Sue played four games of golf the modal score was 98 and the mean 100 and the range 10 What were the scores for the games

Answer:

m = 74.44 kg

Explanation:

p = m × v

670 = m × 9

9m = 670

m = 670/9

m = 74.44 kg

Momentum of an body is defined as the product of mass and velocity.It s unit is kgm/sec.The value of mass will be 74.44 kg.

The measurement of mass in motion is called momentum. Momentum is equal to its mass multiplied by its velocity. Mathematically it is given as;

Momentum = Mass x Velocity

The given data in the problem will be

p is the momentum of runner=670 kgm/sec.

v is the velocity =9m/sec.

m is the mass =?

[tex]\rm p =mv\\\\\rm m =\frac{p}{v} \\\\\rm m =\frac{670}{9} \\\\\rm m =74.44 kg[/tex]

Hence the value of mass will be 74.44 kg.

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Answer:

wtfffff

Explanation:

wtffffff

Answer:

0.5 m/s2

Explanation:

accelration formula : final velocty - starting velocity divided by time

Answer:

my mind cant comprehend a shing word u just said

Explanation:

The probes collect data, they capture the movement in space, therefore, spacecraft that carry instruments to collect data about objects in space are called probes, So the given statement is correct.

Any conductor that is placed into a plasma and gathers current is a probe. Measurements of the current on the probe as a function of the bias potential given to the probe can be used to identify a plasma's properties.

Such probes affect their surroundings while directly detecting electron and ion flows in the plasma. Any conductive material (wire, foil, chamber walls, etc.) can serve as a plasma probe; nevertheless, precise control over the probe's size, shape, and substance is necessary to achieve detectable plasma levels. Astronauts are not present aboard probes. Scientists can analyze the data the probes relay back to Earth. Sputnik 1 was the first spacecraft to launch.

Because probes probably collect data, they capture the movement in space, therefore, spacecraft that carry instruments to collect data about objects in space are called probes.

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Answer:

56000kgm/s

Explanation:

Momentum = Mass * Velocity (both in standard unit)

Therefore, the car's momentum = 1400 * 40 = 56000kgm/s (or 56000Ns)

Answer:

x

Explanation:

The question relates to the energy required to pump water out of the

spout at the top of the triangular prism tank.

The width of the layer of water Δx thick is x m.

Reasons:

The location of the layer Δx m thick = x meters above the bottom

The length of the triangular (cross-section) prism tank = 8 meters

The length of the layer Δx thick layer = 8 meters

From the attached drawing of the cross-section of the tank, we have;

The height of the triangular tank cross-section = 3 meters

The length of the base of the triangular cross-section = 3 meters

By similar triangles, we have;

[tex]\dfrac{Height \ of \ Tank}{Base \ length \ of \ tank}= \dfrac{Height \ of \ layer \ with \ thickness \ \Delta x }{Width \ of \ layer \ with \ thickness \ \Delta x}[/tex]

Which gives;

[tex]\dfrac{3}{3} =\dfrac{x}{Required \ width}[/tex]

[tex]\dfrac{3}{3} =1 = \dfrac{x}{Required \ width}[/tex]

Required width × 1 = x

The required width = x m

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Answer:

i believe it is the microwaves that is closest to the radio waves.

Explanation:

The microwaves are the part of the electromagnetic spectrum is nearest to radio waves.

The range of all EM radiation arranged from high to low frequency (short to long wavelength) according to their frequency or wavelength. is known as the electromagnetic spectrum (EM).

Here,

The visible light from a bulb in your home and the radio waves from a radio station are two examples of electromagnetic radiation. Radiation is energy that travels and disperses as it does.

The properties of the electromagnetic spectrum include the ways in which signals can propagate and how much data they can carry. Signals delivered at higher frequencies typically propagate more quickly but can carry more data in a given amount of space.

The electromagnetic spectrum begins with the lowest frequency radio waves and moves up in frequency and down in wavelength from there. Radio waves are followed by visible light, infrared radiation, and lastly microwaves.

Hence,

The microwaves are the part of the electromagnetic spectrum is nearest to radio waves.

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Answer:

It would be the brick since the mass and the weight is greater than the peanut I might be wrong on this but thats the best answer i can give

Explanation:

Answer:

your answer is C

Explanation:

the spring when fully recoiled is at maximum PE and will then convert to KE

Answer:

A 2.5 g bullet traveling at 350 m/s hits a tree and slows uniformly to a stop while penetrating a distance of 12 cm into the tree's trunk. What is the initial kinetic energy of the bullet? 153.125 J Correct: Your answer is correct.Jan 8, 2020

Explanation:

Answer:12.5

Explanation: Subtract

Answer:

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Explanation:

Answer:

e

Explanation:

Answer:

The charge 3q must be placed at a distance of 1.73d away from the origin.

Explanation:

The electric field at the origin due to q at distance, d from the origin is E = q/4πε₀d².

The electric field at the origin due to 2q at distance, d from the origin is E' = -2q/4πε₀d². (It is negative since it its directed towards the negative x-axis)

Now, the net electric field at the origin due to both charges is thus E" = E + E' = q/4πε₀d² + (-2q/4πε₀d²) = q/4πε₀d² - 2q/4πε₀d² = -q/4πε₀d².

The electric field due to 3q at a distance x from the origin is E₁ = 3q/4πε₀x².

Now, the net electric field E' must cancel out E₁ so that the electric field at the origin is zero.

So, E' + E₁ = 0

E' = -E₁

-q/4πε₀d² = -3q/4πε₀x²

1/d² = 3/x²

taking inverse of both sides, we have

x²/3 = d²

multiplying both sides by 3, we have

x² = 3d²

taking square root of both sides, we have

x = (√3)d

x = ±1.73d

So, the charge 3q must be placed at a distance of 1.73d away from the origin.

Answer:

14 m/s

Explanation:

KE=1/2mv^2

5880J= 1/2(60kg)v^2

5880J=(30)(V^2)

5880/30=V^2

196=v^2

square root of 196 is 14

Answer:

rotates, axis,not sure,day, not sure, night, revolves, not sure, hemisphere, towards, summer, away, winter

Explanation:

Answer:

When something is lifted, work is done on the object against the pull of gravity. This work is converted to a form of potential energy called gravitational potential energy. When the item falls towards Earth like an apple from a tree, gravitational potential energy is converted into kinetic energy.

Answer: False

Explanation:  

Answer:

v = 3.08 km/s

Explanation:

Given that,

The angular velocity of the satellite = [tex]\omega=7.27\times 10^{-5} rad/s[/tex]

A satellite is in orbit 36000km above the surface of the earth.

The radius of the earth is 6400 km

Let v is the velocity of the satellite. It can be calculated as :

[tex]v=r\omega\\\\v=(36000\times 10^3+6400\times 10^3)\times 7.27\times 10^{-5}\\\\v=3082.48\ m/s\\\\v=3.08\ km/s[/tex]

So, the velocity of the satellite is 3.08 km/s.

Answer:

Power = 1471000[W]

Explanation:

In order to solve this problem, we must use the conversion factor of horsepower to Watts.

1 [Hp] → 735.5 [W]

Now using the conversion factor:

[tex]2000[Hp]*\frac{735.5W}{1Hp}=1471000[W][/tex]

Answer:

The car will travel the distance of 15 miles

S= D/T

therefore

D=ST

D = Distance

S = Speed (30 mph)

T = Time (0.5 Hours)

D = 30 * 0.5 = 15

Answer:

-2.1 m/seconds velocity

Answer:

4 to 5

Explanation:

Answer:

Please find the calculated values entered into the following table;

[tex]\left\begin{array}{ccccc}Object & Mass \ (kg)& Weight \ (N) & Height \ (m)&G.P.E.\\A&5&50&2&100\\B&2&20&6&120\\C&8&80&5&400\\D&20&200&0.6&120\\E&5,000&50,000&2&100,000\\F&0.2&2&10&20\\G&67&670&44&29,480\end{array}[/tex]

Explanation:

The given parameters are;

The acceleration due to gravity, g ≈ 10 N/kg

The Gravitational Potential Energy, G.P.E. = m × g × h

The weight of the object, W = m × g

∴ G.P.E. = W × h

The table is filled using the above formula for calculations as follows;

For object A, we have;

Weight = 5 kg × 10 N/kg = 50 N

G.P.E. = 50 N × 2 m = 100 J

For object B, we have;

Weight = 2 kg × 10 N/kg = 20 N

G.P.E. = 20 N × 6 m = 120 J

For object C, we have;

Weight = 8 kg × 10 N/kg = 80N

G.P.E. = 80 N × 5 m = 400 J

For object D, we have;

Weight = 20 kg × 10 N/kg = 200 N

G.P.E. = 200 N × 0.6 m = 120 J

For object E, we have;

Weight = 5,000 kg × 10 N/kg = 50,000 N

G.P.E. = 50,000N × 2 m = 100,000 J

For object F, we have;

Weight = 0.2 kg × 10 N/kg = 2 N

G.P.E. = 2 N × 10 m = 20 J

For object G, we have;

Weight = 67 kg × 10 N/kg = 670 N

G.P.E. = 670 N × 44 m = 29,480 J

Therefore, we have the table filled as follows;

[tex]\left\begin{array}{ccccc}Object & Mass \ (kg)& Weight \ (N) & Height \ (m)&G.P.E.\\A&5&50&2&100\\B&2&20&6&120\\C&8&80&5&400\\D&20&200&0.6&120\\E&5,000&50,000&2&100,000\\F&0.2&2&10&20\\G&67&670&44&29,480\end{array}[/tex]

Answer:

Approximately [tex]2.1\; \rm km[/tex], assuming that [tex]g = -9.8\; \rm m \cdot s^{-2}[/tex].

Explanation:

Let [tex]t[/tex] denote the time required for the package to reach the ground. Let [tex]h(\text{initial})[/tex] and [tex]h(\text{final})[/tex] denote the initial and final height of this package.

[tex]\displaystyle h(\text{final}) = \frac{1}{2}\, g\, t^2 + h(\text{initial})[/tex].

For this package:

Solve for [tex]t[/tex], the time required for the package to reach the ground after being released.

[tex]\displaystyle t^{2} = \frac{2\, (h(\text{final}) - h(\text{initial}))}{g}[/tex].

[tex]\begin{aligned} t &= \sqrt{\frac{2\, (h(\text{final}) - h(\text{initial}))}{g} \\ &\approx \sqrt{\frac{2\times (0\; \rm m - 2500\; \rm m)}{(-9.8\; \rm m \cdot s^{-2})}} \approx 22.588\; \rm s\end{aligned}[/tex].

Assume that the air resistance on this package is negligible. The horizontal ("forward") velocity of this package would be constant (supposedly at [tex]95\; \rm m \cdot s^{-1}[/tex].) From calculations above, the package would travel forward at that speed for about [tex]22.588\; \rm s[/tex]. That corresponds to approximately:[tex]95\; \rm m \cdot s^{-1} \times 22.588\; \rm s \approx 2.1 \times 10^{3}\; \rm m = 2.1\; \rm km[/tex].

Hence, the package would land approximately [tex]2.1\; \rm km[/tex] in front of where the plane released the package.

Answer:

He traveled 9km

Explanation:

To do this problem you need to use the equation which is Speed= distance/time and this problem gives you the speed which is 18 km/h and it gives you the time 1/2 hour so you write the equation 18= d/ 1/2 which his distance is 9km