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Free_Kalibri [48]
3 years ago
10

A population of deer is defined by:

Physics
2 answers:
Nastasia [14]3 years ago
5 0

Answer:

the last one is false

Explanation:

got it right

nika2105 [10]3 years ago
4 0
1. A
2.B
3. True
4. False
5.B
6.B
7.B
8.True
9 true
10 true
11 true
12 C
13 D
15true
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Which terrestrial ecosystem or life zone produces the highest net primary productivity per year? a. temperate forest b. Savanna
horsena [70]

a. hope this helps...

8 0
3 years ago
During Mr. Nye's science class, students were expected to identify various substances using physical properties they could easil
nydimaria [60]
Density is the best property to use, as while multiple different metals could create cubes with the same color, mass, or volume, no different metal could create a cube with the same mass and volume.  Density is based on mass and volume, and as a result no two different metals will have the same density.
4 0
3 years ago
Read 2 more answers
A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting
ololo11 [35]

Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is

T=\dfrac{38}{8}\\\\T=4.75\ s

After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

For an oscillating object, the time period is given by :

T=2\pi \sqrt{\dfrac{m}{k}}

k = spring stiffness constant

So,

k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m

When the cord is in air,

mg=kx

x = the extension in the cord

x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m

So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m

5 0
3 years ago
A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800
Arada [10]

Explanation:

The given data is as follows.

            m = 5000 kg,            h = 800 km = 0.8 \times 10^{6} m

    R_{e} = 6.37 \times 10^{6} m,    r = R + h = 7.17 \times 10^{6} m

   M_{e} = 5.98 \times 10^{24} kg,   G = 6.67 \times 10^{-11} Nm^{2}/kg^{2}

As we know that,

              \frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}

                      v = \sqrt{\frac{GM_{e}}{r^{2}}}

And, it is known that formula to calculate angular velocity is as follows.

               \omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}

                            v = \sqrt{\frac{GM_{e}}{r^{3}}}

                               = \sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}

                                = 1.0402 \times 10^{-3} rad/s

Thus, we can conclude that speed of the satellite is 1.0402 \times 10^{-3} rad/s.

6 0
3 years ago
At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
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