2.Suppose you lifted the ring when the counter-clockwise orbiting marble was at the top of the circle. Describe the direction of the path that the marble would take.

3. Explain why the marble moved the way it did with the ring in place.

4. Explain why the marble moved the way it did when the ring was removed.

Answer:

about 10 milliamperes

Explanation:

currents above 10ma can paralize or "freeze bodies"

Answer:

i love devon bostick

Explanation:

cause....... i do

Answer: It was going 4 m/s

Explanation:

Answer:

It helps melt the ice away on their walkway.

Explanation:

The salt makes the ice to freeze at a lower temperature. So when a solute (salt) comes into contact with the ice the freezing point is lowered.

Answer:

The resistivity of an exceedingly good electrical conductor, such as hard-drawn copper, at 20° C (68° F) is 1.77 × 10-8 ohm-metre, or 1.77 × 10-6 ohm-centimetre. At the other extreme, electrical insulators have resistivities in the range 1012 to 1020 ohm-metres.

Answer:

In general there are three sub-segments of "alternative" energy investment: solar energy, wind energy and hybrid electric vehicles

Hurricanes and tropical storms can also spawn tornadoes and microbursts, create storm surges ... If you are unable to evacuate, go to your wind-safe room.

Answer:

2.125 m/s^2

Explanation:

Use Newton's second law: Force equals mass times acceleration, F = ma. You know the force (51 N) and you know the mass of the mower (24 kg). Remembering that a Newton is the same as kg-m/s2, solve for acceleration:

F = ma

51 kg-m/s2 = 24 kg * a

2.125 m/s2 =a

If a net external force of 51N is extorted on a 24kg lawnmower parallel to the ground, the acceleration of the mower will be 2.125 m/s².

Given the data in the question;

Net External Force; [tex]F = 51N[/tex]

Mass of the Mower; [tex]m = 24kg[/tex]

Acceleration; [tex]a = ?[/tex]

To determine the acceleration of the mower, we make use of the equation from the Newton's Second Law of Motion:

[tex]F = m\ * \ a[/tex]

Where F is the force exerted, m is the mass and a is the acceleration.

But then, we know that; ( A newton is the same thing as  [tex]1 kg.m/s^2[/tex] )

Hence, Our Net External Force; [tex]F = 51kg.m/s^2[/tex]

We substitute our values into the equation to find acceleration "a"

[tex]51kg.m/s^2 = 24kg\ *\ a\\\\a = \frac{51kg.m/s^2}{24kg} \\\\a = 2.125 m/s^2[/tex]

Therefore, If a net external force of 51N is extorted on a 24kg lawnmower parallel to the ground, the acceleration of the mower will be 2.125 m/s².

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Answer:

(a). The rocket's velocity is 117.6 m/s.

(b). The rocket can reach at maximum height is 940.8 m.

(c). The velocity the instant before the rocket crashes on the ground is 135.7 m/s.

Explanation:

Given that,

Acceleration = 29.4 m/s²

Time = 4 sec

(a). We need to calculate the rocket's velocity

Using equation of motion

[tex]v=u+at[/tex]

Put the value into the formula

[tex]v=0+29.4\times4[/tex]

[tex]v=117.6\ m/s[/tex]

We need to calculate the maximum height at the moment fuel ends

For the value of x₁

Using equation of motion

[tex]x_{1}=ut+\dfrac{1}{2}at^2[/tex]

Put the value into the formula

[tex]x_{1}=0+\dfrac{1}{2}\times29.4\times4^2[/tex]

[tex]x_{1}=235.2\ m[/tex]

We need to calculate the value of x₂

Using equation of motion

[tex]v^2=u^2-2gx_{2}[/tex]

Put the value into the formula

[tex]0=u^2-2gx_{2}[/tex]

[tex]x_{2}=\dfrac{u^2}{2g}[/tex]

Put the value in to the formula

[tex]x_{2}=\dfrac{117.6^2}{2\times9.8}[/tex]

[tex]x_{2}=705.6\ m[/tex]

(b). We need to calculate the maximum this rocket can reach

Using formula for height

[tex]H=x_{1}+x_{2}[/tex]

Put the value into the formula

[tex]H=235.2+705.6[/tex]

[tex]H=940.8\ m[/tex]

(c). We need to calculate the velocity the instant before the rocket crashes on the ground

Using equation of motion

[tex]v^2=u^2+2gh[/tex]

Put the value into the formula

[tex]v=\sqrt{2\times9.8\times940.8}[/tex]

[tex]v=135.7\ m/s[/tex]

Hence, (a). The rocket's velocity is 117.6 m/s.

(b). The rocket can reach at maximum height is 940.8 m.

(c). The velocity the instant before the rocket crashes on the ground is 135.7 m/s.

Answer:

C!

Explanation:

I just did the test !

Please stay safe and wear a mask , we are in the mist of a pandemic ! !

The graph shows the change in velocity of an object in free fall is according to option C.

When an object is released from rest in free air considering no friction, the motion is depend only on the acceleration due to gravity, g.

When an object is made to fall freely, it's velocity is only depended on the acceleration due to gravity, g. The velocity decreases till it become zero.

Thus, the correct option is C.

Learn more about free falling.

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#SPJ5

Answer:

there are six types of joints in the human body

Answer: The answer is D. 6

Explanation: Learned in health class a while ago.

Answer:

However, to mention a few, there are other units of speed such as meters per second, feet per second, light-years per millennium, and knots. Feet per second, but specially meters per second are usually used to measure speed of animals, humans, and free fall objects. Knots is used to measure the speed of ship and/or boats.

Explanation:

Answer:

Mathematically: D = m/v. If you know what liquid you have, you can look up its density in a table. Once you know that, all you have to do to find the mass of the liquid is to measure its volume. Once you know density and volume, calculate mass using this relationship: mass = density • volume

Explanation:

Answer: Explanation:

Locations above the Arctic Circle (north of 66.5 degrees latitude; 90 degrees minus the tilt of Earth's axis) receive 24 hours of sunlight.

Orbits: Sun

Answer:

B

Explanation:

Answer:I litterly need a picture of this put  the pic and Ill edit this question

Explanation:

Answer:

[tex]a=190\ m/s^2[/tex]

Explanation:

Mass of a hockey puck, m = 0.17 kg

Force exerted by the hockey puck, F' = 35 N

The force of friction, f = 2.7 N

We need to find the acceleration of the hockey puck.

Net force, F=F'-f

F=35-2.7

F=32.3 N

Now, using second law of motion,

F = ma

a is the acceleration of the hockey puck

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2[/tex]

So, the acceleration of the hockey puck is [tex]190\ m/s^2[/tex].

The magnitude of acceleration of the hockey puck is 190 m/s² .

Given data:

The mass of hockey puck is, m = 0.17 kg.

the magnitude of external force on the puck is, F = 35.0 N.

The magnitude of force of friction is, f = 2.7 N.

We need to find the net force acting on the puck first. For that consider the applied force in the x-direction. Then net force is,

F' = F - f

Solving as,

F' = 35.0 - 2.7

F' = 32.3 N

Now, use the Newton's Second law to obtain the magnitude of acceleration (a) as,

F' = ma

Solving as,

32.3 = 0.17 (a)

a = 32.3/0.17

a = 190 m/s²

Thus, we can conclude that the magnitude of acceleration of the hockey puck is 190 m/s² .

Learn more about the Newton's law of motion here:

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Answer:

Hydrogen is the most abundant element in the Universe; helium is second. However, after this, the rank of abundance does not continue to correspond to the atomic number; oxygen has abundance rank 3, but atomic number 8. All others are substantially less common.

Explanation:

Answer:

The surface with fewer impact craters is a newer surface. It hasn’t been impacted as many times as the older surfaces

Explanation:

Sample Answer

Answer:

Newer surfaces are less damaged by impact craters, because they are less damaged by impact.

Explanation:

This is my answer.

Answer:

p.e=mgh

where m=10kg

g=10m/s(constant)

h=20m

p.e=10x20x10

p.e=2000joules

Answer:

The acceleration of the snowball is 0.3125

Explanation:

The initial speed of the snowball up the hill, u = 0

The speed the snowball reaches, v = 5 m/s

The length of the hill, s = 40 m

The equation of motion of the snowball given the above parameters is therefore;

v² = u² + 2·a·s

Where;

a = The acceleration of the snowball

Plugging in the values, we have;

5² = 0² + 2 × a × 40

∴ 2 ×  40 × a  = 5² = 25

80 × a = 25

a = 25/80 = 5/16

a = The acceleration of the snowball = 5/16 m/s².

The acceleration of the snowball = 5/16 m/s² = 0.3125 m/s² .

Answer:

I know you may not be reading this but for number 13 the answer is 2 hours because t=d/s which is 4000/2000 which equals to 2.

Explanation:I've currently doing that problem as of the time i'm writing this.

Answer:

1) Jumping Jacks (their a great way to get blood into your muscles and keep your blood running)

2) Toe touches (helps warm up your hamstring so you won‘t get injured easily)

3) Jog in place (Warm up leg muscles and whole body)

4) Arm circles (Warms up arm muscles)

5) 10,10,9,9. (This is something I like, you hop on 1 foot for 10 times and then switch to your other foot ten times without stopping, but after the ten you go back foryour previous hopping leg and hop 9 times this time and same for your other foot, THEN hop 8 times with the foot you started with and, keep going in that order in decending order until you reach 1, (a great way to warm up leg muscle)

Answer:

1.0 A

Explanation:

Answer:

the answer is C

I hope I helped (◍•ᴗ•◍)❤

Answer:D:normal spiral

Explanation:

Answer:

D

Explanation: