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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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
Where do inherited traits come from
Answer:
Your parents or anyone in your ancestry.
Explanation:
A spring in a dart gun is compresscht a distance of 0.05 m. The spring has a spring constant
of 1,115 N/m. If the dart has a mass of 0.025 kg, determine the velocity of the dart as it
leaves the dart gun.
Answer:
Explanation:
ASSUMING that the dart is fired horizontally so that gravity potential energy considerations are not needed. Also ignoring friction work.
The spring potential will convert to kinetic.
KE = PS
½mv² = ½kx²
v = [tex]\sqrt{kx^2/m}[/tex]
v = [tex]\sqrt{1115(0.05^2)/0.025}[/tex]
v = 10.55935...
v = 11 m/s
During what time period does the balloon drift in a counterclockwise direction?
A 5.0 m length of rope, with a mass of 0.52 kg, is pulled taut with a tension of 46 N. Find the speed of waves on the rope
Answer:
Speed of waves on the rope is 21 m/s
Explanation:
Length of the rope (l) = 5.0 m
Mass of the rope (m) = 0.52 kg
Tension in the rope (T) = 46 N
Formula of speed of waves on the rope:
[tex] \bold{v = \sqrt{\dfrac{T}{\mu}}} [/tex]
[tex] \mu [/tex] = Mass per unit length of the rope (m/l)
By substituting the values in the formula we get:
[tex] \implies \rm v = \sqrt{\dfrac{T}{ \dfrac{m}{l} }} \\ \\ \implies \rm v = \sqrt{\dfrac{Tl}{m}} \\ \\ \implies \rm v = \sqrt{ \dfrac{46 \times 5}{0.52} } \\ \\ \implies \rm v = \sqrt{ \dfrac{230}{0.52} } \\ \\ \implies \rm v = \sqrt{442.3} \\ \\ \implies \rm v = 21 \: m {s}^{ - 1} [/tex]
Speed of waves on the rope (v) = 21 m/s
7. What is the velocity of an object with a distance of 90m south and a time of
5s?
Answer:
Explanation:
v= s/t
V =90m/5s
V = 8m/s
"A soccer ball is kicked horizontally off a 22 m high hill and lands a distance of 35 m from the edge of the hill." Which variable is 35 m? *
Answer:Let’s assume that, after the soccer ball is kicked and moves through its trajectory, it first makes contact with level ground a horizontal distance of 35 meters from where it was kicked. Let’s also assume that we can neglect air resistance. The time, t, that the soccer ball is in the air until it first contacts the ground can be found from the equation h = (1/2)gt^2 which can be rewritten as t = sqrt(2h/g) where h is the vertical distance the ball falls which is the height of the hill since the ball was kicked horizontally, and g is the acceleration of gravity which is 9.8 m/s^2. So t = sqrt(2(22)/9.8) = 2.12 seconds. In that time, the ball travelled 35 meters so its horizontal velocity was 35 meters/2.12 seconds = 16.5 meters/second.
Explanation:
a 4kg sample of water absorbs 500 joules of energy.How much did the water temp change.The specific heat of water is 4200 J/(kg C)
Answer:
0.02976°C
Explanation:
Heat Supplied = mcø
ø = 500/(4*4200)
ø = 0.02976 °C
True or False If the mass of the object increases, then the potential energy of the object decreases.
One potato plant makes 7.5
potatoes. There are 10.000
people on a ship and 3
potatoes per person. How
many potato plants are
needed per person.
If all conditions are ideal, you may harvest about five to 10 potatoes per plant for your gardening efforts. Yields are based on both the care your give your plants during the growing season and the variety of potatoes you choose to grow.
answer
7.5×10.000=75÷3=25
CAN SOMEONE PLZ HELP
Answer:
magnetic force.
Explanation:bc it makes sense, and can i please get brainliest answer i never asked. its ok if you say no. have a great day <3.
Find the net torque .
Answer:
Explanation:
I will ASSUME this means torque about the dot.
3) 20(3) + 10(6) - 30(4) = 0 N•m
4) 10(0.5) - 6sin45(1) = -0.7573593... or about 0.76 N•m CCW
5) 25(3) - 40sin30(4) = -5 N•m or 5 N•m CCW
6) 15(3) - 12(2) - 10sin45(6) = -21.4264068... or about 21 N•m CCW
what torque required stopping awheel of moment of inertia 6 × 10^-3kgm2 from speed of 40rad/s in 20 sec.
solution:
the formula is T = F * r * sin(theta) so just input the numbers and solve it.
Explanation:
Torque is the twisting force that tends to cause rotation. The point where the object rotates is known as the axis of rotation. Mathematically, torque can be written as T = F * r * sin(theta), and it has units of Newton-meters
What is the net force here?
11 N left
6 N right
1 N right
4 N right
answer = 6n to the right
Explanation:
2n plus 4n equals 6n
since 6n is more than 5n it goes 6n to the right
A 2 kg object being pulled across the floor with a speed of 10 m/sec is suddenly
released and slides to rest in 5 sec. What is the magnitude of the frictional force
producing this deceleration?
Answer:
The frictional force producing this deceleration would have a magnitude of [tex]4\; \rm N[/tex].
Explanation:
The velocity of this object changed by [tex]\Delta v = (-10\; \rm m\cdot s^{-1})[/tex] in [tex]\Delta t = 5\; \rm s[/tex]. The acceleration of this object would be:
[tex]\begin{aligned}a &= \frac{\Delta v}{\Delta t} \\ &= \frac{-10\; \rm m\cdot s^{-1}}{5\; \rm s} = -2\; \rm m\cdot s^{-2}\end{aligned}[/tex].
Let [tex]m[/tex] denote the mass of this object. By Newton's Second Law of Motion, the net force on this object would be:
[tex]\begin{aligned}F &= m \, a \\ &= 2\; \rm kg \times (-2\; \rm m\cdot s^{-2}) \\ &= -4\; \rm N\end{aligned}[/tex].
([tex]1\; {\rm kg \cdot m \cdot s^{-2} = 1\; {\rm N}[/tex].)
If the floor is level, friction would be the only unbalanced force on this object. Thus, the magnitude of the frictional force on this object would also be [tex]4\; {\rm N}[/tex], same as the magnitude of the net force on this object.
Based on the law of conservation of mass, if approximately 20g of water reacts in the following equation,
producing about 15g of oxygen, what mass of hydrogen (H2) was produced in this reaction?
2H20
2H2 + O2
Answer: H2O2
Explanation:
First, calculate the moles of H2O2 reacting. In order to do this, we must evaluate the relative molecular mass ( Mr ) of hydrogen peroxide
You are on the roof of a building 30 m above the ground. Using hand crank 3 points
you are to lift a 300 kg dresser on to the roof. How much work must to
generated to lift the dresser? *
Answer:
The work will increase the potential energy.
W = PE = mgh = 300(9.8)(30) = 88200 J
You will create your own circuit , discuss how electricity flows and describe how circuits are made in
Answer:
An electric current flows in a loop,powering bulbs or other electric COMPONENTS. The loop is an electric circuit. A circuit is made up of various components linked together by wires. The current is driven around the circuit by a power source, such as a BATTERY.
Voltage: is the energy given to each unit of charge that flows in a circuit
Current: is the amount of electric charge flowing past a point in a circuit each second
Wattage: is the amount of electrical energy a circuit uses each second
3. a car takes off with initial velocity of 20m/s and move with uniform acceleration of 12m/s2 for 20s. It maintained a constant velocity for 30s and come to rest with uniform acceleration of 2m/s2. Calculate the total distance covered by the car.
The total distance traveled by the car at the given velocity and time is 900 m.
The given parameters:
initial velocity of the car, u = 20 m/sacceleration of the car, a = 12 m/s²time of motion of the car, t = 20 sfinal time = 30 sfinal acceleration = 2 m/s²The final time of motion of car before coming to rest is calculated as follows;
[tex]v_f = v_0 -at\\\\0 = 20 - 2t\\\\t = 10 \ s[/tex]
The graph of the car's motion is in the image uploaded.
The total distance traveled by the car is calculated as follows;
[tex]total \ distance = A \ + B \ + C\\\\total \ distance = (\frac{1 }{2} \times 20 \times 20) \ + (20 \times 30) \ + (\frac{1}{2}\times 10 \times 20)\\\\total \ distance = 900 \ m[/tex]
Thus, the total distance traveled by the car at the given velocity and time is 900 m.
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A roll of kitchen aluminum foil is 30 cm wide by 22 m long (if you unroll it). If the foil is 0.15 mm thick, and the specific weight of aluminum is 26460 N/m3, how much does the roll of aluminum foil weigh
The weight of the aluminum foil is 26.20 N
To find the weight of the aluminum foil when it is unrolled, we need to find its volume.
The volume V = lwt where
l = length of aluminum foil = 22 m, w = width of aluminum foil = 30 cm = 0.30 m and t = thickness of aluminum foil = 0.15 mm = 0.15 × 10⁻³ m.So, V = lwt
= 22 m × 0.30 m × 0.15 × 10⁻³ m
= 0.99 × 10⁻³ m³.
So, its weight W = ρV where
ρ = specific weight of aluminum = 26460 N/m³ and V = volume of aluminum foil = 0.99 × 10⁻³ m³So, W = ρV
W = 26460 N/m³ × 0.99 × 10⁻³ m³
W = 26195.4 × 10⁻³ N
W = 26.1954 N
W ≅ 26.20 N
So, the weight of the aluminum foil is 26.20 N
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AnswAnswer This!!!!!!
I'll give brainliest to whoever gets it right.
I need ideas of what kind of simple motor i can build and how i can build it. The simple motor MUST spin without using your own force. What materials would i use and how would i create it. what would i create
Answer:
i don't know but my father i think he can't answer this
what is the force that every mass experts on every other mass called?
Answer: The forces of gravity
Explanation: The consequence of this phenomenon is that every mass exerts a so-called "force of mutual attraction" on every other mass. The attractive force that the celestial bodies exert on other masses by virtue of their total mass is called the force of gravity.
Hope this helps
Dagmar says that diffusion happens really quickly. Is he right or wrong? Explain.
Answer:
Diffusion in gases is quick because the particles in a gas move quickly. It happens even faster in hot gases because the particles of gas move faster.
define parking orbit?
Answer:
An orbit of a spacecraft from which the spacecraft or another vehicle may be launched on a new trajectory.
Hey there!
having a question which is from ( JEE Advanced 2014 )
[tex]\huge\color{black}\boxed{\colorbox{pink}{Question♡}}[/tex]
A rocket is moving in a gravity free space with a constant acceleration of 2 −2 along +x direction (see figure). The length of a chamber inside the rocket is 4 m. A ball is thrown from the left end of the chamber in +x direction with a speed of 0.3 −1 relative to the rocket. At the same time, another ball is thrown in –x direction with a speed of 0.2 −1 from its right end relative to the rocket. The time in seconds when the two balls hit each other is?
[tex] \\ \\ [/tex]
proper explanation needed~
thankyou~
The time taken for the two balls to hit each other is 8 s.
The given parameters:
Acceleration of the rocket, a = 2 m/s²Length of the chamber, s = 4 mSpeed of the first ball, = V1 = 0.3 m/sSpeed of the second ball, V2 = 0.2 m/sThe time taken for the two balls to hit each other is calculated by applying relative velocity formula as shown below;
[tex](V_1 - (-V_2) )t = s\\\\(V_1 + V_2) t = s\\\\(0.3 + 0.2) t = 4\\\\0.5t = 4\\\\t = \frac{4}{0.5} \\\\t = 8 \ s[/tex]
Thus, the time taken for the two balls to hit each other is 8 s.
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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
PLEASE ANSWER THIS, I'LL MARK YOU AS BRAINLIEST
Carlito was observing an an that crawled along a table. With a piece of chalk, he followed his path. He determined the ant’s displacements by using a ruler and protractor. The displacements were as follows: 2 cm east, 3.5 cm, 32° north of east and 2.4 cm, 22° west of north. Find the resultant vector using graphical method.
Answer:
this is the answer
Explanation:
hope it's clear
pls Mark me as the brainliest pls
An object of mass 6.36 kg is released from rest and drops 2.05 m to the floor. The collision is completely inelastic. How much kinetic energy is lost during the collision
Answer:
Essentially all of it
Explanation:
The potential energy was
PE = mgh = 6.36(9.81)(2.05) = 127.90278 = 128 J
ignoring air resistance, this PE converts to KE
With no rebound final velocity is zero, so Kinetic energy lost = 128 J
A disgruntled physics student, frustrated with
finals, releases his tensions by bombarding the
adjacent building, 13.5 m away, with water
balloons. He fires one at 38◦
from the horizontal with an initial speed of 23.6 m/s.
The acceleration of gravity is 9.8 m/s
2
.
For how long is the balloon in the air?
Answer:
Explanation:
The balloon would require a time of
t = d/v = 13.5/ (23.6cos38) = 0.7259...s
to travel the horizontal distance.
the vertical position relative to the throw point at that time is
h = 0 + (23.6sin38)(0.7259) + ½(-9.8)(0.7259²)
h = 7.9652...
so as long as the adjacent building is at least 8.0 m higher than the student position, the balloon is in the air for 0.726 s.
If the building is shorter than 8.0 m above the student, the balloon will land on the building roof and will be in the air for a longer period of time