Answer:
c) It decreases.
Explanation:
If you treat this as a case of momentum of the sled while sledding on a friction-less ice, then mass and velocity are directly proportional. This means increasing either mass or velocity , the momentum of the object increases proportionally.
In this question, when the child falls sideways off the sled, the mass reduces, the velocity experiences the same causing a decrease in momentum .
what problems might it cause if a company tried to recycle materials without sorting them first? at least 2 or 3 sentences please!!!!
Answer:
They would probably not be looking good.. like if u don't sort it out they won't be shaped well and there are many types of plastic and if we combine them together they won't give the correct form..
hope this helps :)
A hockey puck attached to a horizontal spring oscillates on a frictionless, horizontal surface. The spring has force constant 4.50 N/m and the oscillation period is 1.20 s. What is the mass of the puck
Answer:
m = 0.164 kg
Explanation:
T (period)
k (force/spring constant)
m (mass)
T = 2*Pi*sqrt(m/k)
T/(2*Pi) = sqrt(m)/sqrt(k)
(T/(2*Pi))*sqrt(k) = sqrt(m)
m = ((T/(2*Pi))*sqrt(k))^2
m = 4.5*((1.2/(2*Pi)))^2
m = 0.1641403175
Please help with this question
Answer:
I believe the answer is 5718.75. Respond if it is wrong please.
Explanation:
I used a calculator.
The velocity of a Bus is reduced uniformly from 15 m/s to 7 m/s while traveling a distance of 90 m.
i. Compute the acceleration.
ii. How much further will the Bus travel before coming to rest, provided the acceleration remain constant?
Answer:
i) a = 0.977 [m/s²]
ii) x = 115.06 [m]
Explanation:
In order to determine the acceleration, we must use the following equation of kinematics.
[tex]v_{f} ^{2} =v_{o} ^{2} -2*a*x[/tex]
where:
Vf = final velocity = 7 [m/s]
Vo = initial velocity = 15 [m/s]
a = acceleration [m/s²]
x = displaciment = 90 [m]
Now replacing:
[tex](7)^{2} =(15)^{2} -2*a*90\\2*a*90 = 15^{2} - 7^{2} \\180*a=176\\a=0.977[m/s^{2}][/tex]
When the bus coming to rest.
[tex]v_{f}^{2} =v_{o} ^{2} -2*a*x\\0 = 15^{2} -2*0.977*x\\x = 115.06[m][/tex]
A girl swings a 0.250 kg rock attached to a taut string in a circle around her head. Her hand holds the end of the string above her head, and the string angles down slightly (11.9° below the horizontal.). The string is massless and 0.75m long.
A coordinate system lies with its origin a the location where the string comes out of the girl's hand. The positive z-axis points vertically. the projection onto the horizontal plane is such that the string makes an angle of 34.6° with the x-axis.
At a certain instant, the rock makes 2.50 revolutions per second (rev/s) in a counterclockwise direction as seen from above.
What is the tangential velocity vector at this instant?
Complete Question
The diagram of with this question is shown on the first uploaded image
Answer:
The value is [tex]v = -6.543 \^ i + 9.47 \^ j + 0 \^ k[/tex]
Explanation:
From the question we are told that
The mass of the rock is [tex]m = 0.250 \ kg[/tex]
The length of the string is [tex]L = 0.75 \ m[/tex]
The angle the string makes horizontal is [tex]\theta = 11.9^o[/tex]
The angle which the projection of the string onto the xy -plane makes with the positive x-axis is [tex]\phi = 34.6^o[/tex]
The angular velocity of the rock is [tex]w = 2.50 rev/s = 2.50 * 2\pi =15.7 \ rad/s[/tex]
Generally the radius of the circle made by the length of the string is mathematically represented as
[tex]r = L cos(\theta )[/tex]
=> [tex]r = 0.75 cos(11.9 )[/tex]
=> [tex]r = 0.734 \ m[/tex]
Generally the resultant tangential velocity is mathematically represented as
[tex]v__{R}} = w * r[/tex]
=> [tex]v__{R}} = 15.7 *0.734[/tex]
=> [tex]v__{R}} = 11.5 \ m/s[/tex]
Generally the tangential velocity along the x-axis is
[tex]v_x = -v__{R}} * sin(\phi)[/tex]
=> [tex]v_x =- 11.5 * sin(34.6)[/tex]
=> [tex]v_x = -6.543 \ m/s[/tex]
The negative sign show that the velocity is directed toward the negative x-axis
Generally the tangential velocity along the y-axis is
[tex]v_y = v__{R}} * cos(\phi)[/tex]
=> [tex]v_y = 11.5 * cos(34.6)[/tex]
=> [tex]v_y = 9.47 \ m/s[/tex]
Generally the tangential velocity along the y-axis is
[tex]v_z = v__{R}} * cos(90)[/tex]
=> [tex]v_z = 0 \ m/s[/tex]
Generally the tangential velocity at that instant is mathematically represented as
[tex]v = -6.543 \^ i + 9.47 \^ j + 0 \^ k[/tex]
Two protons are on either side of an electron as shown below:
Diagram
The electron is 30 µm away from the proton on its left and 10 µm away from the proton on its right. What is the magnitude and direction of the net electric force acting on the electron? Note that
P=1.6×10(-19) c=-e
Select one:
a. 2.0×10−18N to the right
b. 7.0×10−24N to the right
c. 2.0×10−18N to the left
d. −9.59×10−24N to the right
e. 7.0×10−24N to the
Answer:
Choice a. approximately [tex]2 \times 10^{-18}\; \rm N[/tex] to the right.
Explanation:
Look up the Coulomb constant, [tex]k[/tex]:
[tex]k \approx 8.987552 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}[/tex].
Coulomb's Law may be used to find the magnitude of the electric force between the electron and each proton.
Let [tex]q_1[/tex] and [tex]q_2[/tex] denote the magnitude of electric charge on two point charges. Let [tex]r[/tex] denote the distance between these two charges.
By Coulomb's Law, the magnitude of the force between these two point charges would be:
[tex]\displaystyle F = \frac{k\, q_1\, q_2}{r^2}[/tex].
The unit of distance in the Coulomb's constant here[tex]k \approx 8.987552 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}[/tex] is meters. However, the distance between the electron and each proton is given in micrometers, [tex]\rm \mu m[/tex]. Note that [tex]1\; \rm \mu m = 10^{-6}\; \rm m[/tex].
Convert these units to meters:
[tex]30 \; \rm \mu m = 30 \times 10^{-6}\; \rm m[/tex].
[tex]10 \; \rm \mu m = 10 \times 10^{-6}\; \rm m[/tex].
The question states that the magnitude of electric charge on each proton and each electron is [tex]1.6\times 10^{-19}\; \rm C[/tex].
Using these information, find the magnitude of the electric force between the electron and each proton:
Between the electron and the proton on the left-hand side of the electron: [tex]\begin{aligned}F &= \frac{k\, q_1\, q_2}{r^2}\\ &\approx \frac{\left(8.987552\times 10^{9}\; \rm N \cdot m^2\cdot C^{-2}\right) \times \left(1.6 \times 10^{-19}\; \rm C\right) \times \left(1.6 \times 10^{-19}\; \rm C\right)}{\left(30 \times 10^{-6}\; \rm m\right)} \\ &\approx 0.26 \times 10^{-18}\; \rm N\end{aligned}[/tex].Between the electron and the proton on the right-hand side of the electron:[tex]\begin{aligned}F &= \frac{k\, q_1\, q_2}{r^2}\\ &\approx \frac{\left(8.987552\times 10^{9}\; \rm N \cdot m^2\cdot C^{-2}\right) \times \left(1.6 \times 10^{-19}\; \rm C\right) \times \left(1.6 \times 10^{-19}\; \rm C\right)}{\left(10 \times 10^{-6}\; \rm m\right)} \\ &\approx 2.56 \times 10^{-18}\; \rm N\end{aligned}[/tex].The two protons in this question will both attract the electron. Therefore, the force between the electron and the proton on the left-hand side of the electron would point to the left. Similarly, the force between the electron the proton on the right-hand side of the electron would point to the right.
The magnitude of the net electric force on the electron would be:
[tex]2.56 \times 10^{-18}\; \rm N - 0.26\times 10^{-18}\;\rm N \approx 2\times 10^{-18}\; \rm N[/tex].
These two forces act along the same line in opposite direction. Therefore, the resultant force of these two force would be in the direction of the larger force of the two.
In this question, the electric force between the electron and the proton on the right-hand side of the electron is larger. Hence, the net electric force of the protons on the electron should point to the right (towards the proton that is closer to the electron.)
When you use a match to light a candle _______________________________ transformations of energy
Answer:
thermal or heat energy
7. How many forces are acting on this object? *
Fnorm
Ffrict
Fapp
Fgray
Å
Answer:
4
Explanation:
As you can see in the free body diagram there are 4 forces acting on the body.
5
b. What is the molecular shape of the molecule? (3 points)
Answer:
tetrahedral shape.
Explanation:
Answer:
Molecular Geometries. The VSEPR theory describes five main shapes of simple molecules: linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral. Apply the VSEPR model to determine the geometry of molecules where the central atom contains one or more lone pairs of electrons.
fun facts are,
Butterflies taste with their hind feet.
A group of crows is called murder.
Tomato sauce was sold in the 1800's as medicine.
A donkey will sink in quicksand, but a mule won't.
Lakes Can Explode. ...
Earth Used to Be Purple. ...
60 Tons of Cosmic Dust Fall to Earth Daily. ...
The Oceans Hold $771 Trillion Worth of Gold. ...
Earth Once Had Two Moons. ...
Earth Might Still Have Two Moons. ...
The Greatest Vertical Drop Is in Canada. ...
The Hottest Spot in the World Is in California.
Answer:
that is cool and i have one interesting fact
Explanation: North Korea and Cuba are the only places you can't buy Coca-Cola