}-mg\sin\theta$

Put the value into the formula

$296a=3106-286-296\times9.8\times\sin26.2$

$a=\dfrac{3106-286-296\times9.8\times\sin26.2}{296}$

$a=5.20\ m/s^2$

Hence, The magnitude of the motorcycle's acceleration is 5.20 m/s².

5 0

3 years ago