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3 years ago

When must scientific theories be changed?

2 answers:

4 0

Surprises are the reason there are new discoveries. We should clap our hands when we get a surprising result from an experiment.

Governments? What do they have to do with science unless they are funding the project. If they are not, science and governments do not really mix.

Well that has happened, but I don't think that's the answer.

D is your answer.

vlada-n [284]3 years ago

3 0

D because when new evidence is found other question will come up about it

You might be interested in

(b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1510 m/s? (c) With wh

makvit [3.9K]

Answers:

(a) $2509.98 m/s$

(b) 397042.215 m

Explanation:

The question is incomplete, please remember to write the whole question :) However, part (a) is written below:

(a) What is the escape speed on a spherical asteroid whose radius is $700 km$ and whose gravitational acceleration at the surface is $a_{g}=4.5 m/s^{2}$

Knowing this, let's begin:

a) In this part we need to find the escape speed $V_{e}$ on the asteroid:

$V_{e}=\sqrt{\frac{2GM}{R}}$ (1)

Where:

$G$ is the universal gravitational constant

$M$ is the mass of the asteroid

$R=700 km=700(10)^{3} m$ is the radius of the asteroid

On the other hand we know the gravitational acceleration is $a_{g}=4.5 m/s^{2}$ , which is given by:

$a_{g}=\frac{GM}{R^{2}}$ (2)

Isolating $GM$ :

$GM=a_{g}R^{2}$ (3)

Substituting (3) in (1):

$V_{e}=\sqrt{\frac{2a_{g}R^{2}}{R}}=\sqrt{2a_{g}R}$ (4)

$V_{e}=\sqrt{2(4.5 m/s^{2})(700(10)^{3} m)}$ (5)

$V_{e}=2509.98 m/s$ (6) This is the escape velocity

b) In this part we will use the Conservation of mechanical energy principle:

$E_{o}=E_{f}$ (7)

Being:

$E_{o}=K_{o}+U_{o}=\frac{1}{2}m V^{2} - \frac{GMm}{R}$ (8)

$E_{f}=K_{f}+U_{f}=0 - \frac{GMm}{R+h}$ (9)

Where:

$E_{o}$ is the initial mechanical energy

$E_{f}$ is the final mechanical energy

$K_{o}$ is the initial kinetic energy

$K_{f}=0$ is the final kinetic energy

$U_{o}$ is the initial gravitational potential energy

$U_{f}$ is the final gravitational potential energy

$m$ is the mass of the object

$V=1510 m/s$ is the radial speed of the object

$h$ is the distance above the surface of the object

Then:

$\frac{1}{2}m V^{2} - \frac{GMm}{R}=- \frac{GMm}{R+h}$ (10)

Isolating $h$ :

$h=\frac{2 a_{g} R^{2}}{2a_{g}R-V^{2}}-R$ (11)

$h=\frac{2 (4.5 m/s^{2}) (700(10)^{3} m)^{2}}{2(4.5 m/s^{2})(700(10)^{3} m)-(1510 m/s)^{2}}-700(10)^{3} m$ (11)

$h=397042.215 m$ (12) This is the distance above the asteroid's surface

c) We will use the Conservation of mechanical energy principle again, but now the condition is that the object is dropped at a distance $h=981.8 km=981.8(10)^{3} m$ . This means that at the begining the object only has gravitational potential energy and then it has kinetic energy and gravitational potential energy:

$\frac{-GMm}{R+h}=\frac{-GMm}{R}+\frac{1}{2}mV^{2}$ (13)

Isolating $V$ :

$V=\sqrt{2a_{g} R(1-\frac{R}{R+h})}$ (14)

$V=\sqrt{2(4.5 m/s^{2}) (700(10)^{3} m)(1-\frac{700(10)^{3} m}{700(10)^{3} m+981.8(10)^{3} m})}$ (15)

Finally:

$V=1917.76 m/s$

7 0

2 years ago

An 8 kg mass moving at 8 m/s collides with a 6 kg mass

steposvetlana [31]

Answer:

10 m/s

Explanation:

Momentum before collision = momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(8 kg)(8 m/s) + (6 kg)(6 m/s) = (8 kg)(5 m/s) + (6 kg) v

64 kg m/s + 36 kg m/s = 40 kg m/s + (6 kg) v

60 kg m/s = (6 kg) v

v = 10 m/s

3 0

3 years ago

How does a sound wave transfer energy to your ears ?

KiRa [710]

It transfers energy through the source of the sound. Your ear detects sound waves when vibrating air particles cause your ear drum to vibrate

4 0

2 years ago

Read 2 more answers

Please answer the following question

aleksklad [387]

Answer:

1. Light Bulb

2. A switch

3. A battery

4. Ammeter

5. Resistor

6. Voltmeter

4 0

2 years ago

A tennis player smashes a ball of mass m horizontally at a vertical wall. The ball rebounds at the same speed v with which it st

Morgarella [4.7K]

Answer:

B.0

Explanation:

Change in momentum: This is defined as the product of mass and change in velocity of a body. or it can be defined as the product of force and time of a body. The fundamental unit of change in momentum is kg.m/s

Change in momentum = M(V-U)......................... Equation 1

where M = mass of the ball, V = final velocity of the ball, U = initial velocity of the ball.

Let: M = m kg and V = U = v m/s

Substituting these values into equation 1

Change in momentum = m(v-v)

Change in momentum = m(0)

Change in momentum = 0 kg.m/s

Therefore the momentum of the ball has not changed.

The right option is B.0

5 0

2 years ago