a coconut falls from the top of a tree and takes 3.5 seconds to reach the ground. How tall is the tree?

The volume of a material is the total amount of matter that it can contain. The volume of the given coin has been determined to be 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex]. Since the gold doubloon do not absorb water, then its volume remains constant at the ocean floor.

The volume of the gold doubloon can be determined by;

volume = [tex]\pi r^{2}[/tex] + h

where r is the radius of the coin and h is its thickness.

Such that; diameter = 6.1 cm (61 mm) and h = 2.0 mm

r = [tex]\frac{diameter}{2}[/tex]

 = [tex]\frac{61}{2}[/tex]

r = 30.5 mm

Thus,

volume of the coin = [tex]\frac{22}{7}[/tex] x [tex](30.5)^{2}[/tex] x 2

                               = 5847.2857

Therefore, the volume of the gold doubloon is 5847.3 [tex]mm^{3}[/tex]. This can also be expressed as 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex].

Since the gold doubloon is  not miscible with water, thus its volume at a depth of 770 m at the ocean floor is the same as its initial volume.

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Answer:

ma*XsinB

option 2 is correct

Answer:

c) cornea being wavy or not spherical

Answer:

F12 = G M1 M2 / R12^2

F12' = G M1 M2 / R12'^2

F12' / F12 = R12'^2 / R12^2 = (1/3)^2

F12' = 1/9 F12

The new force is 1/9 the of the old force

In this case, it is possible to solve this problem by using the widely-known steam tables which show that at 90 °C, the pressure that produces a vapor-liquid mixture at equilibrium is about 70.183 kPa (Cengel, Thermodynamics 5th edition).

Moreover, for the calculation of the volume, it is necessary to calculate the volume of the vapor-liquid mixture, given the quality (x) it has:

[tex]x=\frac{m_{steam}}{m_{total}}[/tex]

Thus, since 8 kg correspond to liquid water, 2 kg must correspond to steam, so that the quality turns out:

[tex]x=\frac{2kg}{10kg} =0.20[/tex]

Now, at this temperature and pressure, the volume of a saturated vapor is  2.3593 m³/kg whereas that of the saturated liquid is 0.001036 m³/kg and therefore, the volume of the mixture is:

[tex]v=0.001036m^3/kg+0.2(2.3593-0.001036 )m^3/kg=0.4727m^3/kg[/tex]

This means that the volume of the container will be:

[tex]V=10kg*0.4727m^3/kg\\\\V=4.73m^3[/tex]

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Answer:

Why is community action important? ... Involving communities in the design and delivery of services can help to achieve a number of objectives, including: Building community and social capacity – helping the community to share knowledge, skills and ideas. Community resilience – helping the community to support itself

The work done by the man against friction is 4,192.86 J.

The given parameters;

The work done by the man against friction is calculated as follows;

[tex]W = F \times d \times cos(75)\\\\W = 540 \times 30 \times cos(75)\\\\W = 4,192.86 \ J[/tex]

Thus, the work done by the man against friction is 4,192.86 J.

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Answer:

0.067 births per chipmunk per year

Explanation:

I took the test

Answer:

I feel like its the second one but I'm not completely sure..

Explanation:

L=1m

=2mm²=(2/1000²)=2(10^-6)m

y=4x10¹¹N/m²

∆l=2mm=2/10.00=0.002m

=(4x10¹¹x2x10^-6x2x10^-3)÷1m

=16x10¹¹-⁶-³

=16x10¹¹-⁹

=16x10²

=1600N

where a=cross sectional area=2x10^-6m

C=2mm= 2x10^-3m

L=1m

hope it helps

Answer:

d = -33.1 m and Vf = -25.5 m/s

Explanation:

a = -9.8 m

t = 2.6 s

Vᵢ = 0 m/s

d = ?

Vf = ?

Now,

d = Vᵢ × t + 0.5 × a × t²

d = (0 m/s) × (2.60 s) + 0.5 × (-9.8 m/s²) × (2.60 s)²

d = -33.1 m (- indicates direction)

Vf = Vᵢ + a × t

Vf = 0 + (-9.8 m/s²) × (2.60 s)

Vf = -25.5 m/s (- indicates direction)

Thus, d = -33.1 m and Vf = -25.5 m/s

-TheUnknownScientist 72

Answer:

Explanation:

The work of the brakes will equal the initial kinetic energy of the car

Fd = ½mv²

F = mv²/2d

F = 2457(18²) / (2(62))

F = 6,419.903...

F = 6.4 kN

Answer: increases

Explanation:

Temperature is a measure of the average velocity of the molecular particles. The faster they go, the higher the temperature.

Answer:

Approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex], assuming that no external force (e.g., gravitational pull) was acting on this rocket.

Explanation:

Assume that no external force is acting on this rocket. The system of the rocket and the fuel on the rocket would be isolated (an isolated system.) The momentum within this system would be conserved.

Let [tex]v_{0}\; \rm m\cdot s^{-1}[/tex] be the initial velocity of the rocket.

The velocity of the exhaust gas would be [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] since the gas is ejected away from the rocket.

Let [tex]\Delta v\; \rm m\cdot s^{-1}[/tex] denote the increase in the velocity of the rocket. The velocity of the rocket after ejecting the gas would be [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex].

The momentum [tex]p[/tex] of an object of velocity [tex]v[/tex] and mass [tex]m[/tex] is [tex]p = m \cdot v[/tex].

The combined mass of the rocket and the fuel was [tex]10000\; \rm kg[/tex]. The initial momentum of this rocket-fuel system would be:

[tex]\begin{aligned}p_{0} &= m \cdot v\\ &= 10000\; {\rm kg} \times v_{0}\; {\rm m \cdot s^{-1}} \\ &= (10000\; v_{0})\; \rm {kg \cdot m\cdot s^{-1}}\end{aligned}[/tex].

The momentum of the [tex]5.0\; \rm kg[/tex] of fuel ejected at [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] would be:

[tex]\begin{aligned} & 5.0 \; {\rm kg} \times (v_{0} - 5000)\; {\rm m\cdot s^{-1}}\\ =\; & (5.0\, v_{0} - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

After ejecting the [tex]5.0\; \rm kg[/tex] of the fuel, the mass of the rocket would be [tex]10000\; \rm kg - 5.0\; \rm kg = 9995\; \rm kg[/tex]. At a velocity of [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex], the momentum of the rocket would be:

[tex]\begin{aligned} & 9995 \; {\rm kg} \times (v_{0} + \Delta v)\; {\rm m\cdot s^{-1}}\\ =\; & (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

Take the sum of these two quantities to find the momentum of the rocket-fuel system after the fuel was ejected:

[tex]\begin{aligned}p_{1} &= (5.0\, v_{0} - 25000)\; {\rm kg \cdot m\cdot s^{-1}\\ &\quad\quad + (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}} \\ &= (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

The momentum of the rocket-fuel system would be conserved. Thus [tex]p_{0} = p_{1}[/tex].

[tex](10000\, v_{0})\; {\rm kg \cdot m\cdot s^{-1}} = (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

Solve this equation for [tex]\Delta v[/tex], the increase in the velocity of the rocket.

[tex]10000\, v_{0} = 10000\, v_{0} + 9995\, \Delta v - 25000[/tex].

[tex]9995\, \Delta v = 25000[/tex].

[tex]\begin{aligned}\Delta v &= \frac{25000}{9995} \approx 2.5\end{aligned}[/tex].

Thus, the velocity of the rocket would increase by approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex] after ejecting the [tex]5.0\; \rm kg[/tex] of fuel.

Answer:

C

Explanation:

Definition of drug: a medicine or other substance which has a physiological effect when ingested or otherwise introduced into the body

Explanation:

The orbiting period of a satellite at a height h from earth' surface is

T=2πr32gR2

where r=R+h.

Then, T=2π(R+h)R(R+hg)−−−−−−−−√

Here, R=6400km,h=1600km=R/4

T=2πR+R4−−−−−−√R(R+R4g)−−−−−−−−−⎷=2π(1.25)32Rg−−√

Putting the given values,

T=2×3.14×(6.4×106m9.8ms−2)−−−−−−−−−−−−√(1.25)32=7092s=1.97h

Now, a satellite will appear stationary in the sky over a point on the earth's equator if its period of revolution around the earthh is equal to the period of revolution of the earth up around its own axis whichh is 24h. Let us find the height h of such a satellite above the earth's suface in terms of the earth,'s radius.

Let it be nR.Then

T=2π(R+nR)R(R+nRg)−−−−−−−−−−√

=2π(Rg)−−−−−√(1+n)32

=2×3.14(6.4×106m/s9.8m/s2)−−−−−−−−−−−−−−−⎷(1+n)32

(5075s)(1+n)32=(1.41h)(1+n)32

For T=24h, we have (24h)=(1.41h)(1+n)32

or (1+n)32=241.41=17

or 1+n(17)23=6.61

or n=5.61

The height of the geostationary satellite above the earth's surface is nR=5.61×6400km=6.59×104km.

Answer:

The box arrives first.

Explanation:

Hope this helps!! :))

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According to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.

Friction may be defined as the resistance that is offered by the surfaces that are in contact when they move past each other. It is a type of force that opposes the motion of a solid object over another.

There are mainly four types of friction: static friction, sliding friction, rolling friction, and fluid friction. According to the context of this question, the sphere possesses less friction as compared to the box. This is because the box has an irregular surface that possesses high friction over the inclined surface.

Therefore, according to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.

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Answer:

she will eventually slow down and come to a stop

Answer:

All organisms begin their lives as single cells.Overtime,these organisms grow and take on the characteristics of their species...All organisms grow,and different parts of organisms may grow at different rates.Organisims made out of only one cell

may change little during their lives, but they do grow

Explanation:

brainlest me please

Answer:

L=mvr

α=0

v=ωr

L=mωr

2

2

1

mv

2

=k

L=Iω

k=

2

1

Iω

2

ω

1

=2ω

k

1

=1/2k

∴I

1

=

8

1

I

L

1

=

8

1

I×2ω

=

4

1

Iω

=L/4

Physical motion that occurs when an item rotates or spins on an axis is known as “rotatory motion,” sometimes known as “rotational motion” or “circular motion.”

Angular displacement per unit of time is measured by angular frequency, also referred to as radial or circular frequency (). Therefore, it has degrees (or radians) per second as its units. 1 Hz = 6.28 rad/sec, thus. 1 radian equals 57.3° because 2 radians equal 360°.

The frequency of rotation is the rate at which an item rotates around an axis, sometimes referred to as rotational speed or rate of rotation.

However, the linear velocity also depends on how far the particles are from the axis of rotation. Because every particle distances are different, so too will the linear velocities for each particle.

Therefore, All the particles in a rigid body's rotational motion move in the same direction in a given interval. Therefore, all the particles' angular velocities will be the same.

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Answer:

0

Explanation:

0 is the answer