Velocity and Acceleration Quick Check
Item 1
Use this graph of velocity vs. time for two objects to answer the question.
Item 2
Item 3
С
Item 4
Item 5
D
velocity
time
Which statement makes an accurate comparison of the motions for objects C and D?
(1 point)
lol

Hi there!

We can use the equation:

v = at, where in this instance:

v = velocity (m/s)

a = acceleration due to gravity (m/s²)

t = time (s)

g ≈ 9.8 m/s², so:

v = 9.8(3) = 29.4 m/s

A solid such as Sugar molecules or a liquid like in water molecules or gas molecules such as oxygen and nitrogen molecules in air. Since gas molecules have the weakest intermolecular forces than other molecules in the other two states then they will be the fastest.

Answer:

gas

Explanation:

since gas is in the air I think the answer is gas

Answer:

78

Explanation:

x=xi+vi(t)+1/2a(t)^2

x=10+16(4)+1/2(0.50)(4)^2

x=74+4

x=78 m

Answer:

You can jump 1.5 feet on Earth.

Explanation:

Because the moons gravity is weaker that earth so it would be easier to jump further on the moon.

Answer:

Thermal Conductivity Easily Transmits Heat Among Fine Ceramics

Answer:

Light can travel in three ways from a source to another location: (1) directly from the source through empty space; (2) through various media; (3) after being reflected from a mirror.

Explanation:

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, C represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

[tex]tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}[/tex]

The steepness of the slope is therefore;

[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100[/tex]

Where;

[tex]\overline{AM}[/tex] = Half the width of the truck = [tex]\dfrac{2.4 \, m}{2}[/tex] = 1.2 m

[tex]\overline{CM}[/tex] = The elevation of the center of gravity above the ground = 2.2 m

[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%[/tex]

[tex]tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}[/tex]

[tex]Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right) \approx 28.6 ^{\circ}[/tex]

The maximum steepness of the slope where the truck can be parked is 54.55 %.

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Answer:

everything will float up and go up to space and die

Explanation:

gravity keeps us down and once it stops everything will float up. And if it were to stop everywhere everything and everyone will die and everything will be destroyed.

Answer:

b Shiny

Explanation:

Trust me it's right

Answer:

22Ω

Explanation:

if    V ⇒ voltage

      I ⇒ current

      R ⇒ resistance

V = IR

220 = 10 x R

220 / 10 = R

22 = R

Answer:

Explanation:

Without friction, a roller coaster continuously converts potential energy to kinetic energy and back again. Total energy will be constant.

Let m be the mass of the car and ground level is the origin.

on the 5.5 m hill, total energy is

E = PE + KE

E = mgh + ½mv²  

E = m(9.8)(5.5) + ½m(9.3)² = 97m J

a) The maximum height will occur when the total energy is all potential energy.

E = mgh

h = E/mg

h = 97m/m(9.8) = 9.9 m  

As this value is greater than the height of the third hill at 5.5 + 4.0 = 9.5 m The car will cross the last hill with some remaining velocity in kinetic energy.

b) As 9.5 m is greater than 9.3 m, the 9.5 m hill will have more of the total energy of the system as potential energy, This mean there is less kinetic energy and therefore less velocity (and speed) on top of the 9.5 m hill.

c) KE = E - PE

KE = 97m - m(9.8)(9.5 - 1.0)

KE = 97m = 83.3m

KE = 13.7m = ½mv²

v² = √(2(13.7)

v = 5.2345...

v = 5.2 m/s

Answer:

D. Tungsten

Explanation:

W - Wolfram

Answer:

What does Newton's third law describe?

Hi there!

The formula for velocity given acceleration:

v = at

Plug in given values:

v = 6.4(7) = 44.8 m/s

Answer:

At the highest point when you toss a ball into the air.

Explanation:

At the higest point of a trajectory of a ball, the velocity is zero for a split second and there is no speed and direction. However, there still is acceleration of -10 m/s^2 because the force of gravity is still acting upon it at that point.

Hi there!

An example of this could be when a ball is thrown vertically into the air and reaches the TOP of its trajectory.

When an object is thrown with a vertical velocity, the acceleration due to gravity results in a decrease in its positive (upward) velocity until it reaches its highest point, where the instantaneous velocity = 0 m/s and the object begins to fall back down (negative velocity).

Additionally, throughout its entire trajectory, the ball experiences an acceleration due to gravity of g = 9.8 m/s², even at its highest point where there is a velocity = 0 m/s.

Explanation:

The artificial satellite experiences a centripetal force [tex]F_c[/tex] as it moves around the earth and it is defined as

[tex]F_c = m\dfrac{v^2}{r} = m\left(\dfrac{2\pi r}{T}\right)^2\left(\dfrac{1}{r}\right) = \dfrac{4\pi^2mr}{T^2}[/tex]

where m is the mass of the satellite, r is its orbital radius and T is its orbital period. But we need to find the radius first.

Recall that the satellite is orbiting at a height where its acceleration due to gravity is 6.44 m/s^2. Since we know that the weight mg of the satellite is equal to the gravitational force [tex]F_G[/tex] between the earth and the satellite, we can write

[tex]mg = F_G = G\dfrac{mM}{r^2}[/tex]

[tex]\Rightarrow g = G\dfrac{M}{r^2}[/tex]

where M is the mass of the earth (=[tex]5.972×10^{24}\:\text{kg}[/tex]) and G is the universal gravitational constant (=[tex]6.674×10^{-11}\:\text{N-m}^2\text{/kg}[/tex]). Plugging in the values, we find that the radius of the satellite's orbit is

[tex]r = \sqrt{\dfrac{GM}{g}} = \sqrt{\dfrac{(6.674×10^{-11}\:\text{N-m}^2\text{/kg})(5.972×10^{24}\:\text{kg})}{6.44\:\text{m/s}^2}}[/tex]

[tex]\:\:\:\:\:= 7.87×10^6\:\text{m}[/tex]

Now that we have the value for the radius, we can now calculate the orbital period T. Recall that the centripetal force is equal to the weight of the satellite at its orbital radius. Therefore,

[tex]F_c = mg \Rightarrow \dfrac{4\pi^2mr}{T^2} = mg[/tex]

or

[tex]4\pi^2r = gT^2[/tex]

Solving for T, we get

[tex]T^2 = \dfrac{4\pi^2r}{g} \Rightarrow T = \sqrt{\dfrac{4\pi^2r}{g}}[/tex]

We can further simplify the above expression into

[tex]T = 2\pi\sqrt{\dfrac{r}{g}}[/tex]

Plugging in the values for r and g, we get

[tex]T = 2\pi\sqrt{\dfrac{(7.87×10^6\:\text{m})}{(6.44\:\text{m/s}^2)}}[/tex]

[tex]\:\:\:\:\:= 6945\:\text{s} = 1.93\:\text{hrs}[/tex]

Answer:

I feel like its the second one but I'm not completely sure..

Explanation:

Answer:

she will move in the same direction at the same speed forever.

Explanation:

If there are no outside forces like gravity the net force will never change, she will just keep flying for forever and ever! poor lady

Explanation:

The range R of a projectile is given the equation

[tex]R = \dfrac{v_0^2}{g}\sin{2\theta}[/tex]

The maximum range is achieved when [tex]\theta = 45°[/tex] so our equation reduces to

[tex]R_{max} = \dfrac{v_0^2}{g}[/tex]

We can solve for the initial velocity [tex]v_0[/tex] as follows:

[tex]v_0^2 = gR_{max} \Rightarrow v_0 = \sqrt{gR_{max}}[/tex]

or

[tex]v_0 = \sqrt{(9.8\:\text{m/s}^2)(9.5×10^6\:\text{m})}[/tex]

[tex]\:\:\:\:\:\:\:=9.6×10^3\:\text{m/s}[/tex]

To find the maximum altitude H reached by the missile, we can use the equation

[tex]v_y^2 = v_{0y}^2 - 2gy = (v_0\sin{45°})^2 - 2gy[/tex]

At its maximum height H, [tex]v_y = 0[/tex] so we can write

[tex]0 = (v_0\sin{45°})^2 - 2gH[/tex]

or

[tex]H = \dfrac{(v_0\sin{45°})^2}{2g}[/tex]

[tex]\:\:\:\:\:\:= \dfrac{[(9.6×10^3\:\text{m/s})\sin{45°}]^2}{2(9.8\:\text{m/s}^2)}[/tex]

[tex]\:\:\:\:\:\:= 2.4×10^6\:\text{m}[/tex]

The distance traveled by the body in the given time is 1,250 m.

The given parameters;

The distance traveled by the body in the given time is calculated as follows;

[tex]s =ut + \frac{1}{2} at^2\\\\s = 0 \ + \ \frac{1}{2} (4)(25^2)\\\\s =1,250 \ m[/tex]

Thus, the distance traveled by the body in the given time is 1,250 m.

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Answer:

The box arrives first.

Explanation:

Hope this helps!! :))

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According to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.

Friction may be defined as the resistance that is offered by the surfaces that are in contact when they move past each other. It is a type of force that opposes the motion of a solid object over another.

There are mainly four types of friction: static friction, sliding friction, rolling friction, and fluid friction. According to the context of this question, the sphere possesses less friction as compared to the box. This is because the box has an irregular surface that possesses high friction over the inclined surface.

Therefore, according to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.

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Answer:

c) cornea being wavy or not spherical

Answer:

192

Explanation:

Answer:

Explanation:

Initial vertical velocity

vy₀ = 50.0sin60 = 43.3 m/s

This initial velocity is reduced to zero by gravity in a time of

t = v/a = 43.3/9.81 = 4.41 s

h(max) = ½gt² = ½(9.81)4.41² = 95.6 m

The ball will return to earth in the same amount of time

t(max) = 2(4.41) = 8.82 s

The horizontal velocity is

vx = 50.0cos60 = 25.0 m/s

d = vt = 25.0(8.82) = 221 m

That 's one heck of a kick! No air resistance of course.

Answer:

3) 311 m

Explanation:

Circumference = 2πR = π m/rev

99 rev(π m/rev) = 99π m or about 311 meters

Answer:

Explanation:

Radio waves

Radio waves have photons with the lowest energies. Microwaves have a little more energy than radio waves. Infrared has still more, followed by visible, ultraviolet, X-rays and gamma rays.