Answer:
Explanation:
General formula
m1 * vi + m2*v2 = m1*v3 + m2*v4
Givens
m1 = 5
m2 = 2.5
v1 = 8 m/s
v2 = - 4 m/s
v3 = -4 m/s
v4 = x
Solution
5 * 8 - 2.5 * 4 = 5 * -4 + 2.5*x
40 - 10 = -20 + 2.5x
30 = - 20 + 2.5x
50 = 2.5x
x = 50/2.5
x = 20 m/s in the positive direction
Remark
Does this answer make sense? It should. You have 5 kg moving 8m/s in the plus direction. That's a lot of momentum. In addition after the collision, it turns around which is more momentum needed.
It has to give up that extra momentum to the 2.5 kg mass.
Which of the following is another name for a convex lans?
O Diverging liens
3. Converging lens
O c Shrinking lans
OD Security lans
Answer:
3. converging lens
Explanation:
When the rays of light coming parallel to principle axis after refraction through the lens passes through a point called focus, since it converge all the ray at one point, that is why it is said to be converging lens.
A pumpkin is thrown horizontally off of a building at a speed of 2.5 — and travels a horizontal distance of
12 m before hitting the ground. We can ignore air resistance.
What is the vertical velocity when it hits the ground
Answer:
Explanation:
This is a missile throwing exercise, let's find the distance
x = v₀ₓ / t
t = v₀ₓ / x
let's calculate
t = 2.5 / 12
t = 0.2083 s
as time is a scalar this is the same value for descends to the ground
y = v_{oy} t - 1/2 g t²
we calculate
v_y = 0 - 9.8 0,2083
v_y = - 2.04 m / s
the negative sign indicates that the speed is down
A pumpkin is thrown horizontally off of a building at a speed of 2.5 m/s and travels a horizontal distance of 12 m before hitting the ground. We can ignore air resistance.
What is the pumpkin's vertical velocity when it hits the ground?
Answer: -47.04
How many people does it take to change a lightbulb?
This force on compass dials is an example of a force that _______.
Give Reason.......Why we use life jackets when we go in the water ???
Answer:
so we don't drown and die.
Explanation:
How much work is done on a pumpkin with a force of 24 newtons when you lift it 15 meters? *
Answer:
I'm not that busy solving but I'll tell you the formula that Force x distance is equal to work done
The work is done on a pumpkin when we lift it by 15 m with 24 N is 360 J
What is Work ?Work done is the amount energy gained (loosed) in bringing the body from initial position to final position. It is denoted by W and its SI unit is joule(J).
i.e. Work(W) is force(F) times displacement(s).
W=F× s
When a body is displaced with 1 newton of force by 1 m, then we can say that work has been done on the body by 1 joule.
Writing for it's dimension,
W=F× s
Force has dimension [L¹ M¹ T²]
Displacement has dimension [L¹]
multiplying both the dimensions Force and Displacement
we get,
dimension of Work [L² M¹ T²]
According to newton's second law of motion,
Force(F) is mass(M) times acceleration(a).
i.e. F=ma
Given,
Force = 24 N
Displacement = 15 m
W=F.s= 24*15 = 360 J
Hence work done on pumpkin is 360 J
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The base of a box is .45 m by .65 m. It weighs 8694 N. What is the pressure exerted on the floor by the box?
Answer:
[tex]Pressure = 29723.1\ N/m^2[/tex]
Explanation:
Given
[tex]Force = 8694N[/tex]
[tex]Length = 0.45m[/tex]
[tex]Width = 0.65m[/tex]
Required
The force exerted on the floor by the box
First, calculate the area covered by the box (i.e. the base area)
[tex]Base\ Area = Length * Width[/tex]
[tex]Base\ Area = 0.45m * 0.65m[/tex]
[tex]Base\ Area = 0.2925m^2[/tex]
Pressure is calculated as:
[tex]Pressure = \frac{Force}{Area}[/tex]
[tex]Pressure = \frac{8694N}{0.2925m^2}[/tex]
[tex]Pressure = 29723.0769231\ N/m^2[/tex]
[tex]Pressure = 29723.1\ N/m^2[/tex] --- approximated
PLZZ HELP
If you have two objects moving at the same velocity, would the object with bigger mass have higher or lower kinetic energy?
Answer:
The kinetic energy of a moving object is directly proportional to its mass and directly proportional to the square of its velocity. This means that an object with twice the mass and equal speed will have twice the kinetic energy while an object with equal mass and twice the speed will have quadruple the kinetic energy.
HELP ME
PLZZZ
ASAP
HELP
LIFE
OR
DEATH
SITUATION
The half-life of carbon-14 is 5730 years.
Carbon-14 is used for carbon dating. Carbon dating can tell us how old some
objects are
A skeleton was carbon dated. The results showed that there was only 12.5% of the
original amount of carbon-14 left in the skeleton.
Calculate the age of the skeleton.
HELP ASAP PLZZZ I BEG YOUUUU
Answer:
13,308 MAYBE IF IT ISN'T IM SO SORRY
Explanation:
1) Si un mango cae a una velocidad de 75m/s y tarda 26 seg. en caer. ¿ Cuál habrá sido la velocidad con qué el mango llegó al suelo?
Answer:
El mango llega al suelo a una velocidad de 329.982 metros por segundo.
Explanation:
El mango experimenta un movimiento de caída libre, es decir, un movimiento uniformemente acelerado debido a la gravedad terrestre, despreciando los efectos de la viscosidad del aire y la rotación planetaria. Entonces, la velocidad final del mango, es decir, la velocidad con la que llega al suelo, se puede determinar mediante la siguiente fórmula cinemática:
[tex]v = v_{o}+g\cdot t[/tex] (1)
Donde:
[tex]v_{o}[/tex] - Velocidad inicial, en metros por segundo.
[tex]v[/tex] - Velocidad final, en metros por segundo.
[tex]g[/tex] - Aceleración gravitacional, en metros por segundo al cuadrado.
[tex]t[/tex] - Tiempo, en segundos.
Si sabemos que [tex]v_{o} = -75\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex] y [tex]t = 26\,s[/tex], entonces la velocidad final del mango es:
[tex]v = v_{o}+g\cdot t[/tex]
[tex]v = -75\,\frac{m}{s}+\left(-9.807\,\frac{m}{s} \right)\cdot (26\,s)[/tex]
[tex]v = -329.982\,\frac{m}{s}[/tex]
El mango llega al suelo a una velocidad de 329.982 metros por segundo.
LOTS OF BRAINLIST WILL BE GIVING TO THOSE WHO HELP
chemical reaction for fossil fuels:combustion of fuels.
Difference between biomass and fossil fuels:how much carbon dioxide is produced
Comparing biomass with other renewable energy sources:does not have as much energy potential as fossil fuels.
Making energy choices locally: biomass
What is the magnitude of the electrostatic force between two electrons each having a charge of 1.6 x 10-19 C separated by a distance of 1.00 × l0– 8 meter?
Answer:
[tex]fe = \frac{9 \times 10 {}^{9} \times 1.6 \times 10 {}^{ - 19} \times 1.6 \times 10 { - 19}^{?} }{(1 \times 10 { }^{ - 8}) {}^{2} } \\ fe = 23.04 \times 10 {}^{ - 13} n[/tex]
how long would it take a leopard, running at an average speed of 20 m/s to travel 500 m?
Answer:
25 seconds
Explanation:
500/20
Show that liquid pressure is directly
proportional to height of liquid in a vessel.
Answer:
P=F/A where F is the weight of the water and A is the area on which it is resting. The weight of the water is mg. The mass of the water is dv where d is the density and v is the volume. Finally, the volume of the water in a vessel is equal to the area of the base of the vessel times the height of the vessel. (v=Ah)
Plugging everything in we get:
P = dAhg/A
So
P=dhg
So we have shown that liquid pressure is directly proportional to height of liquid in a vessel.
Initial velocity vector vA has a magnitude of 3.00 meters per second and points 20.0o north of east, while final velocity vector vB has a magnitude of 6.00 meters per second and points 40.0o south of east. Find the magnitude and the direction of the change in velocity vector Δv (which is the vector subtraction of the two vectors: final velocity vector minus initial velocity vector).
Answer:
[tex]5.2\ \text{m/s}[/tex]
[tex]70^{\circ}[/tex] south of east
Explanation:
[tex]v_a[/tex] = 3 m/s
[tex]\theta_a[/tex] = [tex]20^{\circ}[/tex] north of east
[tex]v_b[/tex] = 6 m/s
[tex]\theta_b[/tex] = [tex]40^{\circ}[/tex] south of east = [tex]360-40=320^{\circ}[/tex] north of east
x and y component of [tex]v_a[/tex]
[tex]v_{ax}=v_a\cos \theta\\\Rightarrow v_{ax}=3\times \cos 20^{\circ}\\\Rightarrow v_{ax}=2.82\ \text{m/s}[/tex]
[tex]v_{ay}=v_a\sin\theta\\\Rightarrow v_{ay}=3\times \sin20^{\circ}\\\Rightarrow v_{ay}=1.03\ \text{m/s}[/tex]
x and y component of [tex]v_b[/tex]
[tex]v_{bx}=v_b\cos \theta\\\Rightarrow v_{bx}=6\times \cos 320^{\circ}\\\Rightarrow v_{bx}=4.6\ \text{m/s}[/tex]
[tex]v_{by}=v_b\sin\theta\\\Rightarrow v_{by}=6\times \sin320^{\circ}\\\Rightarrow v_{by}=-3.86\ \text{m/s}[/tex]
[tex]\Delta v=v_b-v_a\\\Rightarrow \Delta v=(4.6-2.82)\hat{i}+(-3.86-1.03)\hat{j}\\\Rightarrow \Delta v=1.78\hat[i}-4.89\hat{j}[/tex]
Magnitude
[tex]|\Delta v|=\sqrt{(-4.89)^2+1.78^2}\\\Rightarrow \Delta v=5.2\ \text{m/s}[/tex]
Direction
[tex]\theta=\tan{-1}|\dfrac{-4.89}{1.78}|\\\Rightarrow \theta=70^{\circ}[/tex]
The magnitude of the change in velocity vector is [tex]5.2\ \text{m/s}[/tex] and the direction is [tex]70^{\circ}[/tex] south of east.
The change in velocity will be [tex]\Delta V=5.2\ \frac{m}{s}[/tex] and the direction will be [tex]70^o[/tex] South to east.
What are vector quantities?Any quantity which is defined by its magnitude and direction both are called as the vector quantities.
Now the data given in the question will be given as:
[tex]V_a[/tex] = 3 m/s
[tex]\theta[/tex] = [tex]20^o[/tex] north of east
[tex]V_b[/tex] = 6 m/s
[tex]\theta[/tex] = [tex]40^o[/tex]south of east = 360-40=320 north of east
Now we will find the x and y component of [tex]V_a[/tex]
[tex]V_{ax}=V_acos\theta[/tex]
[tex]V_{ax}=3\times Cos20[/tex]
[tex]V_{ax}=2.82\ \frac{m}{s}[/tex]
[tex]V_{ay}=V_aSin\theta[/tex]
[tex]V_{ay}=3\times Sin20[/tex]
[tex]V_{ay}=1.03\ \frac{m}{s}[/tex]
Now we will find the x and y component of [tex]V_b[/tex]
[tex]V_{bx}=V_bcos\theta[/tex]
[tex]V_{bx}=6\times cos\320[/tex]
[tex]V_{bx}=4.6\ \frac{m}{s}[/tex]
[tex]V_{by}=V_bSin\theta[/tex]
[tex]V_{by}=6\times Sin320[/tex]
[tex]V_{by}=-3.86\ \frac{m}{s}[/tex]
Now change in velocity will be
[tex]\Delta V=V_b-V_a[/tex]
[tex]\Delta V=(4.6-2.82)i+(-3.86-1.03)j[/tex]
[tex]\Delta V=1.78i-4.89j[/tex]
The magnitude can be find out as follows:
[tex]\Delta V=\sqrt{(-4.89^2+(1.78^2)}[/tex]
[tex]\Delta V=5.2\ \frac{m}{s}[/tex]
The direction of the vector will be
[tex]\theta= tan^{-1}(\dfrac{-4.89}{1.78})[/tex]
[tex]\theta=70^o[/tex]
Thus the change in velocity will be [tex]\Delta V=5.2\ \frac{m}{s}[/tex] and the direction will be [tex]70^o[/tex] South to east.
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How does creativity affect scientific work?
Answer & Explanation:
In science, rationality and creativity work together. Creativity allows us to view and solve problems with innovation and openness. Scientific theories often came from sparks of creative thinking and bold yet logical processes.
Pls help with all questions dew in 10 minutes!
Gravity attracts an object to
A.
Earth's magnetic poles
B.
Earth's surface
C.
Earth's equator
D.
Earth's center
Answer:
D
Explanation:
A conveyer belt carries a load of mass 180kg n lift it up in 1
Answer:
Cool.
Explanation:
What's the question..? :|
El monoxido de carbono reacciona con el hidrogeno gaseoso para producir metanol (ch3oh) calcule el reactivo limite y el reactivo en exceso si la reaccion inicia con 2,0 g de cada reactivo calcule cuantos gramos de metanol se obtiene
Answer:
Se obtienen 2,27 gramos de metanol.
Explanation:
La reacción entre monóxido de carbono e hidrógeno para producir metanol es la siguiente:
CO + 2H₂ → CH₃OH
Para encontrar el reactivo limitante y el reactivo en exceso, debemos calcular el número de moles de CO y H₂:
[tex]\eta_{CO} = \frac{m}{M} [/tex]
En donde:
m: es la masa
M: es el peso molecular
[tex]\eta_{CO} = \frac{m}{M_{CO}} = \frac{2,0 g}{28,01 g/mol} = 0,071 moles [/tex]
[tex]\eta_{H_{2}} = \frac{2,0 g}{2,02 g/mol} = 0,99 moles [/tex]
Dado que la relación estequiométrica entre CO y H₂ es 1:2, el número de moles de hidrógeno gaseoso que reaccionan con el monóxido de carbono es:
[tex] \eta_{H_{2}} = \frac{2}{1}*0,071 = 0,142 moles [/tex]
Entonces, se necesitan 0,142 moles de H₂ para reaccionar con 0,071 moles de CO y debido a que se tienen más moles de H₂ (0,99 moles) entonces el reactivo limitante es CO y el reactivo en exceso es H₂.
Ahora podemos encontar la masa de metanol obtenida usando el reactivo limitante (CO) y sabiendo que la realcion estequiométrica entre CO y CH₃OH es 1:1.
[tex] \eta_{CH_{3}OH} = \eta_{CO} = 0,071 moles [/tex]
[tex] m = 0,071 moles*32,04 g/mol = 2,27 g [/tex]
Por lo tanto, se obtienen 2,27 gramos de metanol.
Espero que te sea de utilidad!
phase difference formula
Answer:
Δx is the path difference between the two waves.
...
Phase Difference And Path Difference Equation.
Formula Unit
Phase Difference \Delta \phi=\frac{2\pi\Delta x}{\lambda } Radian or degree
Path Difference \Delta x=\frac{\lambda }{2\pi }\Delta \phi meter
A material kept at high temperature is seen to emit photons with energies of 0.3 eV, 0.5eV, 0.8 eV, 2.0 eV, 2.5 eV, and 2.8 eV. These are the only photon energies observed. It is now cooled down to a very low temperature so that it is not emitting photons anymore. If a beam of light with a continuous range of energies from 0.01 eV to 10 eV shines on the material, what photon energies in this beam will correspond to dark absorption lines?
1. 0.3 eV, 0.5 eV, 2.8 eV
2. 0.3 eV, 0.5 eV, 0.8 eV
3. 0.3 eV, 0.8 eV, 2.8 eV
4. 0.8 eV, 2.0 eV, 2.8 eV
5. 0.5 eV, 0.8 eV, 2.8 eV
6. 0.3 eV, 2.5 eV, 2.8 eV
Answer:
the correct one is 2
Explanation:
The emission and arcion spectra are similar, in the emission spectra a heated material emits photons for the internal transitions of the electrons, in the absorption spectrum the continuous emission of a lamp is absorbed at the energies that the material has transitions.
In our case the material is low early, so all the electrons are in their base states.
As the light beam has all wavelengths, those corresponding to the first emission of the material will be observed.
When reviewing the answers, the correct one is 2
Please help, I do NOT need any links.
Answer:
P since without a host the parasite won't be able to survive they would start decreasing as well but if the hosts were no more then they would go extinct. But since it is just decreasing then it should be P
"The diagram below shows three kettles with their powers and the time they take to boil 500cm of
water. How many units (kWh) does the 3kW kettle use to boil the water?"
The question above is about a diagram which shows 3 kettles. The one on the left is the on I have to focus on. The power of it is 3kW (KiloWatts) and its takes 3 minutes to boil the given amount of water. Can someone please answer this question? Don't give me an IP grabber or virus download link. If you give me a URL link of any kind, I will report your answer. Don't try it. The answer must be in kWh (KiloWatt Hours).
Answer:
3kWx(3/60)h=0.15kwh
Which best describes the energy of a sound wave as it travels through a medium?
It increases.
lt decreases.
It remains the sam.
It depends on the medium,
Answer:
it depends on the medium :D
If the average human has a density of 1.01g/cm3
and the density of the Dead Sea is 1.23g/mL, why
can't we swim in the dead sea? Must have the
words (Mass, volume, and density)
Answer:The Dead Sea water has a density of 1.24 kg/litre
Explanation:
which makes swimming similar to floating.
Is O2 classified as a compound?
Answer: NO.
Explanation: Oxygen is not a compound. It has only one element in it.
An athletes heart beats 62 times per minute. What is the frequency of her heart beat?
Answer:
22
Explanation:
bc it just is
the moon's mass is ____?
Answer:
7.35..kg
Explanation:
Hope this will help you
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How many calories are released when 6 grams of 100°C steam turns to 0°C ice?
Answer: 4276.2 calories
Explanation:
Given
mass of steam is 6 gm at [tex]100^{\circ}C[/tex]
Conversion of steam to ice involves
steam to water at [tex]100^{\circ}C[/tex]water at [tex]100^{\circ}C[/tex] to water at [tex]0^{\circ}C[/tex]water to the ice at [tex]0^{\circ}C[/tex]Calories released during the conversion of steam to water at [tex]100^{\circ}C[/tex]
[tex]E_1=mL_v\quad [L_v=\text{latent heat of vaporisation}]\\E_1=6\times 533=3198\ cal.[/tex]
Calories released during the conversion of water at [tex]100^{\circ}C[/tex] to water at [tex]0^{\circ}C[/tex]
[tex]E_2=mc(\Delta T)\quad [c=\text{specific heat of water,}1\ cal./gm.^{\circ}C]\\E_2=6\times 1\times 100=600\ cal.[/tex]
Calories released during the conversion of water to the ice at [tex]0^{\circ}C[/tex]
[tex]E_3=mL_f\quad [L_f=\text{latent heat of fusion}]\\E_3=6\times 79.7=478.2\ cal.[/tex]
The total energy released is
[tex]E=E_1+E_2+E_3\\E=3198+600+478.2=4276.2\ cal.[/tex]