A particle moves with a velocity v in a circle of radius r, then its angular velocity is equal

to………….. and acts along the…………..​

Answers

Answer 1

Answer:

Given that,  

Speed = v  

Radius = r  

We have to ascertain the precise speed  

Utilizing equation of speed  

Where, v = velocity/speed

r = radius  

= precise speed/ angular velocity

Angular Velocity :  

The precise speed is characterized by the speed of turn.  

The precise speed is straightforwardly corresponding to the direct speed and contrarily relative to the range of the molecule.  

Subsequently, The precise speed is v/r


Related Questions

The Michelson interferometer can be used to measure the index of refraction of a gas by placing an evacuated transparent tube in the light path along one arm of the device. Fringe shifts occur as the gas is slowly added to the tube. Assume 560-nm light is used, the tube is 5.10 cm long, and 168 bright fringes pass on the screen as the pressure of the gas in the tube increases to atmospheric pressure. What is the index of refraction of the gas

Answers

Answer:

The index of refraction of the gas is  [tex]n_g = 1.000922[/tex]

Explanation:

From the question we are told that

   The wavelength of light is  [tex]\lambda = 560 \ nm = 560 *10^{-9} \ m[/tex]

   The length of tube is  [tex]L = 5.10 \ cm = \frac{5.10 }{100} =0.0510 \ m[/tex]

    The number of fringes is  [tex]N = 168[/tex]

Generally the  index of refraction of the gas  is mathematically repreented as

      [tex]n_g = 1 + \frac{N \lambda }{2 L }[/tex]

substituting values

       [tex]n_g = 1 + \frac{168 *[ 560* 10^{-9}] }{2 * 0.0510 }[/tex]

       [tex]n_g = 1.000922[/tex]

A laser beam is incident from the air at an angle of 30.0° to the vertical onto a solution of Karo syrup in water. If the beam is refracted to 19.24° to the vertical, what is the index of refraction of the syrup solution?

Answers

Answer:

Index of Refraction = 1.52

Explanation:

The index of refraction or the refractive index is given as:

Index of Refraction = Sin i/Sin r

where,

i = angle of incidence

r = angle of refraction

In this case of karo syrup, we have the following data:

i = angle of incidence  = 30°

r = angle of refraction = 19.24°

Therefore, substituting these values in the equation, we get:

Index of Refraction = Sin 30°/Sin 19.24°

Index of Refraction = 0.5/0.3295

Index of Refraction = 1.52

Hence, the refractive index or the index of refraction of the Karo Syrup is found to be 1.52.

You are working in a biology lab and learning to use a new ultracentrifuge for blood ts. The specifications for the centrifuge say that a red blood cell rotating in the ultracentrifuge moves at 470 m/s and has a radial acceleration of 150,000 g's (that is, 150,000 times 9.8 m/s2). The radius of the centrifuge is 0.15 m. You wonder if this claim is correct.
What is the radial acceleration of the ultra-centrifuge using calculations ?

Answers

Answer:

The radial acceleration is  [tex]a_r = 1472667 \ m/s^2[/tex]

Explanation:

From the question we are told that

      The velocity of the red blood cell is  [tex]v_r = 470 \ m/s[/tex]

       The radial acceleration of the red blood cell is [tex]a_r = 150000g = 150000*9.8 = 1470000 \ m/s^2[/tex]

       The radius of the centrifuge is  [tex]r = 0.15 \ m[/tex]

Generally radial acceleration is  mathematically represented as

      [tex]a_r = \frac{v^2}{r}[/tex]

substituting values

     [tex]a_r = \frac{470^2}{0.15}[/tex]

     [tex]a_r = 1472667 \ m/s^2[/tex]

You are at the controls of a particle accelerator, sending a beam of 2.10×107 m/s protons (mass m) at a gas target of an unknown element. Your detector tells you that some protons bounce straight back after a collision with one of the nuclei of the unknown element. All such protons rebound with a speed of 1.80×107 m/s. Assume that the initial speed of the target nucleus is negligible and the collision is elastic.
A) Find the mass of one nucleus of the unknown element.
B) What is the speed of the unknown nucleus immediately after such a collision?

Answers

Answer:

a

The mass is  [tex]m_2 =21.75*10^{-27} \ kg[/tex]

b

The velocity is  [tex]v_2 = 3.0*10^{6} m/s[/tex]

Explanation:

From the question we are told that

     The speed of the protons is  [tex]u_1 = 2.10*10^{7} m/s[/tex]

     The mass of the protons is  [tex]m[/tex]

     The speed of the rebounding protons are [tex]v_1 = -1.80 * 10^{7} \ m/s[/tex]

The negative sign shows that it is moving in the opposite direction

     

Now according to the law of energy conservation mass of one nucleus of the unknown element. is mathematically represented as

        [tex]m_2 = [\frac{u_1 -v_1}{u_1 + v_1} ] m_1[/tex]

Where [tex]m_1[/tex] is the mass of a single proton

          So substituting values

       [tex]m_2 = \frac{2.10 *10^{7} - (-1.80 *10^{7})} {(2.10 *10^7) + (-1.80 *10^{7})} m_1[/tex]

        [tex]m_2 =13 m_1[/tex]

The mass of on proton is  [tex]m_1 = 1.673 * 10^{-27} \ kg[/tex]

So     [tex]m_2 =13 ( 1.673 * 10^{-27} )[/tex]

        [tex]m_2 =21.75*10^{-27} \ kg[/tex]

Now according to the law of linear momentum conservation the speed of the unknown nucleus immediately after such a collision is mathematically evaluated as

      [tex]m_1 u_1 + m_2u_2 = m_1 v_1 + m_2v_2[/tex]

Now  [tex]u_2[/tex] because before collision the the nucleus was at rest

So

        [tex]m_1 u_1 = m_1 v_1 + m_2v_2[/tex]

=>    [tex]v_2 = \frac{m_1(u_1 -v_1)}{m_2}[/tex]

Recall that [tex]m_2 =13 m_1[/tex]

So

       [tex]v_2 = \frac{m_1(u_1 -v_1)}{13m_1}[/tex]

=>         [tex]v_2 = \frac{(u_1 -v_1)}{13}[/tex]

substituting values

              [tex]v_2 = \frac{( 2.10*10^{7} -(-1.80 *10^{7}))}{13}[/tex]

              [tex]v_2 = 3.0*10^{6} m/s[/tex]

   

a block (mass 0.6kg) is released from rest at point A at the top of a ramp inclined at 36.9 degress above the horizontal. The block moves a distance of 4m along the ramp to point b at the bottom of the ramp. How much work is done by gravity as the block goes from a to b

Answers

14.136 J as shown on the photo with two thought processes but overall same calculation

Suppose the kicker launches the ball at 60∘ instead of 30∘. Assuming that the goal is 4.55 m high and 40 m away, what minimum initial speed v0 would the ball need to have in order to just clear the goal?

Answers

Answer:22 m/s

Explanation:

Given

launch angle [tex]\theta =60^{\circ}[/tex]

height of goal [tex]h=4.55\ m[/tex]

and horizontal distance [tex]x=40\ m[/tex]

Suppose initial speed is [tex]u[/tex]

Trajectory of a Projectile is given by

[tex]y=x\tan \theta -\frac{1}{2}\frac{gx^2}{u^2\cos ^2\theta }[/tex]

substituting the values we get

[tex]4.55=40\tan (60)-0.5\times \frac{9.8\times (40)^2}{u^2\cdot \cos ^260 }[/tex]

[tex]4.55=69.28-0.5\times \frac{15,680}{u^2\cdot 0.25}[/tex]

[tex]\frac{31,360}{u^2}=69.28-4.55[/tex]

[tex]\frac{31,360}{64.73}=u^2[/tex]

[tex]u^2=484.47[/tex]

[tex]u=22.01\ m/s[/tex]

So, initial launch speed is [tex]22\ m/s[/tex]

Having successfully ransomed a nearby city with her rail gun, Prof. Marcia Grail is looking for new and exciting ways to wreck science based havok for profit. She constructs a device capable of creating a constant intensity of electromagnetic radiation of any frequency over a circular space of radius 50cm. In order to do the most damage to biological systems (e.g. make those filthy normal humans pay), what type of electromagnetic wave does she need to set the device for

Answers

Answer:

Gamma radiation

Explanation:

Electromagnetic radiations are waves that do not require material medium for their propagation, and they travel at the same velocity. When these waves are arranged with respect to increasing frequency or decreasing wavelength, it forms a spectrum.

Since Prof. Marcia Grail's device can create an electromagnetic radiation of any frequency, the appropriate radiation to achieve her goal would be gamma radiation.

Gamma radiation is a high penetrating radiation with lethal effects on exposed biological molecules, causing induction of cancer and genetic mutations even at low level.

What are the 2 types of electricity

Answers

The two are AC and DC.

Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of these seeds can be determined with a high-speed camera. In an experiment on one type of plant, seeds are projected at 20 cm above ground level with initial speeds between 2.3 m/s and 4.6 m/s. The launch angle is measured from the horizontal, with +90∘ corresponding to an initial velocity straight up and -90∘ straight down. The experiment is designed so that the seeds move no more than 0.20 mm between photographic frames. What minimum frame rate for the high-speed camera is needed to achieve this?

a. 250 frames/s
b. 2500 frames/s
c. 25,000 frames/s
d. 250,000 frames/s.

Answers

Answer:

c. 25,000 frames/s

Explanation:

For computing the minimum frame rate for high speed first we have to determine the time by applying the following equation

[tex]t = \frac{d}{s}[/tex]

[tex]= \frac{0.2\ mm}{4.6\ m/s }[/tex]

[tex]= \frac{0.2 \times 10 ^{-3}}{4.6\ m/s }[/tex]

[tex]= 4.347 \times 10^{-5} sec[/tex]

Now the frame rate is

[tex]Frame\ rate = \frac{1}{t}[/tex]

[tex]= \frac{1}{4.347 \times 10^{-5} sec}[/tex]

= 23,000 frame per sec

≈ 25,000 frame per sec

First we have find the time then after finding out the time we calculate the frame time by applying the above formula so that the minimum frame rate could come

Your friend says your body is made up of more than 99.9999% empty space. What do you think

Answers

Answer:false

Explanation:

standing on a hill that is 3.5 m high above level ground. A wall that is on level ground is 15 m away from the student. The student decides to throw a ball horizontally at a combined height of 4.5 m, at the wall. The student measures the time it takes to make an impact and finds it to be 0.65 s. At what height, from the ground, did the student hit the side of the wal

Answers

Answer:

The height will be "2.42 m".

Explanation:

The given values are:

time, t = 0.65 seconds

g = 9.8

As we know,

⇒  [tex]x=vt+\frac{1}{2}gt^2[/tex]

∴ v = 0

On putting the estimated values, we get

⇒    [tex]=0+\frac{1}{2}\times 9.8\times (0.65)^2[/tex]

⇒    [tex]=\frac{1}{2}\times 9.8\times 0.4225[/tex]

⇒    [tex]=\frac{1}{2}\times 4.1405[/tex]

⇒    [tex]=2.07025 \ m[/tex]

Now,

Height, [tex]h=4.5-2.07025[/tex]

⇒             [tex]=2.42 \ m[/tex]

Which of the following statements about this system of lumps must be true? A. The momentum of the system is conserved during the collision. B. The kinetic energy of the system is conserved during the collision. C. The two masses lose all their kinetic energy during the collision. D. The velocity of the center of mass of the system is the same after the collision as it was before the collisio

Answers

Answer:

option A.

Explanation:

Since the two lumps collide together, it is an inelastic collision.  An inelastic collision, in contrast to an elastic collision, is a collision in which kinetic energy is not conserved due to the action of internal friction. Part of the kinetic energy is changed to some other form of energy in the collision

Momentum is conserved in inelastic collisions. This means the correct option is option A.


The initial mass of a radioisotope is 10.0 g. If the radioisotope has a half life of 2.75 years, how much remains after four half lives?

Answers

After four half-lives, the amount that remains is (1/2)⁴ of the original sample.

That's (1/16) of 10 g, or 0.625 g.

It doesn't matter how long the half-life is.

A radioisotope has a starting mass of 10.0 g. Half lives of the isotopes are 0-10, 1-5, 2-2.25, 3-1.25, and 4-0.625 if the radioisotope has a half life of 2.75 years.

Who or what are radioisotopes?

A chemical element in an unstable state that emits radiation as it decomposes and becomes more stable. Both the natural environment and a laboratory can produce radioisotopes. They are utilized in imaging studies and therapy in medicine.

Why do you use the term "half-life"?

In radioactivity, the half-life is the amount of time needed for half of a radioactive sample's atomic nuclei to spontaneously decay (convert into different nuclear species by emitting particles and energy), or, more precisely, the amount of time needed for the number of disintegrations per second.

To know more about Radioisotope, visit-

https://brainly.com/question/19472025

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How much would it cost to cover the entire land area of United States with dollar bills? To answer this question, you may find the following information useful: The contiguous United States (all of the states minus Alaska and Hawaii) can be approximated as a rectangle that measures 1000 miles north to south and 3000 miles east to west, while Alaska has about one-fifth the area of the contiguous states and Hawaii is small enough to be ignored for this rough calculation. Also, note that a dollar bill measures around 6.5cm×15.5cm6.5cm×15.5cm.

Answers

Answer:

It would cost approximately $925,455,484 million to cover continental US and Alaska with $1 bills.

Explanation:

the area of a one dollar bill = 6.5 cm x 15.5 cm = 100.75 sq cm

the approximate area of continental US + Alaska = (1,000 miles x 3,000 miles) x 1.2 = 3,600,000 sq miles

each sq mile is roughly 2.58999 sq km, so the total area in sq km = 9,323,964 sq km

1 sq km = 1,000,000 sq meters

each sq meter = 10,000 sq cm

1 sq m = 10,000 / 100.75 = 99.25558 bills

1 sq km = 99,255,583.13 bills

9,323,964 sq km = 925,455,483,900,000 bills

If the US includes the area of the Hawaii and the Alaska. It would be costing approx. $925,455,484 mn to cover continental with the $1 bills.  

As the area of 1 dollar bill = 6.5 cm x 15.5 cm = 100.75 cm2 The area of US + Alaska = (1,000 miles x 3,000 miles) x 1.2 = 3,600,000 sq Thus the total area in km2  = 9,323,964 sq km  1 km2 equals to 1,000,000 sq meters each sq meter = 10,000 sq cm 1 sq m = 10,000 / 100.75 = 99.25558 bills  1 km2 equals to 99,255,583.13 bills 9,323,964 sq km =  925,455,483,900,000 bills .

Learn more about the would it cost to cover the entire land area.

brainly.com/question/17333335.

Based on the information in the table, which two elements are most likely in the same group, and why?

A table with 5 columns and 5 rows labeled facts about 4 elements. The first column labeled element has entries bismuth (B i), nitrogen (N), oxygen (O), sodium (N a), thallium (T l). The second column labeled atomic mass (a m u) has entries 209, 14, 16, 23, 204. The third column labeled total electrons has entries 83, 7, 8, 11, 81. The fourth column labeled valence electrons has entries 5, 5, 6, 1, 3. The fifth column labeled year isolated has entries 1753, 1772, 1772, 1807, 1861.

bismuth and thallium, because their atomic masses are very similar
nitrogen and oxygen, because they were both first isolated in the same year
sodium and thallium, because their names both end in the same suffix: -ium
bismuth and nitrogen, because they have the same number of valence electrons

Answers

Answer: bismuth and nitrogen, because they have the same number of valence electrons

Explanation:

Elements are distributed in groups and periods in a periodic table.

Elements that belong to same groups will show similar chemical properties because they have same number of valence electrons.

The number of valence electrons in Bismuth and nitrogen are 5 and thus thus they will show similar chemical properties and thus belong to the same group.

The atomic masses of elements in a group will differ drastically.

The group number has got nothing to be the isolation year.

Thus bismuth and nitrogen belong to same group because they have the same number of valence electrons

Answer:

D

Explanation:

i got it right in my test

The frames for a pair of eyeglasses have a radius of 2.14 cm at 20.0°C. Lenses with radius of 2.16 cm have to be inserted into these frames. To what temperature must the technician heat the frames to accommodate the lenses? The frames are made of a material whose thermal expansion coefficient is 1.30 10-4/°C.

Answers

Answer:

The technician must heat the frame to a temperature of T₂ = 0.0072°C

Explanation:

Change in one-dimension of an object upon heating is given by the following equation:

ΔL = α L ΔT

For the radius this equation can be re-written as follows:

ΔR = α R ΔT

where,

ΔR = Change in Radius =  2.16 cm - 2.14 cm = 0.02 cm

α = Thermal Expansion Coefficient = 1.3 x 10⁻⁴ /°C

R = Original Radius = 2.14 cm

ΔT = Change in Temperature = T₂ - 20°C

Therefore,

0.02 cm = (1.3 x 10⁻⁴ /°C)(2.14 cm)(T₂ - 20°C)

T₂ - 20°C = (0.02 cm)/(1.3 x 10⁻⁴ /°C)(2.14 cm)

T₂ = 0.0072°C + 20°C

T₂ = 20.0072°C

4. An electric iron has a
power rating of 750W
a. How many joules of
electric energy does it
change into heat energy
every second?
b. How many joules of
work can it do in 3
seconds
c. How long does it take
the iron to do 1500J of
work?
5. Use the kinetic particle
theory to explain why a
solid has a definite shape
and liquid has none.​

Answers

Answer:

4.

a) W = 750 J

b) W = 2250 J

c) t = 2 sec

5. Answered in explanation

Explanation:

4.

The formula of power is given as:

P = W/t

where,

P = Power

W = Work Done

t = Time Taken

a)

Here,

P = 750 W

t = 1 sec

W = ?

Therefore,

750 W = W/1 sec

W = 750 J

b)

Here,

P = 750 W

t = 3 sec

W = ?

Therefore,

750 W = W/3 sec

W = (750 W)(3 sec)

W = 2250 J

c)

Here,

P = 750 W

t = ?

W = 1500 J

Therefore,

750 W = 1500 J/t

t = 1500 J/750 W

t = 2 sec

5.

According to Kinetic Particle Theory, the molecules are tightly packed with each other, by strong inter-molecular forces and they can only vibrate at their position. While, molecules or particles in liquids have lesser attractive forces among them. They can move in layers and can take the shape of any container. This is the reason why solid has a definite shape and liquid has none.

The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal chain on the scale, it read 02.1kg. The physicist, then decided to test Einstein’s equation, and compressed the spring and tied it with the chain and placed it on the scale. It read 02.7kg. Which of the following conclusions is the most likely the physicist will come to?

a. Einstein's equation has an error
b. The scale is broken
c. Compressing the spring didn't add energy
d. The scale's resolution is too low to read the change in mass

Answers

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

F = kx

since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

Δm =  4.9 x 10⁻¹³ kg

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

d. The scale's resolution is too low to read the change in mass

Learning Goal: To understand the behavior ofthe electric field at the surface of a conductor, and itsrelationship to surface charge on the conductor.

A conductor is placed in an external electrostatic field. Theexternal field is uniform before the conductor is placed within it.The conductor is completely isolated from any source of current orcharge.

PART A)
Which of the following describes the electricfield inside this conductor?

It is in thesame direction as the original external field.
It is in theopposite direction from that of the original externalfield.
It has adirection determined entirely by the charge on itssurface.
It is alwayszero.
PART B)
The charge density inside theconductor is:

0
non-zero;but uniform
non-zero;non-uniform
infinite
PART C)
Assume that at some point just outside thesurface of the conductor, the electric field has magnitudeE and is directed toward thesurface of the conductor. What is the charge density eta on the surface of the conductor at thatpoint?
Express your answer in terms ofE and epsilon_0.
eta =

Answers

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a   it is always zero

b  0

c  [tex]\eta = -\epsilon _o E[/tex]

Explanation:ss

Here the  net charge is  on the outer surface of the conductor thus this means that the net charge inside the conductor is zero

Generally the charge density of a conductor is dependent on the charge per unit area  which implies that the charge density is dependent on the net charge  so this  means that the charge density inside the conductor is zero

 

Generally the direction of electric field this from the  positive charge to the negative charge  so from the question we can deduce  that the negative charge is located on the surface of the conductor

    So We can mathematically define the charge density on the surface of the electric field as

             ∮[tex]E \cdot dA = \frac{-Q}{\epsilon _o}[/tex]

Where E is the electric field

          [tex]dA[/tex] change in unit area

           [tex]-Q[/tex] is the negative charge

          [tex]\epsilon _o[/tex]  is the permittivity of free space

So

          [tex]EA = \frac{-Q}{\epsilon _o }[/tex]

           [tex]\frac{Q}{A} = -\epsilon _o E[/tex]

          [tex]\eta = -\epsilon _o E[/tex]

Where [tex]\eta[/tex] is the charge density

   

A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm, and two probes are separated by 1.8 cm. Voltage applied at the two ends is 5 V. A pulse arrives at the collection point at 0.608 ms, and the separation of the pulse is 180 sec. Calculate mobility and diffusion coefficient for minority carriers. Verify it from the Einstein relation.

Answers

Answer:

Mobility of the minority carriers, [tex]\mu_{n} =1184.21 cm^{2} /V-sec[/tex]

Diffusion coefficient for minority carriers,[tex]D_{n} = 29.20 cm^2 /s[/tex]

Verified from Einstein relation as  [tex]\frac{D_{n} }{\mu_{n} } = 25 mV[/tex]

Explanation:

Length of sample, [tex]l_{s} = 2 cm[/tex]

Separation between the two probes, L = 1.8 cm

Drift time, [tex]t_{d} = 0.608 ms[/tex]

Applied voltage, V = 5 V

Mobility of the minority carriers ( electrons), [tex]\mu_{n} = \frac{V_{d} }{E}[/tex]

Where the drift velocity, [tex]V_{d} = \frac{L}{t_{d} }[/tex]

[tex]V_{d} = \frac{1.8}{0.608 * 10^{-3} } \\V_{d} = 2960.53 cm/s[/tex]

and the Electric field strength, [tex]E = \frac{V}{l_{s} }[/tex]

E = 5/2

E = 2.5 V/cm

Mobility of the minority carriers:

[tex]\mu_{n} = 2960.53/2.5\\\mu_{n} =1184.21 cm^{2} /V-sec[/tex]

The electron diffusion coefficient, [tex]D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }[/tex]

[tex]\triangle x = (\triangle t )V_{d}[/tex], where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)

[tex]\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3} }\\\triangle x =0.533 cm[/tex]

[tex]D_{n} = \frac{0.533^{2} }{16 * 0.608 * 10^{-3} }\\D_{n} = 29.20 cm^2 /s[/tex]

For the Einstein equation to be satisfied, [tex]\frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V[/tex]

[tex]\frac{D_{n} }{\mu_{n} } = \frac{29.20}{1184.21} \\\frac{D_{n} }{\mu_{n} } = 0.025 = 25 mV[/tex]

Verified.

A substance that produces H+ ions in solution is a.

A)soap

B)acid

C)salt

D)base

Answers

The answer is D)base

Write the formula for the Newton’s Law of Gravitation.​

please answer this question!!!

Answers

Answer:

F = G(m1m2)/R2.

Explanation:

Newton's law of universal gravitation:

Gravitational Force, in Newtons, between two objects =

(a constant)·(one mass)·(the other mass)/(distance between them)²

acting on EACH object, in the direction of the other object.

If the masses are in kilograms and the distance is in meters, then the constant is 6.67 x 10⁻¹¹ m³/kg-sec² .

I may be wrong, but I don't think Newton had any number to use for the constant.  It had not been measured yet, the kilogram and the meter had not been invented yet, and there certainly was no unit called "a Newton" during his lifetime.

He might have been able to calculate the value of the constant by applying his law of gravity to the motion of one or two  planets. But he would have needed to know the mass of the sun and the planets he used, and I don't think those were known yet in Newton's time.  

The density of gasoline is 730 kg/m3 at 00C. Its volume expansion coefficient is 0.000960C-1. If one gallon of gasoline occupies 0.0038 m3, how many extra kilograms of gasoline are obtained when 10 gallons of gasoline are bought at 00C rather than at 200C

Answers

Answer:

Explanation:

volume expansion coefficient = .000960 C⁻¹

d₀ = d ( 1 + .000960 x t )  , d is density at t and d₀ is density at 0 degree Celsius .

if t = 200

d₀= d ( 1 + .00096 x 200 )

= d ( 1 + .00096 x 200 )

= 1.192d

10 gallons of gasolene

= 10 x .0038 m³ of gasolene

= .038 m³

difference of mass = volume x difference of density

= .038 x ( d₀ - d )

.038 x ( d₀ - d₀ / 1.192 )

= .038 x ( 1 - 1 / 1.192 ) d₀

= .00612 x 730

= 4.468 kg .

In order for an external force to do work on a system, Question 9 options: a) there must be a component of the force perpendicular to the motion. b) there must be a component of force parallel to the motion of the object. c) the force can be at any angle relative to the motion of the object.

Answers

Answer:

b) there must be a component of force parallel to the motion of the object.

Explanation:

We know that work done on a body by an external force is calculated by the formula given below:

W = F.d = Fd Cos θ

where,

W = Work Done by the force on the body

F = Magnitude of force

d = displacement of the body

θ = The angle between the direction of motion of the body and the force applied

It is clear from the formula of the work done, that "F Cosθ" represents the component of the force, that is acting in the direction of motion of the object or parallel to the direction of motion of the object. So, if there is no component of force parallel to motion of object, then this factor will become zero. As a result, the work done will also be zero.

Therefore, the correct option will be:

b) there must be a component of force parallel to the motion of object.

50 POINTS
What happens to kinetic energy of a moving object of the velocity decreases? (Look at photo multiple choice)

Answers

Answer:   A    Answers. Assuming that the terminal velocity doesn't change during the fall, then the kinetic energy would remain constant. However the terminal velocity decreases during the fall since the air becomes denser at lower altitudes.

Explanation:

What happens to the KE of an object when it slows down and heats up? - Quora. The kinetic energy goes down and the loss of the kinetic energy is through the production of heat energy. In real world this is due to friction, or an opposing force that decelerates the object, or a combination of both.

Answer:

Kinetic energy stays the same

A girl walks South at 2.7 m/s. What is the y component of her velocity​

Answers

Answer:

-2.7m/s

Explanation:

...............

A boat crosses a 200m wide river at 3m/s, north relative to water. The river flows at 1m/s as shown. What is the velocity of the boat as observed by a stationary observer on the river back from the boat departed?

Answers

Answer:

The resultant velocity = 3.16 m/s

Explanation:

Since the boat is moving North of the direction of the riverflow, the river would either be flowing westward or Eastward. The two motions form a right angle triangle with the resultant velocity being the hypotenuse of the traingle.

The resultant velocity will be given as ;

R² = B² + r²

Where B is the velocity of the boat and r is the velocity of the river

R² = 3² + 1²

R² = 10

R = √10 = 3.16 m/s

Therefore, the resultant velocity = 3.16 m/s

A television weighs 8.50 pounds. How many grams is this? (Hint: You need to

use two unit conversion fractions. 1 pound equals about 0.454 kg.)

Answers

Answer:

3859 grams

Explanation:

Given: Weight of a television = 8.50 pounds

To find: Weight of a television in grams

Solution:

1 pound = 0.454 kg and 1 kg = 1000 g

So,

1 pound = 0.454 × 1000 = 454 grams

8.50 pounds = 8.50 × 454 = 3859 grams

Therefore,

Weight of television in grams = 3859 grams

Whale sharks swim forward while ascending or descending. They swim along a straight-line path at a shallow angle as they move from the surface to deep water or from the depths to the surface. In one recorded dive, a shark started 50 m below the surface and swam at 0.85 m/s along a path tipped at a 13° angle above the horizontal until reaching the surface.

Required:
a. What was the horizontal distance between the shark's starting and ending positions?
b. What was the total distance that the shark swam?
c. How much time did this motion take?

Answers

Answer:

a)217m

b)222m

c)261s  

Explanation:

Considering the figure in the attachment

depth 'h'=50m

speed 'v'=0.85m/s

angle 'θ'=13°

a) we have

[tex]\frac{h}{x}[/tex]=tanθ

x = h / tanθ => 50/tan13°

x=216.57≈217m

b)we have

[tex]\frac{h}{d}[/tex]=sinθ

d=h / sinθ => 50 / sin13°

d=222.27≈222m

c)Time needed = d / v = 222/0.85

                          261.18≈261s

What is the gravitational force between mars and Phobos

Answers

Answer:

[tex]F=5.16\times 10^{15}\ N[/tex]

Explanation:

We have,

Mass of Mars is, [tex]m_M=6.42\times 10^{23}\ kg[/tex]

Mass of its moon Phobos, [tex]m_P=1.06\times 10^{16}\ kg[/tex]

Distance between Mars and Phobos, d = 9378 km

It is required to find the gravitational force between Mars and Phobos. The force between two masses is given by

[tex]F=G\dfrac{m_Mm_P}{d^2}[/tex]

Plugging all values, we get :

[tex]F=6.67\times 10^{-11}\times \dfrac{6.42\times 10^{23}\times 1.06\times 10^{16}}{(9378\times 10^3)^2}\\\\F=5.16\times 10^{15}\ N[/tex]

So, the gravitational force is [tex]5.16\times 10^{15}\ N[/tex].

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