What is the average speed of a bird that flies 1 km in 30 minutes?

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P

Natasha_Volkova [10]

Answer

given,

distance of first satellite = 48,000 Km

distance of second satellite = 64,000 Km

orbital period = 6.39 day

Using equation of time period

$T = \dfrac{2\pi r^{3/2}}{\sqrt{Gm_{pluto}}}$

now, from the above equation we can say that only variable is Time period and r is the radii of orbit.

from the first satellite

$\dfrac{T_{charon}}{r^{3/2}_{charon}}=\dfrac{T_{sat1}}{r^{3/2}_{sat1}}$

$T_{sat1}=\dfrac{T_{charon}\ r^{3/2}}{r^{3/2}_{charon}}$

$T_{sat1}=\dfrac{6.39\times (48000)^{3/2}}{19600^{3/2}}$

$T_{sat1}=24.5\ days$

for second satellite

$T_{sat2}=\dfrac{T_{charon}\ r^{3/2}_{sat2}}{r^{3/2}_{charon}}$

$T_{sat1}=\dfrac{6.39\times (64000)^{3/2}}{19600^{3/2}}$

$T_{sat1}=37.7\ days$

7 0

3 years ago

A block with mass M = 3 kg is moving on a flat surface with constant speed v1 = 12 m/s. A bullet with mass m = 0,1 kg is shot at

adelina 88 [10]

The speed does the block move after it is hit by the bullet that remains stuck inside the block will be 23.7 m/sec and it takes 12.07 seconds to stop.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

Apply the law of conservation of momentum principle;

m₁v₁+m₂v₂cosΘ =(m₁+m₂)V

3 kg × 12 m/s + 0,1 kg × 400 m/s cos 20° = (3+0.1)V

36 + 40 cos 20° = 3.1 V

V=23.7 m/sec

The time it takes to stop when the friction coefficient between the block and the surface is 0.2 is found as;

V = u +at

V = 0+ μgt

23..7=0.2× 9.81 ×t

t=12.07 sec

Hence, it takes 12.07 seconds to stop.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

4 0

2 years ago

A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.65 m/s2 for 17.0 s. 2

nlexa [21]

Hello

Let's solve the problem in the three different steps

1) Uniformly accelerated motion, with acceleration $a_1 = 2.65~m/s^2$ and for a total time of $t_1=17~s$ . The body is initially at rest, so the distance covered is given by
$S= \frac{1}{2}a_1t_1^2=382.9~m$
Calling $v_f$ and $v_i$ the final and initial velocity, and since the $v_i=0~m/s$ because the body starts from rest, we can use
$a= \frac{v_f-v_i}{t}$
to find the final velocity after this first leg:
$v_{f}=v_i+a_1t_1=45~m/s$
And the average velocity in this first leg is
$v_1= \frac{v_f+v_i}{2}=22.5~m/s$

2) Uniform motion. The velocity is constant and it is equal to the final velocity of the first leg: $v_2=45~m/s$ . This is also the average velocity of the second leg.
The total time of this second leg is $t_2=1.60~min = 96~s$ . The distance covered is given by
$S_2=v_2t_2=45~m/s \cdot 96~s=4320~m$

3) Uniformly decelerated motion, with constant deceleration of $a_3=-9.39~m/s^2$ and for a total time of $t_3=4.8~s$ . Here, the initial velocity of the body is the final velocity of the previous leg, i.e. $v_i=45~m/s$ . Therefore, the distance covered in this leg is given by
$S_3=v_i t_3 + \frac{1}{2} a_3 t^2 =107.8~m$
The final velocity in this leg is given by
$v_f=v_i+at=45~m/s-9.39~m/s^2 \cdot 4.8~s = -0.07~m/s$
The negative sign means that after decelerating, the body has started to go in the opposite direction. Similarly to step 1, the average velocity in this leg is given by
$v_3 = \frac{1}{2}(v_f+v_i)= \frac{1}{2}(-0.07~m/s+45~m/s)= 22.5~m/s$

4) Finally, the total distance covered in the motion is
$S=S_1+S_2+S_3=382.9~m+4320~m+107.8~m=4810.7~m$
To find the average velocity, we must "weigh" the average velocity of each leg for the correspondent time of that leg:
$v_{ave}= \frac{v_1t_1+v_2t_2+v_3t_3}{t_1+t_2+t_3}=40.8~m/s$

8 0

3 years ago

The following are the Earth–Sun distance at the equinoxes and solstices: March equinox 149.0 million km June solstice 152.0 mill

Mice21 [21]

Answer:

During <u>winter (late December/early January)</u> the Earth is closest to the Sun and during <u>summer (late June/early July)</u> the Earth is farthest from the Sun.

Explanation:

In the northern hemisphere, the earth usually comes closer to the sun during the time of winter season, mostly in late December or early January.

On the other hand, the earth is farthest from the sun during the time of summer season, mostly in late June or early July.

When the earth is closer to the sun, during the winter, it is comparatively cold. It is due to the absorption of a lesser amount of incoming solar radiation. The tilt of the earth is also responsible for this low temperature.

But, when the earth is farthest from the sun, during the summer, it is comparatively hot. It is due to the absorption of a large amount of incoming solar radiation.

5 0

3 years ago

List some ways that the sun's energy affects our atmosphere

julia-pushkina [17]

Answer:

Nothing is more important to us on Earth than the Sun. Without the Sun's heat and light, the Earth would be a lifeless ball of ice-coated rock. The Sun warms our seas, stirs our atmosphere, generates our weather patterns, and gives energy to the growing green plants that provide the food and oxygen for life on Earth.

Explanation:

3 0

3 years ago