Answer:
Explanation:
For a person at about 20° North latitude, an object 70° above the celestial equator would never set. It's arc path would touch the horizon be never sink below it. Observers north of 20° see it all night. Observers south of 20° an object 70° above the celestial equator would spend the greatest amount of time above the horizon.
For southern hemisphere observers, the object 40" below the celestial equator will spend the most time above the horizon. Nearly 12 hours per day. Did you mean 40°? 40 seconds is very close to the equator itself. However, the result is the same.
For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.
What is the equator?The Equator is an imaginary line passing through the middle of a globe. It is equidistant from the North Pole and the South Pole, Its is a horizontal line residing at 0 degrees latitude.
For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.
One that is 40" below the celestial equator would be above the horizon for the greatest amount of time for southern hemisphere observers.
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Your friend's Frisbee has become stuck 19 m above the ground in a tree. You want to dislodge the Frisbee by throwing a rock at it. The Frisbee is stuck pretty tight, so you figure the rock needs to be traveling at least 4.1 m/s when it hits the Frisbee.
If you release the rock 1.8 m above the ground, with what minimum speed must you throw it?
Answer:
18.36 m/s
Explanation:
We can solve this using conservation of energy. The energy in the system will be conserved since there are no outside forces acting upon it so the potential energy and kinetic energy will be equal. Giving us this formula to start:
1/2mv^2=mgh
m=mass
g=gravity
h=height
v=velocity
We can start by figuring out the total height the rock travels which we can do by subtracting the height of the frisbee by the height the rock started at.
19m-1.8m=17.2m
Now we can plug in our variables to solve for velocity.
First we negate mass since its on both sides and cancels out leaving us with.
1/2v^2=gh
Plug in.
1/2v^2=(9.8)(17.2)
1/2v^2=168.56
v^2=337.12
v=18.36m/s
If you apply a net force of 100 N to the hoverboard, and it accelerates
2m/s/s, how much mass does it have?
Answer:
50 kgExplanation:
The mass of an object given only the force acting on it and it's acceleration can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force in N
a is the acceleration in m/s²
From the question
f = 100 N
a = 2 m/s²
We have
[tex]m = \frac{100}{2} = 50 \\ [/tex]
We have the final answer as
50 kgHope this helps you
A student catches a can of soda dropped from some unknown height by Mr. Fineman. If
the can was dropped from rest, and it is traveling at 8.9 m/s before it arrives in the
student's hand...
Answer:
v² = u² + 2gh
8.9² = 0² + 2(9.8)h
h = 4.0 m
v = at
t = 8.9/9.8
t = 0.91 s
Challenge
Marks: 1
How much energy is transferred by a 500 watt heater that is on for 2
minutes?
Answer:
Explanation:
500 W = 500 J/s
500 J/s(2 min(60 s/min) = 60000 J
In a police ballistics test, 2.00-g bullet traveling at 700 m/s suddenly hits and becomes embedded in a stationary 5.00-kg wood block. What is the speed of the block immediately after the bullet has stopped moving relative to the block
Answer:
Here we use the conservation of momentum theorem.m stands for mass, and v stands for velocity. The numbers refer to the respective objects.
m1v1 + m2v2 = m1vf1 + m2vf2
Since the equation is perfectly inelastic, the final velocity of both masses is the same. Let’s account for this in our formula.
m1v1 + m2v2 = vf(m1 + m2)
Let’s substitute in our givens.
(0.002 kg)(700 m/s) + (5 kg)(0 m/s) = vf(0.002 kg + 5 kg)
I assume you are proficient in algebra I, so I will not include the steps to simplify this equation.
Note that I have considered the bullet’s velocity to be in the positive direction,
The answer is vf = 0.280 m/s
A wheel with radius 41.5 cm rotates 5.13 times every second.
Find the period of this motion.
What is the tangential speed of a wad of chewing gum stick to the rim of the wheel?
The tangential speed of a wad of chewing gum to the rim of the wheel is approximately 1337.659 centimeters per second.
Let suppose that the wheel rotates at constant angular speed ([tex]\omega[/tex]), in radians per second, the tangential speed of a wad of chewing gum to the rim of the wheel ([tex]v[/tex]), in centimeters per second, is:
[tex]v = 2\pi\cdot r\cdot f[/tex] (1)
Where:
[tex]r[/tex] - Radius of the wheel, in centimeters[tex]f[/tex] - Frequency, in hertzIf we know that [tex]f = 5.13\,hz[/tex] and [tex]r = 41.5\,cm[/tex], then the tangential speed of the chewing gum is:
[tex]v = 2\pi\cdot (41.5\,cm)\cdot (5.13\,hz)[/tex]
[tex]v \approx 1337.659\,\frac{cm}{s}[/tex]
The tangential speed of a wad of chewing gum to the rim of the wheel is approximately 1337.659 centimeters per second.
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A ball is launched from ground level at 30 m/s at an angle of 35° above the horizontal. how far does it go before it is at ground level again
Answer:
Explanation:
Ignoring air resistance
Initial vertical velocity is 30sin35 = 17.2 m/s
Gravity reduces this velocity to zero in a time of
t = v/g =17.2 / 9.8 = 1.755 s
it takes the same time to come back down to ground level for a total flight time of 2(1.755) = 3.51 s
The horizontal velocity is 30cos35 = 24.57 m/s
the distance traveled horizontally is
d = vt = 24.57(3.51) = 86.298... = 86 m
The primary evidence that has led astronomers to conclude that the expansion of the universe is accelerating comes from __________.
The Universe is often studied by Scientist. The primary form of evidence that has led astronomers to conclude that the expansion of the universe is accelerating is the Observations of white dwarf supernovae.
There was the discovery of the faintness of high-redshift Type I supernovae by a team of scientist which showed that the expansion of the universe is accelerating.
The key evidence for that made scientist to talk about the expansion of the universe is accelerating comes from viewing and studying of white dwarf supernovae. It has been found that most distant stars all orbit at approximate speed as stars found about 30,000 light-years from their center.
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If you have to measure the temperature > above 80°c, do you use an alcohol or mercury thermometer? why?
Answer:
I use mercury thermometer to measure the temperature above 80°C because the boiling point of alcohol is 78°C . So, it can't measure the temperature above 78°C.
3. A car travelling at 12 m/s into a stationary truck of about 10 times the cars mass. a. If the collision was completely inelastic, what velocity would the two travel at if the stuck together? b. If the collision was completely elastic, what would be the velocities of the car and truck after the collision? c. In order to exert a force of only 3500N on the truck during the collision, how much time would the collision have to take?
(a) The final velocity of the two vehicles if the collision was inelastic is 1.1 m/s.
(b) For the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.
(c) The time taken to exert the given force is 0.00625 m (s).
The given parameters;
Initial velocity of the car, u₁ = 12 m/sInitial velocity of the truck, u₂ = 0Mass of the car, = mMass of the truck, = 10m(a) The final velocity of the two vehicles if the collision was inelastic is calculated as follows;
[tex]m_1 u_1 + m_2u_2 = v(m_1+ m_2)\\\\12m + 10m(0) = v(m + 10m)\\\\12m = v(11m)\\\\v = \frac{12m}{11m} \\\\v = 1.1 \ m/s[/tex]
(b) The final velocity of the two vehicles if the collision was elastic is calculated as follows;
[tex]m_1 u_1 + m_2u_2 = m_1v_1 + m_2v_2\\\\\12m \ + \ 10m(0) = mv_1 + 10mv_2\\\\12m = m(v_1 + 10v_2)\\\\12 = v_1 + 10 v_2\ \ - --(1)[/tex]
Apply one-directional velocity equation:
[tex]u_1 +v_1 = u_2 + v_2\\\\12 + v_1 = 0 + v_2\\\\12+ v_1 = v_2 \ \ --- (2)[/tex]
Substitute the value of [tex]v_2[/tex] into equation (1);
[tex]12 = v_1 + 10(12 + v_1)\\\\12= v_1 + 120 + 10v_1\\\\12- 120 = 11v_1\\\\-108 = 11v_1\\\\v_1 = \frac{-108}{11} \\\\v_1 = -9.81 \ m/s\\\\[/tex]
Solve for [tex]v_2[/tex];
[tex]v_2 = 12 + v_1\\\\v_2 = 12 - 9.81\\\\v_2 = 2.19 \ m/s[/tex]
Thus, for the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.
(c)
The change in the momentum of the truck is calculated as;
[tex]\Delta P = m_2(v_2 - u_2)\\\\\Delta P = 10m(2.19)\\\\\Delta P = 21.9m[/tex]
The time taken to exert the given force is calculated as follows;
[tex]Ft = \Delta P\\\\t = \frac{\Delta P}{F} \\\\t = \frac{21.9 \ m}{3500} \\\\t = 0.00625 \ m (seconds)[/tex]
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A person sitting in a chair with wheels stands up, causing the chair to roll backward across the floor. The momentum of the chair
a. was zero while stationary and increased when the person stood.
b. was greatest while the person sat in the chair.
c. remained the same.
d. was zero when the person got out of the chair and increased while the person sat.
Answer:
a. was zero while stationary and increased when the person stood.
Explanation:
momentum is mass times velocity.
initial velocity was zero
final velocity was NOT zero.
is it true that playing badmenton help you to become a better person?
Answer:
There is no scientific evidence that playing specifically l badminton makes you a better person, but sport and exercise in general release hormones which can make you feel more happy therefore making you nicer to the people around you and 'a better person'.
Answer:
It's true because playing any sport makes a person happy. So a happy person is a better person.
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half-life questionnnnn:
I need help. please look at the image below and let me know I need this by 7:20 am pst.
Answer:
3(1.5) = 4.5 V
Explanation:
WILL GIVE BRAINLIEST!!! NO LINKS PLZ!!!!
A 225 g hockey puck is sliding on ice in an arena towards the end boards that are 15.7 m away. The puck is travelling 12.0 m/s when it slides into some rough ice (coefficient of kinetic friction= 0.550).
Determine:
a) the acceleration of the puck on the rough ice.
b) the distance from the end boards the puck is when it comes to a stop.
Please show work.
Answer:
Explanation:
a) the acceleration of the puck on the rough ice.
a = μg = 0.550(9.81) = 5.3955 = 5.40 m/s²
(comes from μ = F/N = ma/mg = a/g)
b) the distance from the end boards the puck is when it comes to a stop.
v² = u² + 2as
0² = 12.0² + 2(-5.40)s
s = 13.3 ft
so distance from the boards is
15.7 - 13.3 = 2.4 m
by the way...that's some VERY rough ice...more like sand.
Help me outtttt jejjejejeje
Answer:
do it got a picture
Explanation:
HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!pls
Which car is experiencing negative acceleration?
A. a car sitting at a red light
B. a car slowing down as it approaches a red light
C. a car driving in a circle at a constant speed
D. a car changing its speed from 0 km/hr to 10 km/hr
Answer:B
Explanation:
Answer: b
Explanation: b as it slows down and decelerates
2. An object is falling under gravity with terminal velocity. What is happening to its speed?
A. It is decreasing to a lower value.
B. It is increasing
C. It is decreasing to zero.
D. It is staying constant.
Answer:
the speed of the object has become constant.
Explanation:
At terminal velocity, air resistance equals in magnitude the weight of the falling object. Because the two are oppositely directed forces, the total force on the object is zero, and the speed of the object has become constant.
A 10kg object is 15 meters up a hill. Find its potential energy
Answer:
Explanation:
Relative to an origin at the bottom of the hill,
PE = mgh = 10(9.8)(15) = 1470 J
Explain how the linear rate spring operates?
How might "Roller Coaster Physics" have been different if the author's purpose had been to warn
readers of the dangers of roller coasters?
Use at least two details from the passage to support your answer.
Answer: The answer is hupa loopa
Explanation:
A solid, uniform sphere with a mass of 2.5 kg rolls without slipping down an incline plane starting from rest at a vertical height of 19 m. If the sphere has a radius of 0.60 m, what is the angular speed of the sphere at the bottom of the incline plane
Answer:
1/2 m v^2 + 1/2 I ω^2 = m g h conservation of energy
I = 2/5 m R^2 inertia of solid sphere
1/2 m v^2 + 1/5 m ω^2 R^2 = m g h
1/2 v^2 + 1/5 v^2 = g h
v^2 = 10 g h / 7 = 1.43 * 9.80 * 19 m^2/s^2 = 266 m^2/s^2
v = 16.3 m/s
v = R ω
ω = 16.3 / .6 = 27.2 / sec
At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11, how much time does it take the puck to come to rest?
At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11, time it take the puck to come to rest is 4.51 sec.
What is speed?The speed of an item, which is a scalar quantity in everyday usage and kinematics, is the size of the change in that object's position over time or the size of the change in that object's position per unit of time.
Given in the question at the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11,
force = ma = μmg, putting the value,
a = - 1.175 m/sec²
now using equation of motion
v = u + at
t = 4.51 sec
So, time taken by the puck to come to rest is 4.51 sec.
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a heat engine is a device that uses to produce useful work
A heat engine is a device that uses to produce useful work.
Definition - a device for producing motive power from heat, such as a gasoline engine or steam engine.
So..
If this is a true or false question.. Your answer is:
TRUE
Answer:
HEAT
Explanation:
A circular disk of radius 0.200 m rotates at a constant angular speed of 2.50 rev/s. What is the centripetal acceleration (in m/s2) of a point on the edge of the disk?
[tex]a_c = 3.14\:\text{m/s}^2[/tex]
Explanation:
First, we need to convert the given angular speed [tex]\omega[/tex] from rev/s to rad/s:
[tex]2.50\:\dfrac{\text{rev}}{\text{s}}×\dfrac{2\pi\:\text{rad}}{1\:\text{rev}} = 15.7\:\text{rad/s}[/tex]
The centripetal acceleration [tex]a_c[/tex] is defined as
[tex]a_c = \dfrac{v^2}{r}[/tex]
Recall that [tex]v = r\omega[/tex] so we can write [tex]a_c[/tex] as
[tex]a_c = \dfrac{(r\omega)^2}{r} = \omega^2r[/tex]
[tex]\;\;\;\;\;=(15.7\:\text{rad/s})^2(0.200\:\text{m}) = 3.14\:\text{m/s}^2[/tex]
A 200-kg, 2.0-m-radius, merry-go-round in the shape of a flat, uniform, circular disk parallel to level ground is rotating at 1.2 cycles/second about an axis through its center of mass and perpendicular to the ground. A 50-kg boy jumps onto the edge of the merry-go round and lands at a fixed point. What is the angular velocity of the merry-go-round after the boy lands on it
Answer:
Explanation:
Conservation of angular momentum.
Disk I = ½MR²
Point mass I = mR² (boy)
Initial angular momentum
L₀ = Iω = ½MR²ω₀
Final angular momentum
L₁ = Iω = (½MR² + mR²)ω₁
as momentum is conserved, these are equal
(½MR² + mR²)ω₁ = ½MR²ω₀
ω₁ = ω₀(½MR²/ (½MR² + mR²))
ω₁ = ω₀(½M/ (½M + m))
ω₁ = 1.2(½(200)/ (½(200) + 50))
ω₁ = 1.2(⅔)
ω₁ = 0.8 cycles/second or 0.8(2π) = 1.6π rad/s
A wagon of dog treats (combined mass 55 kg) is rolling at 2.1 m/s. A dog with mass 21 kg dives into the wagon, colliding with just enough momentum to make both stop. If the collision between the dog and the wagon lasts 0.1 s, what is the magnitude of the average force that will be exerted on the dog by the collision with the wagon
Answer:
Explanation:
An impulse results in a change of momentum
If the wagon and dog both stop, they must have had equal and opposite momentums
FΔt = mΔv
F = mΔv/Δt = m(v₁ - v₀)/(t₁ - t₀)
v₁ = t₀ = 0
F = m(v₀)/t₁
F = 55(2.1)/0.1 = 1155 N
We could have also figured the dog's initial velocity and used the dog's mass in the equation as well. Result would be identical.
which one would cool down faster, water or a block of concrete? why?
Answer:
If you mean by temperature
Water, cause say the water just sits there, it'll take a few minutes to cool down. But for concrete having all the hot sun rays on it during summer, can make it very hot, if you put 'cold water' there, it'll only last there for a moments (and that's by effect not waiting), if you wait for it to cooldown by itself, it'll take quite awhile.
If you mean solidify:
The stuff required to make concretes takes HOURS and HOURS to finish solidifying, unlike water being turn into ice normally only takes about half an hour or an hour (maybe 2) in the freezer (I don't know how long for ice, but not long at all in comparison to concrete).
Explanation:
Explanation was mostly part of the answer
A 12-kg block on a horizontal frictionless surface is attached to a light
spring (force constant = 0.80 kN/m). The block is initially at rest at its
equilibrium position when a force (magnitude P = 80 N) acting parallel to
the surface is applied to the block, as shown. What is the speed of the
block when it is 13 cm from its equilibrium position?"
The speed of the block at the displacement from the equilibrium position is 1.062 m/s.
The given parameters:
Mass of the block, m = 12 kgSpring constant, k = 0.8 kN/mExtension of the spring, x = 13 cm = 0.13 mApplied parallel force, F = 80 NThe speed of the block is calculated by applying the principle of conservation of mechanical energy as shown below;
[tex]\frac{1}{2} mv^2 = \frac{1}{2}kx^2\\\\mv^2 = kx^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m} } \\\\v = \sqrt{\frac{800 \times 0.13^2}{12} } \\\\v = 1.062 \ m/s[/tex]
Thus, the speed of the block at the displacement from the equilibrium position is 1.062 m/s.
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Answer:
The speed of the block at the displacement from the equilibrium position is 1.1266 m/s.
Step-by-step explanation:
Solution :
Using principle of conservation of mechanical energy formula to find the speed of the block :
[tex]\begin{gathered} \longrightarrow{\pmb{\sf{\frac{1}{2} mv^2 = \frac{1}{2}kx^2}}}\end{gathered}[/tex]
»» m = Mass of the block, »» k = Spring constant,»» x = Extension of the spring»» F = Applied parallel forceAs per given data information in the question we have :
✧ Mass of the block = 12 kg✧ Spring constant = 0.8 kN/m✧ Extension of the spring = 0.13 m✧ Applied parallel force = 80 NSubstituting all the given values in the formula to find the speed of the block
[tex]\longrightarrow{\sf{ \: \:\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: \cancel{\dfrac{1}{2}}mv^2 = \cancel{\dfrac{1}{2}}kx^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: mv^2 = kx^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v^2 = \dfrac{kx^2}{m}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: \sqrt{{v}^{2} } = \sqrt{ \dfrac{kx^2}{m}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{kx^2}{m}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{800 \times {0.13}^{2}}{12}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{800 \times {0.13} \times 0.13}{12}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{800 \times 0.0169}{12}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{13.52}{12}}}}[/tex]
[tex]{\star{\underline{\boxed{\rm{\red{ v \approx 1.1266 \: m/s}}}}}}[/tex]
Hence, the speed of block is 1.1266 m/s.
[tex] \rule{300}{1.5}[/tex]
