Answer:
The objects outside the reference frame aren't moving. It appears this way since the vehicle you are inside is moving, but unless the objects are people, animals, or other vehicles, the objects aren't moving.
What causes an astigmatism?
A. damaged lens
B. retina not focusing the image
C. cornea being wavy or not spherical
D. sclera not refracting light properly
Answer:
c) cornea being wavy or not spherical
A clothes dryer in a home draws a current of 10 amps when connected on a special 220-volts household circuit.what is the resistance of the dryer?
Answer:
22Ω
Explanation:
if V ⇒ voltage
I ⇒ current
R ⇒ resistance
V = IR
220 = 10 x R
220 / 10 = R
22 = R
Read the sentence from the text. “They are as glossy as satin or sunlight reflecting off water!" What does the word glossy mean in the sentence? O A. pointed o B. shiny O C. small O D. strong
Answer:
b Shiny
Explanation:
Trust me it's right
An object accelerates from rest, with a constant acceleration of 6.4 m/s2, what will its velocity be after 7s?
I also need the Formula
Hi there!
The formula for velocity given acceleration:
v = at
Plug in given values:
v = 6.4(7) = 44.8 m/s
In which state of matter are molecules fastest?
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
A solid sphere rolls without slipping down an incline, starting from rest. At the same time, a box starts from rest at the same altitude and slides down the same incline, with negligible friction. Which object arrives at the bottom first
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.
What is Friction?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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What is the best description of the distribution of the galaxies that lie within about 200 Mpc of Earth
PLEASE HELP I DONT GET THISS
Answer:
I feel like its the second one but I'm not completely sure..
Explanation:
The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.2 m above the ground, how steep a slope can the truck be parked on without tipping over
The slope of the road can be given as the ratio of the change in vertical
distance per unit change in horizontal distance.
The maximum steepness of the slope where the truck can be parked without tipping over is approximately 54.55 %.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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a man can jump 9meteres on the moon.how high can he jump on the earth.
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.
A car with an initial position of 10.0 m
and an initial velocity of 16.0 m/s accelerates at an average rate of 0.50 m/s2 for 4.0 s. What is the car’s position after 4.0 s?
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
What does Newton's third law describe?
A) the tendency of stationary objects to remain at rest.
B) How the force applied to an object is related to a change in its motion.
C) The four fundamental forces of nature.
D) The forces between two objects
Thanks!!
Answer:
What does Newton's third law describe?
D) The forces between two objectsAn artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.44 m/s2. Determine the orbital period of the satellite.
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]
A football is kicked with an initial velocity of 50.0 m/s, 60° above the horizontal line. Find the following: The time it takes to reach the maximum height; The maximum height reached by the projectile; The time of flight; and The range of projectile.
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.
If you drop your keys from the tallest building in San Antonio, how fast will
they be falling after 3 seconds?
9.8 m/s
0 14,7 m/s
29.4 m/s
44,1 m/s
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
The linear distance traveled by a wheel of radius 50cm after 99 complete revolutions is?
1)99m
2)210m
3)311
4)433
Answer:
3) 311 m
Explanation:
Circumference = 2πR = π m/rev
99 rev(π m/rev) = 99π m or about 311 meters
A rollercoaster car passes the hill which is 5.5m above the ground at speed 9.3m/s, and rolls over the second hill which is 2.5m above the ground, and heads toward the third hill which is 4.0 m higher than the first one. If the track is frictionless,
a. What maximum height will the car climb on the third hill? [h max = 9.9m, so car will climb the entire 9.5m hill]
b. Will the speed of the car on top of the hill 3 be lower or higher than its speed on the top of the hill one? [lower]
c. Calculate the speed of the car when it is 1m lower than the top of the third hill. [5.3m/s]
Would somebody kindly go over the questions :D
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
Why metals have thermoconductivity higher than ceramic?
Answer:
Thermal Conductivity Easily Transmits Heat Among Fine Ceramics
For the ballistic missile aimed to achieve the maximum range of 9500 km, what is the maximum altitude reached in the trajectory
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]
A 64 kg student is standing atop a spring in an
elevator that is accelerating upward at 3.0 m/s2
The spring constant is 3000 N/m.
A) by how much is the spring compressed?
Answer:
192
Explanation:
Nikolai is using a hand-operated grain mill to grind wheatberries into flour. The mill is operated by spinning a fly-wheel with radiusR= 23 cm, which has a handle attachedto the outer edge. After grinding for a few minutes at a con-stant angular speedωi, he lets go of the handle and allows themechanism to come to rest as it undergoes constant angularacceleration. This happens over the course oft= 0.50 s, andthe flywheel undergoes a quarter of a rotation during this time.What is the linear tangential accelerationaof the handle as itcomes to rest? For the limits check, investigate what happenstoaas the time required to stop the flywheel becomes small(t→0).
Answer:
Explanation:
α = Δω/t = (0 - ωi)/0.50 = -2ωi rad/s²
ωf² = ωi² + 2αθ
θ = (ωf² - ωi²) / 2α
2π/4 = (0² - ωi²) / (2(-2ωi))
2π/4 = ωi / 4
ωi = 2π rad/s
α = -2(2π) = -4π rad/s²
a = rα = 0.23(-4π) = 0.92π m/s² ≈ -2.89 m/s²
as the time to stop approaches zero, acceleration goes toward infinity.
A ball is dropped from an 80.0 m building. What is the ball's velocity after 3.00 s? Use an order-of-magnitude estimation to identify the correct choice.
A. -2.9 m/s
B. -29.4 m/s
C -8.8 m/s
D. -88.3 m/s
Answer:b
Explanation:
-29.4 m/s
The velocity of the ball dropped from 80 m if it reaches the ground within 3 seconds is 26.6 m/s. If it is in midway within this time, then the velocity will be 29.4 m/s.
What is velocity ?Velocity of a moving object is the measure of its distance travelled per unit time. Velocity is a vector quantity having both magnitude and direction. Acceleration is the rate of change in velocity.
Given that, height of the building = 80 m
the ball is moving downwards by acceleration due to gravity g = 9.8 m/s².
Then after 3 seconds, the velocity of the ball is calculated as follows:
velocity = acceleration × time
v = 9.8 m/s² × 3 s = -29.4 m/s
If the ball reaches the ground within the time of 3 s, then, the velocity is:
v = 80 m/3s = 26.6 m/s.
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give with an example a cause where the velocity of an object is zero but its acceleration is not zero .
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.
2 W' is the symbol of a) antimony b) gold c) polonium d) tungsten
Answer:
D. Tungsten
Explanation:
W - Wolfram
How do light travels
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:
PLEASE HELP ME WITH THISSSS
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
1.25 is the closest to 1.04 or not I want to answer please. I think it's true, but I want to prove it scientifically, please.
Answer:
in general context yes it is closest to 1.04
Explanation:
theres no right or wrong way to scientifically prove this though.
Overall in scale its closest to 1.04 hope that helped
what would happen if gravity were to stop everywhere?
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.
Which region of electromagnetic spectrum will provide photons of the least energy
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.
1. Explain who is doing more work and why: a bricklayer carrying bricks and placing them on the wallof a building being
constructed, or a project supervisor observing and recording the progress of the
workers from an observation booth.