The number of protons plus the number of neutrons decreases
Beta decay occurs when, in a nucleus with too many protons or too many neutrons, one of the protons or neutrons is transformed into the other. In beta minus decay, a neutron decays into a proton, an electron, and an antineutrino: n Æ p + e - +.
This should be pretty easy i think.
Describe the components within the system involved in the collision between pilot and landing zone.
Answer:
its a sure for not but i think its a 192583
Interpreting Graphics
11. Look at the model of an atom below. Does the
model represent a metal atom or a nonmetal
atom? Explain your answer.
ng
By looking at the model shown in the problem one can clearly say that this model represents the metal because there are a total of 12 electrons revolving around the nucleus which clearly represents the Magnesium metal.
What are metalloids?The elements of the periodic tables that behave as metals, as well as the nonmetal in some chemical or physical aspects, are known as metalloids. Some examples of metalloids are Boron, silicon, germanium, arsenic, antimony, tellurium, etc.
As given in the problem, there is a graphical representation of the model of an atom,
One can see that the model in the problem accurately depicts the metal by looking at the total of 12 electrons circulating around the nucleus, which is a clear representation of the magnesium metal.
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What would be the acceleration of a block sliding down an inclined plane that makes an angle of 45° with the horizontal if the coefficient of sliding friction between two surfaces in 0.3? 141
Ans: 4.95 m/sec?
Answer:
unfortunately we have not received a reply to the park yet but
Explanation:
mohamedz and he had killed him
What is the difference between a low tide and a high tide
Answer:
High water level during a tide is called High tide.
Low water level during a tide is called Low tide.
A ball is projected at an immovable wall with a speed vi and bounces back the wall in such a manner that it only has 1/3 of its original linear momentum. a) Determine what fraction of the kinetic energy is lost in the collision.
The fraction of the kinetic energy of the ball lost during the collision is [tex]\frac{1}{3}[/tex].
The given parameters;
initial speed of the ball, = vifinal momentum of the ball, Pf = ¹/₃PiThe initial and final momentum of the ball is calculated as;
[tex]P_i = m_ivi[/tex]
[tex]P_f = m_fv_f = \frac{1}{3} m_iv_i[/tex]
The initial and final kinetic energy of the ball is calculated as;
[tex]K.E_i = \frac{1}{2} m_iv_i^2 = \frac{1}{2} P_iv_i\\\\K.E_f = \frac{1}{2} m_fv_f^2= \frac{1}{2} (\frac{1}{3} P_iv_i)= \frac{1}{6} P_iv_i[/tex]
The change in the kinetic energy is calculated as;
[tex]\Delta K.E = K.E_f - K.E_i \\\\\Delta K.E = \frac{1}{6} P_iv_i - \frac{1}{2} P_iv_i = \frac{1}{3} P_iv_i[/tex]
Thus, the fraction of the kinetic energy of the ball lost during the collision is [tex]\frac{1}{3}[/tex].
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someone throws a rubber ball vertically upward from the roof of a building 8.2m in height. the ball rises, then falls.it just misses the edge of the roof, and strikes the ground.if the ball is in the air for 6.2s, what was its initial velocity?(disregard air resistance. a=-g=-10m/s²
Answer:
H = V t - 1/2 g t^2 V and g are in different directions
-8.2 = V * 6.2 - 1/2 * 10 * 6.2*2 = 6.2 V - 192.2
V = 184 / 6.2 = 29.7 m/s
Check: find time for ball to return to zero height
0 = 29.7 T - 5 T^2
T = 29.7 / 5 = 5.94 sec
The ball must have fallen 8.2 m in (6.2 - 5.94) sec = .26 sec
S = 29.7 * .26 + 5 * .26^2 = 8.1
Déterminer la charge de noyau d’atome d’oxygène en coulomb C
Answer:
determine a large mam I have no idea
6. Explain the importance of the law of conservation of mass.
Answer:
According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants. The law of conservation of mass is useful for a number of calculations and can be used to solve for unknown masses, such the amount of gas consumed or produced during a reaction.
Explanation:
<3
Which of the following processes generated heat during earths formation that is still a source of heat in earths interior
Answer:
There are three main sources of heat in the deep earth: (1) heat from when the planet formed and accreted, which has not yet been lost; (2) frictional heating, caused by denser core material sinking to the center of the planet; and (3) heat from the decay of radioactive elementsHello everyone, I just want to know the formula to solve this problem:
What is the speed of sound at 62°C?
and how to solve this kind of problem? My brain isn't working.. :)
Answer:
367 m/sExplanation:
Use simplified formula:
[tex]v=331.3\sqrt{1+T/273.15}[/tex] , where T- temperature in Celsiuswhat is potentiometer
Answer:
potentiometer is a device used to measure potential difference across the circuit
Why are people so rude???
Answer:
It could be low self-esteem
Explanation:
There probably insecure have problems and issues of their own could be that were abused growing up and weren’t treated right who knows.
A 0.5kg ball falls from a building that Is 50m high. how much kinetic energy will it have when it has fallen half way to the ground?
A 34. 8 n object is in free fall. What is the magnitude of the net force which acts on the object? answer in units of n.
Answer:
-34.8n
Explanation:
when object in free fall, only force is the Force by weight,
which is 34.8 and direction is down.
If force F, volume V and density D are taken as fundamental quantities, then the representation of pressure P will be
Using the principle of dimensional analysis, the representation of Pressure would be [tex] P = F^{1}V^{-\frac{2}{3}}D^{0}[/tex]
Pressure is related to Force, Volume and Density thus :
[tex] Pressure, P = F^{a}V^{b}D^{c}[/tex]
Using dimensional analysis :
Pressure, P = ML¯¹T¯²Substituting the dimension into the relation :
[tex] ML^{-1}T^{-2}= [MLT¯²]^{a} [L³]^{b} [ML¯³]^{c} [/tex]
[tex] ML^{-1}T^{-2} = M^{a+c}L^{a+3b-3c}T^{-2a} [/tex]
Equating the power ;
a + c = 1 - - - (1)
a + 3b - 3c = - 1 - - - (2)
-2a = - 2
a = - 2/-2
a = 1
From (1)
1 + c = 1
c = 1 - 1 = 0
c = 0
From (2) :
1 + 3b - 3(0) = - 1
1 + 3b = - 1
3b = - 2
b = - 2/3
Hence, [tex] P = F^{1}V^{-\frac{2}{3}}D^{0}[/tex]
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The representation of pressure is:
[tex]P=FV^{-\frac{2}{3} }[/tex]
Given that:
Force F, volume V and density D are taken as fundamental quantities, hence pressure P is:
[tex]P=F^aV^bD^c\\\\(ML^{-1}T^{-2})=(MLT^{-2})^a(L^3)^b(ML^{-3})^c\\\\\\(ML^{-1}T^{-2})=(M)^{a+c}(L)^{a+3b-3c}(T)^{-2a}[/tex]
Hence:-
2 = -2a
a = 1
a + c = 1
1 + c = 1
c = 0
a + 3b - 3c = -1
1 + 3b - 3(0) = -1
3b = -2b = -2/3
Hence the representation of pressure is:
[tex]P=FV^{\frac{-2}{3} }D^{0}\\\\P=FV^{\frac{-2}{3} }[/tex]
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Help me quick!!! please!!
Calculate the power of an engine which raises a mass of 1000 kg to a height of 40 m in 10 s.
Answer:
39.240 W
Explanation:
Let's start by calculating the work done by the engine. We can assume that it is the same work done by the weight of the object to bring it from 40m to the surface: as much energy it takes to bring it up, the same ammount it takes to bring it down. Said work is [tex]w= \vec F\cdot \vec{h} = mg h = 1000 \times 9.81\times 40 = 392.400 J[/tex]
At this point we can simply apply the definition of power, that is [tex]P = \frac wt[/tex], to get the power of the engine is [tex]39.240 W[/tex]
Answer:
P=39200W
Explanation:
P=(mgh)/t
P=(40×9.8×1000)/10
P=39200W
what is the formula for calculating potentiometer
28
If it takes 1,697 m to stop a car initially moving at 46 m/s, what distance is required to stop a car moving at a quarter
of the initial speed under the same conditions?
Answer:
Approximately [tex]106\; \rm m[/tex], [tex](1/16)[/tex] of the original distance, assuming that the acceleration is constant.
Explanation:
Let [tex]v[/tex] and [tex]u[/tex] denote the final and initial velocity of the vehicle ([tex]v = 0\; \rm m \cdot s^{-1}[/tex] since this vehicle has come to a stop.)Let [tex]a[/tex] denote the acceleration of the vehicle (a constant under the assumption.) Let [tex]x[/tex] denote the displacement of the vehicle.If the acceleration on this vehicle is constant, the SUVAT equations would apply:
[tex]\displaystyle v^{2} - u^{2} = 2\, a\, x[/tex].
Rearrange this equation to find an expression for [tex]x[/tex], the displacement required for the vehicle to come to a stop:
[tex]\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a} \\ &= \frac{-u^{2}}{2\, a} && (\text{$v = 0\; \rm m \cdot s^{-1}$})\end{aligned}[/tex].
Thus, under these assumptions, [tex]x[/tex] would be proportional to [tex]u^{2}[/tex]. In other words, the distance required for this vehicle to come to a stop would be proportional to the square of the initial velocity of this vehicle.
If the initial velocity [tex]u[/tex] is reduced to [tex](1/4)[/tex] (a quarter) of the initial value, the distance required for stopping this vehicle would be [tex](1/4)^{2} = (1/16)[/tex] of the initial value:
[tex]\begin{aligned}1,\!697\; {\rm m} \times \frac{1}{16} \approx 106\; \rm m\end{aligned}[/tex].
Please help! Question is attached.
Answer:
a. 6
b. 6 m/s²
c. 300 m to the right
d. 30 secs
Explanation:
slope = rise /run
60-0/10-0
= 6
b. slope = acceleration = 6 m/s²
c. d=ut+1/2at²
t=10 (segment A last for 10 secs)
u - initial velocity = 0
so d = 0(10)+1/2*6*10²
=300 m
A child sits on a rotating merry-go-round, 2.1 m from its center. If the speed of the child is 2.2 m/s, what is the minimum coefficient of static friction between the child and the merry-go-round that will prevent the child from slipping?
We have that the the minimum coefficient of static friction is mathematically given as
[tex]\mu= 0.235[/tex]
ForceGenerally the equation for the Force is mathematically given as
[tex]F=n\mu\\\\Where\\\\n = mg\\\\Given\\\\Fc = m(vt2/R)\\\\Therefore\\\\\mu = vt2/gR\\\\\mu = 2.22/(9.8(2.1)) \\\\[/tex]
[tex]\mu= 0.235[/tex]
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I will give 100 Points, 5-stars, and Brainliest
Direct contact is necessary for which type of energy transfer?
A. Conduction
B. Sound Waves
C. Conduction
D. Convection
Answer:
Conduction
Explanation:
Because when a molecule is transferred to another molecule then this in going into direct contact
For example if we have a pan and it is boiling well the heat that is on the pan holder that stays in place and it won`t move anywhere else like there is Radiation and Convection but for this question the answer would be Conduction
Football players boots have spikes at their sole
Give reason
Answer:
it prevents slipping on the grass while playing
Answer:
to keep their grip on the ground so they won't fall
1. Look at this data table
What data seems to be inconsistent with the rest? What are 2 different explanations for this data?
Using visual inspection and knowledge of linear regression , the inconsistent input is the last data point (30, 12.0).
Looking at the data, there is a negative relationship between the depth of water sample and the amount of dissolved oxygen. As the depth increases, the amount of dissolved oxygen decreases. However, the last data point does not concur with this trend. This might have been caused by imputation error or error in the measuring device used.Therefore, the inconsistent data point is (30, 12.0)
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What water pressure must a pump that is located on the first floor supply to have water on the thirteenth of a building with a pressure of 35 lb/in2 Assume that the distance between each floors is 10ft.
The water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.
The given parameters;
Pressure on the 13 th floor, P₁ = 35 PSIDistance between each floor, d = 10 ftThe vertical pressure of the water is calculated as follows;
[tex]P = \rho gh\\\\\frac{P}{h} = \rho g\\\\\frac{P}{h} = k\\\\\frac{P_1}{h_1} = \frac{P_2}{h_2} \\\\[/tex]
The vertical height of the first floor from the 13th floor = 130 ft
The vertical height of the 13 ft floor = 10 ft
[tex]P_1 = \frac{P_2 h_1}{h_2} \\\\P_1 = \frac{35 \times 130}{10} \\\\P_1 = 455 \ PSI[/tex]
Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.
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The horizontal movement of air from an area of high pressure to an area of lower pressure
options:
wind
weather
heat
conduction
Answer:
Explanation:
A wind is the horizontal movement of air from an area of high pressure to an area of lower pressure. Winds are caused by differences in air pressure. Most differences in air pressure are caused by unequal heating of the atmosphere
A student did 24 J of work on a chair. She applied a force of 12 N and moved the chair 2 m. What else do you need to know to determine the amount of power used?
Answer:
time
Explanation:
power = work done ÷ time
People who drink coffee are more likely to develop cancer
Answer:
The answer is False
A copper (Young's modulus 1.1 x 1011 N/m2) cylinder and a brass (Young's modulus 9.0 x 1010 N/m2) cylinder are stacked end to end, as in the drawing. Each cylinder has a radius of 0.24 cm. A compressive force of F = 7900 N is applied to the right end of the brass cylinder. Find the amount by which the length of the stack decreases.
We have that for the Question, it can be said that the amount by which the length of the stack decreases is
[tex]dl'=3.621*10^{-4}m[/tex]
From the question we are told
A copper (Young's modulus 1.1 x 1011 N/m2) cylinder and a brass (Young's modulus 9.0 x 1010 N/m2) cylinder are stacked end to end, as in the drawing. Each cylinder has a radius of 0.24 cm.
A compressive force of F = 7900 N is applied to the right end of the brass cylinder. Find the amount by which the length of the stack decreases.
Generally the equation for copper cylinder is mathematically given as
[tex]dl=\frac{Flo}{yA}[/tex]
[tex]dl=\frac{7900*3*10^-^2}{1.1*10^{11}*\pi(0.24*10^{-2})^2}[/tex]
[tex]dl=1.19064778*10^-^4[/tex]
Generally the equation for brass cylinder is mathematically given as
[tex]dl=\frac{7900*5*10^-^2}{9*10^{10}*\pi(0.24*10^{-2})^2}[/tex]
[tex]dl=2.43*10^{-4}[/tex]
Therefore Total change in length
[tex]dl'=1.191*10^-^4+(2.43*10^{-4})[/tex]
[tex]dl'=3.621*10^{-4}m[/tex]
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Wire coils spinning between two magnets will produce …?
A: cold
B : electricity
C: chemicals
D: Steam
Answer:
B
Explanation:
I don't know but its b
a gas increases in pressure from 2.00 atm to 6.00 atm at a constant volume of 1.00 m3 and then expands at constant pressure to a volume of 3.00 m3 before returning to its initial state as shown in the figure below. how much work is done in one cycle?
The work done on the given gas in one cycle is -405.3 kJ.
The given parameters;
initial pressure of the gas, P₁ = 2 atmfinal pressure of the gas, P₂ = 6 atminitial volume of the gas, V₁ = 1 m³final volume of the gas, V₂ = 3 m³Convert the pressure to Pascal (N/m²);
1 atm = 101325
The work done in one cycle is the area of the triangle and it is calculated as follows
[tex]Area = \frac{1}{2} \times base \times height\\\\Area = \frac{1}{2} \times (3 \ m^3\ -\ 1 \ m^3)\times (6 \ atm \ - \ 2 \ atm)\\\\Area = 4 \ atm -m^3\\\\Area = 4 \ atm -m^3 \ \times \frac{101325 \ Pa}{1 \ atm} \\\\Area = 405,300 \ m^3.Pa\\\\Area = 405,300 \ m^3. (N/m^2)\\\\Area = 405,300 \ Nm\\\\Area = 405,300 \ J\\\\Area = 405.3 \ kJ[/tex]
the net work done on the gas = - 405.3 kJ
Thus, the work done on the given gas in one cycle is -405.3 kJ.
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