Why is figure 5 an unhelpful visualization tool for this data set?

Please help!

Our sun is 27 million times larger than our moon. Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth. If tidal forces were based solely on comparative masses, the sun should have a tide-generating force that is 27 million times greater than that of the moon. However, the sun is 390 times further from the Earth than is the moon. Thus, its tide-generating force is reduced by 3903, or about 59 million times less than the moon. Because of these conditions, the sun’s tide-generating force is about half that of the moon.

Answer:

The acceleration due to gravity at Pluto is 0.0597 m/s^2.

Explanation:

Length, L = 1 m

10 oscillations in 257 seconds

Time period, T = 257/10 = 25.7 s

Let the acceleration due to gravity is g.

Use the formula of time period of simple pendulum

[tex]T = 2\pi\sqrt{\frac{L}{g}}\\\\25.7 = 2 \times 31.4\sqrt{\frac{1}{g}}\\\\g = 0.0597 m/s^2[/tex]

Answer:

Explanation:

Givens

t = 10 seconds

vi = 0

vf = 3600 mph

a = ?

d = 10000 feet

Formula

a = (vf - vi)/t

Solution

vf = 3600 mph * 1 hr / 3600 seconds * 5280 feet / 1 mile = 5280 ft / sec

a = (5280 - 0)/10

a = 528 ft/sec^2

Answer:

28 cm and 32 cm

Explanation:

1. The spring pendulum hangs vertically, oscillates harmonic with amplitude 2cm and angular frequency 20 rad/s. The natural length of

a spring is 30cm. What is the minimum and maximum length of the spring during the oscillation? Take g = 10m/s2.

As the amplitude is 2 cm and the natural length is 30 cm. So, it oscillates between 30 -2 = 28 cm to 30 + 2 = 32 cm.

So, the minimum length is 28 cm and the maximum length is 32 cm.

Answer:

could and largejsjisj and we look like they can get to

Answer:

The orbital velocity an aircraft orbiting around a heavenly body is found as follows;

At the orbital velocity, [tex]F_G[/tex] = [tex]F_C[/tex]

Where;

[tex]F_G[/tex] = The gravitational force = [tex]\dfrac{G \cdot M \cdot m}{R_E^2}[/tex]

[tex]F_C[/tex] = The centripetal force = [tex]\dfrac{m \cdot v_0^2}{R_E}[/tex]

Therefore

[tex]v_0 = \sqrt{\dfrac{G \cdot M}{R_E} }[/tex]

Therefore, at the orbital velocity of the spacecraft, the centripetal force attracting the person away from the central region heavenly body is equal to the gravitational force pulling the person towards the center of the heavenly body (which was felt as her or his weight), and the person feels weightless while inside the orbiting spacecraft

Explanation:

Answer: Remove your foot from the gas pedal and slow down

Explanation:

Hydroplaning also refers to aquaplaning and this refers to the scenario whereby there's a layer of water that builds between the tyres of a car and the surface of the road which then brings about a loss of traction which eventually leads to a situation whereby the car doesn't respond to control inputs.

This lack of traction with the pavement takes away your control of the vehicle and to regain control of the vehicle, one should remove your foot from the gas pedal and slow down.

[tex]\boxed{\sf E=QV^2}[/tex]

[tex]\\ \sf\longmapsto Q=\dfrac{E}{V^2}[/tex]

[tex]\\ \sf\longmapsto Q=\dfrac{3.92\times 10^{-4}}{24^2}[/tex]

[tex]\\ \sf\longmapsto Q=\dfrac{3.92\times 10^{-4}}{576}[/tex]

[tex]\\ \sf\longmapsto Q=0.006\times 10^{-4}C[/tex]

[tex]\\ \sf\longmapsto Q=6\times 10^{-1}C[/tex]

[tex]\\ \sf\longmapsto Q=0.6C[/tex]

Now

[tex]\boxed{\sf Q=CV}[/tex]

[tex]\\ \sf\longmapsto C=\dfrac{Q}{V}[/tex]

[tex]\\ \sf\longmapsto C=\dfrac{0.6}{24}[/tex]

[tex]\\ \sf\longmapsto C=0.025F[/tex]

Note:-

Answer:

The magnitude of the unknown force is 44.8 N.

Explanation:

The force can be found with Newton's second law:                

[tex] F = ma [/tex]

Where:

m: is the mass of the toy car = 5.5 kg

a: is the acceleration

F: is the force =?

We can calculate the acceleration with the following kinematic equation:

[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]

Where:

[tex] v_{f} [/tex]: is the final speed = 7.3 m/s

[tex] v_{0} [/tex]: is the initial speed = 2.1 m/s

d: is the distance traveled = 3 m

Hence, the acceleration is:

[tex] a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} = \frac{(7.3 m/s)^{2} - (2.1 m/s)^{2}}{2*3 m} = 8.15 m/s^{2} [/tex]

Finally, the magnitude of the force is:

[tex]F = ma = 5.5 kg*8.15 m/s^{2} = 44.8 N[/tex]                                  

Therefore, the magnitude of the unknown force is 44.8 N.

I hope it helps you!                                

Answer:

Using physical and chemical clean up methods to remove the oil

Explanation:

Explanation:

The acceleration g varies by about 1/2 of 1 percent with position on Earth's surface, from about 9.78 metres per second per second at the Equator to approximately 9.83 metres per second per second at the poles.

The answer are :

A. Distance = 15.71 m , The displacement = 10 m

B. distance traveled will increase, displacement will decrease

C. The distance = 31.42 m , Displacement = 0

Distance is a scalar quantity, while displacement is a vector quantity. Displacement is the distance travelled in a specific direction.

Given that a child rides a pony on a circular track with a radius of 5 m, after the child has gone halfway around the track

(A)

The distance traveled will be = 2[tex]\pi[/tex]r / 2

The distance = πr

The distance = 22/7 x 5

The distance = 15.71 m

The displacement = 2r

The displacement = 2 x 5

The displacement = 10 m

(B) The distance traveled will increase as the child completes one circle of the track but the displacement will decrease because the displacement in one cycle of a circular motion is zero since it is a vector quantity.

(C) The distance after a complete circuit of the track = 2πr

The distance = 2 × 22/7 × 5

The distance = 31.42 m

The displacement after a complete circuit of the track will be zero.

That is;

Displacement = 0

Therefore, the distance traveled and the displacement after the child has gone halfway around the track are 15.7 m and 10 m respectively. While the distance and displacement after a complete circle of the track are 31.42 m and 0 respectively.

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Answer:

The emf of a cell is the sum of the electric potential differences (PDs) produced by a separation of charges (electrons or ions) that can occur at each phase boundary (or interface) in the cell. The magnitude of each PD depends on the chemical nature of the two contacting phases.

Explanation:

pls mark brainliest

Answer:

E_O = 96J

Explanation:

E_O = E_I*(%e/100%)

E_O = 120J*(80%/100%)

E_O = 96J

Answer:

Remember the relation:

Speed*Time = Distance.

We can estimate that the speed at which an earthquake "moves", in the surface, is:

S = 6km/s  (this is a low estimation actually)

Then if the wave is detected 12.5 minutes after the earthquake, we know that it traveled for 12.5 minutes before reaching the detector.

So we know the speed of the wave and the time it took to reach the detector, then we can use the equation:

Speed*Time = Distance.

to find the distance.

First, we should write the time in seconds

we know that:

1 min = 60 s

then:

12.5 min = 12.5*(60 s) = 750 s

Then, the wave traveled with a speed of 6 km/s for 750 seconds until it reached the detector, then the distance that it traveled is:

(6km/s)*750s = 4500 km

The distance between the detector and the site of the quake is around 4500 km.

Answer:

Explanation:

        16093.4

Explanation:

B. Doubling the heat flow into the engine while halving the work done

hope this helps you

have a nice day

Explanation:

initial velocity(u)=30km/hr = 30*1000/60*60=8.33 m/s

final velocity(v) =60km/hr = 60*1000/60*60 =16.67 m/s

time taken(t) = 1 minutes

= 60 seconds

Now,

Average velocity = u+v/2

= 8.33 m/s + 16.67m/s÷2

=12.5 m/s

Answer:

The velocity at that time would be [tex](-67.975\; \rm s)[/tex]. The velocity of this particle is continuously changing.

Explanation:

Differentiate the expression for position [tex]x[/tex] with respect to time [tex]t[/tex] to find an expression for velocity.

[tex]\begin{aligned}v(t) &= \frac{d}{dt}[x(t)] \\ &= \frac{d}{dt} \left[ 7.4 + 9.2\, t - 2.1\, t^{3}\right]\\ &= 9.2 - 6.3\, t^{2}\end{aligned}[/tex].

Hence, at [tex]t = 3.5\; \rm s[/tex], velocity would be [tex]v(3.5) = 9.2 - 6.3 \times (3.5)^{2} = -67.975\; \rm m[/tex].

Since velocity [tex]v(t)[/tex] changes with time [tex]t[/tex], the velocity of this particle is continuously changing.

Answer:

0.48N

Explanation:

according to the second law of motion

force=mass×acceleration

the mass in this question is 0.025,the acceleration 19

therefore f=0.025×19

=0.48N

I hope this helps

(a) The de Broglie wavelength is approximately 5.175 × 10⁻¹⁵  meters. The wavelength is lesser than the distance of closest approach

(b) It is proper to treat the alpha particle as a particle and not as wave because the distance of closest approach is much larger than and not comparable to its wavelength for the alpha particle for the alpha particle to be treated as a wave

The given parameters are;

The kinetic energy of the alpha particles = 7.70 MeV = 1.23368 × 10⁻¹² J

The distance from the gold nucleus the alpha particles reach = 29.5 fm

(a) The de Broglie wavelength of a particle is given as follows;

[tex]\mathbf{\lambda = \dfrac{h}{p}}[/tex]

Where;

λ = The wavelength

h = Planck's constant = 6.62607004 × 10⁻³⁴ m²·kg/s

p = The momentum of the particle = Mass of an electron, m × Velocity, v

The mass of an alpha particle, m ≈ 6.645 × 10⁻²⁷ kg

Therefore;

[tex]\lambda = \dfrac{h}{m \times v}[/tex]

The kinetic energy of the alpha particle, K.E. = (1/2)·m·v²

∴ v = √(2 × K.E./m)

Therefore;

[tex]\lambda = \dfrac{h}{m \times \sqrt{2 \times \dfrac{K.E.}{m} } } = \dfrac{h}{ \sqrt{2 \times m \times K.E.} }[/tex]

Plugging in the values of the variables gives;

[tex]\lambda = \dfrac{6.62607004 \times 10 ^{-34} }{ \sqrt{2 \times 6.645 \times 10 ^{-27} \times 1.23368 \times 10^{-12} } } \approx 5.175 \times 10^{-15}[/tex]

The de Broglie wavelength of the alpha particle, λ ≈ 5.175 × 10⁻¹⁵ m

The distance of closest approach = 29.5 fm = 29.5 × 10⁻¹⁵ m

Compared to the distance of closest approach, the wavelength of the alpha particle is lesser than the distance of closest approach

(b)  Given that the distance of closest approach is six times larger than the wavelength of the alpha particle, and alpha particle behaving as waves are expected to approach closer to the gold nucleus in the region of their wavelength before deflection, therefore, the larger distance of closest approach is indicative of a charged particle to charged particle interaction, and therefore, particle behavior of alpha particles.

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Answer:

a. 5 batteries b. 1050 mAh

Explanation:

Here is the complete question

A student project is required to be portable and hand held. It requires 6 V DC power at a current of 150 mA. The batteries for the power supply must last for a minimum of 7 hours of continuous operation. NiMH rechargeable batteries in AA size are to be used. A) How many batteries are needed? B) What mAh capacity should the batteries have?

Solution

A) How many batteries are needed?

Since the nominal voltage for a single NiMH battery is 1.2 V per battery and we require 6V DC power, we combine the batteries in series to obtain a total voltage of 6 V. The number of batteries required, n = total voltage/voltage per cell = 6V/1.2V per battery = 5 batteries

So, the number of batteries needed is 5.

B) What mAh capacity should the batteries have?

Since the batteries are in series, they would each deliver a current of 150 mA. Since we require a current of 150 mA for 7 hours, the number of milliampere-hours capacity mAh of batteries required is Q = It where I = current = 150 mA and t = time = 7 hours.

So, Q = It = 150 mA × 7 h = 1050 mAh.

So, the batteries should have a mAh of 1050 mAh

Explanation:

velocity - the speed with a direction. Thus, inertia could be redefined as follows: Inertia: tendency of an object to resist changes in its velocity. ... Such an object will not change its state of motion (i.e., velocity) unless acted upon by an unbalanced force.

Answer:

Physics is the branch of science concerned with the nature and properties of matter and energy. The subject matter of physics, distinguished from that of chemistry and biology, includes mechanics, heat, light and other radiation, sound, electricity, magnetism, and the structure of atoms. And a more detailed working definition of physics may be: The science of nature, or that which pertains to natural objects, which deals with the laws and properties of matter and the forces which act upon them. Quite often, physics concentrates upon the forces having an impact upon matter, that is, gravitation, heat, light, magnetism, electricity, and others.

Physics helps us to understand how the world around us works, from can openers, light bulbs and cell phones to muscles, lungs and brains; from paints, piccolos and pirouettes to cameras, cars and cathedrals; from earthquakes, tsunamis and hurricanes to quarks, DNA and black holes. Physics helps us to organize the universe. It deals with fundamentals, and helps us to see the connections between seemly disparate phenomena.

Physics provides quantitative and analytic skills needed for analyzing data and solving problems in the sciences, engineering and medicine, as well as in economics, finance, management, law and public policy. Physics is the basis for most modern technology, and for the tools and instruments used in scientific, engineering and medical research and development. Manufacturing is dominated by physics-based technology.

To understand the fundamental principles of the universe, physics utilizes many workings from the other natural sciences. Because of this overlap, phenomena studied in physics (conservation of energy for example) are common to all material systems. The specific ways in which they apply to energy (hence, physics) are often referred to as the "laws of physics." Because each of the other natural sciences biology, chemistry, geology, material science, medicine, engineering, and others, work with systems which adhere to the laws of physics, physics is often referred to as the "fundamental science."

Answer:

Physics can be important in understanding the world and how it works. The  main goal of physics is to understand how the universe behaves.

Explanation:

Answer:The answer is A

Explanation:just did it on a test

Answer:

I think deep-water wave is not a wave

Explanation:

but it will be better if u wait for the second answer as I am not sure

Shallow water wave is not a type of wave. Thus, the correct option is C.

A wave is a disturbance in a medium which carries energy without a net movement of particles in the medium. A wave may take the form of elastic deformation, it is a variation of pressure, electric or the magnetic intensity, electric potential, or the temperature of the medium.

Shallow water waves take place in the shallow water, this means that the waves which occur at depths that is shallower than the wavelength of the wave divided by 20. This would mean that a wave with a wavelength of about 60 meters is a shallow water wave at depth range of less than 60 divided by 20, or 3 meters.

Therefore, the correct option is C.

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The approximate temperature of the star as determined is D) 5500 K.

The Wien's displacement law relates the maximum wavelength of a body to its absolute temperature. Wien's displacement law states that:

λ = [tex]\frac{b}{T}[/tex]

where λ is the maximum wavelength of the body, b is the constant of proportionality and T is the absolute temperature.

Thus from the given question, λ = 525 nm (525 x [tex]10^{-9}[/tex]), and b = 2.9 x [tex]10^{-3}[/tex] mK.

So that,

525 x [tex]10^{-9}[/tex] = [tex]\frac{2.9*10^{-3} }{T}[/tex]

Make T the subject of the formula to have;

T = [tex]\frac{2.9*10^{-3} }{525*10^{-9} }[/tex]

  = 5523.81

T = 5523.81 K

T ≅ 5500.00 K

The approximate temperature of the star in Kelvin is 5500 K.

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Answer:

I dont know answer Sorry For that thank u

Explanation:

The three importance of measurement are as follows :-

Hope it is helpful to you

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