Answer:
El valor de la fuerza que debe ejercer la persona C debe ser de -2 para que la caja esté en equilibrio físico.
Explanation:
Si la caja debe hallarse en equilibrio físico, entonces se debe satisfacer la siguiente ecuación:
[tex]F_{A} + F_{B} + F_{C} = 0[/tex] (1)
Si sabemos que [tex]F_{A} = -3[/tex] y [tex]F_{B} = 5[/tex], entonces el valor de la fuerza que debe ejercer la persona C debe ser:
[tex]F_{C} = -F_{A}-F_{B}[/tex]
[tex]F_{C} = -(-3)-5[/tex]
[tex]F_{C} = -2[/tex]
El valor de la fuerza que debe ejercer la persona C debe ser de -2 para que la caja esté en equilibrio físico.
a teacher and two students are mesuring the speed of sound. the teacher makes loud sound by hitting two cymbals together. each students starts a stopwatch then they see the teacher hit the cymbals, they each stop their stop watch when they hear the sound. describe how a sound wave moves through the air.
Answer:
Slowly and smoothly lol
Answer:
S.I. on Rs. 1600 = T.D. on Rs. 1680. Rs. 1600 is the P.W. of Rs. 1680, i.e., Rs. 80 is on Rs. 1600 at 15%. Time =100 x 80year=1year = 4 months.1600 x 153
Explanation:
Is this you are?
5. A child rides a pony on a circular track with a radius of 5 m Find the distance traveled and the placement
after the child has gone halfway around the track (8) Does the distance traveled increase, decrease, or stay the
same when the child completes one circuit of the track? Does the displacement increase, decrease, or stay the
same? Explain. (C) Find the distance and displacement ter a complete circuit of the track
The answer are :
A. Distance = 15.71 m , The displacement = 10 m
B. distance traveled will increase, displacement will decrease
C. The distance = 31.42 m , Displacement = 0
Difference between Distance and DisplacementDistance is a scalar quantity, while displacement is a vector quantity. Displacement is the distance travelled in a specific direction.
Given that a child rides a pony on a circular track with a radius of 5 m, after the child has gone halfway around the track
(A)
The distance traveled will be = 2[tex]\pi[/tex]r / 2
The distance = πr
The distance = 22/7 x 5
The distance = 15.71 m
The displacement = 2r
The displacement = 2 x 5
The displacement = 10 m
(B) The distance traveled will increase as the child completes one circle of the track but the displacement will decrease because the displacement in one cycle of a circular motion is zero since it is a vector quantity.
(C) The distance after a complete circuit of the track = 2πr
The distance = 2 × 22/7 × 5
The distance = 31.42 m
The displacement after a complete circuit of the track will be zero.
That is;
Displacement = 0
Therefore, the distance traveled and the displacement after the child has gone halfway around the track are 15.7 m and 10 m respectively. While the distance and displacement after a complete circle of the track are 31.42 m and 0 respectively.
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b. The role of the moon is greater than that of the sun in the occurrence of tides. ???
Our sun is 27 million times larger than our moon. Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth. If tidal forces were based solely on comparative masses, the sun should have a tide-generating force that is 27 million times greater than that of the moon. However, the sun is 390 times further from the Earth than is the moon. Thus, its tide-generating force is reduced by 3903, or about 59 million times less than the moon. Because of these conditions, the sun’s tide-generating force is about half that of the moon.
It takes Serina 1.81 hours to drive to school. Her route is 36 km long. What is Serina's average speed on her drive to school?
Answer:
60
Explanation:
because it shows he is moving at a low speed
A student project is required to be portable and hand held. It requires 6 V DC power at a current of 150 mA. The batteries for the power supply must last for a minimum of 7 hours of continuous operation. NiMH rechargeable batteries in AA size are to be used. A) How many batteries are needed
Answer:
a. 5 batteries b. 1050 mAh
Explanation:
Here is the complete question
A student project is required to be portable and hand held. It requires 6 V DC power at a current of 150 mA. The batteries for the power supply must last for a minimum of 7 hours of continuous operation. NiMH rechargeable batteries in AA size are to be used. A) How many batteries are needed? B) What mAh capacity should the batteries have?
Solution
A) How many batteries are needed?
Since the nominal voltage for a single NiMH battery is 1.2 V per battery and we require 6V DC power, we combine the batteries in series to obtain a total voltage of 6 V. The number of batteries required, n = total voltage/voltage per cell = 6V/1.2V per battery = 5 batteries
So, the number of batteries needed is 5.
B) What mAh capacity should the batteries have?
Since the batteries are in series, they would each deliver a current of 150 mA. Since we require a current of 150 mA for 7 hours, the number of milliampere-hours capacity mAh of batteries required is Q = It where I = current = 150 mA and t = time = 7 hours.
So, Q = It = 150 mA × 7 h = 1050 mAh.
So, the batteries should have a mAh of 1050 mAh
I need help asap please
Answer:
I dont know answer Sorry For that thank u
how much would a 70 kg mean weight on the Moon what should be its mass on the earth and on the moon
Answer:
I don't know sorry' please
briefly explain what EMF of a cell cell is
Answer:
The emf of a cell is the sum of the electric potential differences (PDs) produced by a separation of charges (electrons or ions) that can occur at each phase boundary (or interface) in the cell. The magnitude of each PD depends on the chemical nature of the two contacting phases.
Explanation:
pls mark brainliest
1. How much heat energy ( Q ) is required to heat 2.0 kg of copper from 30.0 oC to 80.0 oC?
Answer:
38500
Explanation:
I looked it up so it may be wrong
At the end of an investigation, you must__________ ____________. Your results may or may not support your hypothesis.
Answer:
could and largejsjisj and we look like they can get to
A 63 kg kg person starts traveling from rest down a waterslide 6.0 mm above the ground. At the bottom of the waterslide, it then curves upwards by 1.0 mm above the ground such that the person is consequently launched into the air. Ignoring friction, how fast is the person moving upon leaving the waterslide
Answer:
change waterslide according to question. and you are good to go. check photo for solve
The person moving with speed upon leaving the waterslide is equal to 0.313 m/s while ignoring the friction.
What is the law of conservation of energy?If there is a loss of energy in any part of the system, there must be the same amount of a gain of energy in some other part of the isolated system.
Kinetic energy + Potential energy = constant
Therefore, the initial kinetic and potential energy of the person must be equal to the final kinetic and potential energy.
mgh₁ + mv₁²/2 = mgh₂ + mv₂²/2
gh₁ + v₁²/2 = gh₂ + v₂²/2
2gh₁ + v₁² = 2gh₂ + v₂²
2g(h₁ - h₂) = v₂² - v₁²
[tex]v_2=\sqrt{2g(h_2-h_1) + v^2_1}[/tex] .................(1)
Given, the mass of the person, m = 63 Kg
The initial speed of the person as at rest, v₁ = 0
Consider that the speed of the person while leaving the waterslide is v₂
The initial height of the person above the ground, h₁ = 6.0 mm
The height of the person leaving the waterslide, h₂= 1.0 mm
Substitute, the value of m, v₁, h₁ and h₂ in equation (1):
[tex]v_2 = \sqrt{2(9.8m/s^2)(6mm-1mm) + (0)^2}[/tex]
[tex]v_2 =\sqrt{2(9.8)\times 5 \times 10^{-3}} \;m/s[/tex]
[tex]v_2 =\sqrt{0.098}\; m/s[/tex]
v₂ = 0.313 m/s
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When 24.0 V is applied to a
capacitor, it stores 3.92 x 10-4 J of
energy. What is the capacitance?
[?] x 10!? E
[tex]\boxed{\sf E=QV^2}[/tex]
[tex]\\ \sf\longmapsto Q=\dfrac{E}{V^2}[/tex]
[tex]\\ \sf\longmapsto Q=\dfrac{3.92\times 10^{-4}}{24^2}[/tex]
[tex]\\ \sf\longmapsto Q=\dfrac{3.92\times 10^{-4}}{576}[/tex]
[tex]\\ \sf\longmapsto Q=0.006\times 10^{-4}C[/tex]
[tex]\\ \sf\longmapsto Q=6\times 10^{-1}C[/tex]
[tex]\\ \sf\longmapsto Q=0.6C[/tex]
Now
[tex]\boxed{\sf Q=CV}[/tex]
[tex]\\ \sf\longmapsto C=\dfrac{Q}{V}[/tex]
[tex]\\ \sf\longmapsto C=\dfrac{0.6}{24}[/tex]
[tex]\\ \sf\longmapsto C=0.025F[/tex]
Note:-
SI unit of charge is Coulomb(C)SI unitvof Capacitance is Farad(F)what will happen to the gravitational force between two bodies if the distance between them is halved keeping their masses constant ?..
Using deductive reasoning, write the converse, inverse, and contra positive of an if-then statement. Part I: Write an if-then statement below. (1 point)
Converse:
if q , then p
Inverse:
If not p , the not q.
Contra Positive:
If not q, then not p.
If then statement is a conditional statement. In this statement there is some condition at the first part of the sentence and then in the second part there is a conclusion.
The example of if then sentence is as follows,
If this camel is thirsty, then it will drink water from the well.
In this example there is a condition in first part that if camel is thirsty, in the second part there is conclusion which states that then it will drink water from well.
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A scientist measures the light from a distant star
at 525 nm. The constant for Wien's
displacement law is 2.9 x 10-3 m K. What is the
approximate temperature of the star in Kelvins?
A) 1500 K
B) 180,000 K
C) 1.5 K
D) 5500 K
The approximate temperature of the star as determined is D) 5500 K.
The Wien's displacement law relates the maximum wavelength of a body to its absolute temperature. Wien's displacement law states that:
λ = [tex]\frac{b}{T}[/tex]
where λ is the maximum wavelength of the body, b is the constant of proportionality and T is the absolute temperature.
Thus from the given question, λ = 525 nm (525 x [tex]10^{-9}[/tex]), and b = 2.9 x [tex]10^{-3}[/tex] mK.
So that,
525 x [tex]10^{-9}[/tex] = [tex]\frac{2.9*10^{-3} }{T}[/tex]
Make T the subject of the formula to have;
T = [tex]\frac{2.9*10^{-3} }{525*10^{-9} }[/tex]
= 5523.81
T = 5523.81 K
T ≅ 5500.00 K
The approximate temperature of the star in Kelvin is 5500 K.
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Define the term ‘matter’ in physics.
Answer:
Matter is any substance that has mass and takes up space by having volume
How does 'g' vary from place to place?
Explanation:
The acceleration g varies by about 1/2 of 1 percent with position on Earth's surface, from about 9.78 metres per second per second at the Equator to approximately 9.83 metres per second per second at the poles.
A car increase its speed steadily from 30km/hr to 60km/hr in 1 min A what is the average speed during this time
Explanation:
initial velocity(u)=30km/hr = 30*1000/60*60=8.33 m/s
final velocity(v) =60km/hr = 60*1000/60*60 =16.67 m/s
time taken(t) = 1 minutes
= 60 seconds
Now,
Average velocity = u+v/2
= 8.33 m/s + 16.67m/s÷2
=12.5 m/s
PLEASE HELP!!
A set of water waves travels at 20.0 m/s , and 5.0 waves pass you in 4.0 s. What is the wavelengths of the waves?
a. 0.25 m
b. 8.0 m
c. 0.20 m
d. 4.0 m
Answer:
Explanation:
[tex]\lambda\\[/tex] = v/f
^That is the formula we are going to use.
Now, we were given the speed (v), which is 20.
Now we need to find frequency, in order to solve for the wavelength.
Frequency is the amount of waves in a fixed unit of one second, meaning our F value is the value of 5 divided by 4.
5/4 = 1.25
Therefore our F is 1.25
Now lets plug it in
[tex]\lambda\\[/tex] = v/f
[tex]\lambda\\[/tex] = 20/1.25
[tex]\lambda\\[/tex] = 16
Conversion:
[tex]\lambda\\[/tex] = 8
If the wave is detected 12.5 minutes after the earthquake, estimate the distance from the detector to the site of the quake
Answer:
Remember the relation:
Speed*Time = Distance.
We can estimate that the speed at which an earthquake "moves", in the surface, is:
S = 6km/s (this is a low estimation actually)
Then if the wave is detected 12.5 minutes after the earthquake, we know that it traveled for 12.5 minutes before reaching the detector.
So we know the speed of the wave and the time it took to reach the detector, then we can use the equation:
Speed*Time = Distance.
to find the distance.
First, we should write the time in seconds
we know that:
1 min = 60 s
then:
12.5 min = 12.5*(60 s) = 750 s
Then, the wave traveled with a speed of 6 km/s for 750 seconds until it reached the detector, then the distance that it traveled is:
(6km/s)*750s = 4500 km
The distance between the detector and the site of the quake is around 4500 km.
A force of 10 N is making an angle of 30° with the horizontal. Its horizontal component will be:
A. 4N
B. 5N
C. 7N
D. 8.7 N
Answer:The answer is A
Explanation:just did it on a test
Una onda se propaga en un medio P, de tal forma que recorre una distancia 8D en un tiempo 4T. La misma onda cuando se propaga por un medio Q recorre una distancia 4D en el mismo tiempo anterior, es decir, en 4T. Respecto a esta onda es correcto afirmar que *
Sabemos todo material tiene asociado un coeficiente de difracción n.
De tal forma que la velocidad de una onda electromagnetica que viaja a travez de dicho material, será:
v = c/n
donde c es la velocidad de la luz.
Con la información dada, podremos concluir que el coeficiente de difracción del medio Q es dos veces el del medio P.
Ahora veamos como llegamos a esto:
Sabemos que en el medio P, la onda recorre una distancia 8*D en un tiempo 4*T
entonces la velocidad de la onda en el medio P es:
[tex]V_p = \frac{8*D}{4*T} = 2*\frac{D}{T}[/tex]
Mientras que en el medio Q, recorre una distancia 4*D en un tiempo 4*T, por lo que la velocidad en el medio Q será:
[tex]V_Q = \frac{4*D}{4*T} = \frac{D}{T}[/tex]
Podemos ver que:
[tex]V_P = 2*V_Q[/tex]
Reescribiendo las velocidades como el cociente entre la velocidad de la luz y el correspondiente coeficiente de difracción obtenemos:
[tex]\frac{c}{n_P} = 2*\frac{c}{n_Q} \\\\n_Q = 2*n_P[/tex]
Es decir, podemos concluir que el coeficiente de difracción del medio Q es dos veces el del medio P.
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An astronaut on Pluto attaches a small brass ball to a 1.00-m length of string and makes a simple pendulum. She times 10 complete swings in a time of 257 seconds. From this measurement she calculates the acceleration due to gravity on Pluto. What is her result
Answer:
The acceleration due to gravity at Pluto is 0.0597 m/s^2.
Explanation:
Length, L = 1 m
10 oscillations in 257 seconds
Time period, T = 257/10 = 25.7 s
Let the acceleration due to gravity is g.
Use the formula of time period of simple pendulum
[tex]T = 2\pi\sqrt{\frac{L}{g}}\\\\25.7 = 2 \times 31.4\sqrt{\frac{1}{g}}\\\\g = 0.0597 m/s^2[/tex]
In a simple machine the energy input is 120J if the efficiency of the machine is 80% calculate the energy output
Answer:
E_O = 96J
Explanation:
E_O = E_I*(%e/100%)
E_O = 120J*(80%/100%)
E_O = 96J
How many meters are in 10 miles?
Answer:
Explanation:
16093.4
Write any three importance of Measurement.
Stay Safe,Stay Healthy and Stay Happy
Please answer me this question fast, PLEASE
Explanation:
The three importance of measurement are as follows :-
measurement is used to compare the items when barter system takes place .measurement helps in weighing the foods , groceries etc .Measurement are important in laboratory when there experiments is being taken for different purposes .Hope it is helpful to you
✌️✌️✌️✌️✌️✌️✌️
define Neutons 3rd law .
Answer:
that when two object interact, they apply forces to one another that are equal in magnitude and opposite in direction.
Explanation:
Newton's third law: the law of action and reaction
A rocket blasts off. In 10.0 seconds it is at 10,000 ft, traveling at 3600 mph. Assuming the direction is up, calculate the acceleration.
Answer:
Explanation:
Givens
t = 10 seconds
vi = 0
vf = 3600 mph
a = ?
d = 10000 feet
Formula
a = (vf - vi)/t
Solution
vf = 3600 mph * 1 hr / 3600 seconds * 5280 feet / 1 mile = 5280 ft / sec
a = (5280 - 0)/10
a = 528 ft/sec^2
A position of a particle moving on an x axis is given by x=7•4 + 9•2t - 2•1 t^3, with x in meters and t in seconds. what is its velocity at t=3•5s? is the velocity,or is it Continuously changing?
Answer:
The velocity at that time would be [tex](-67.975\; \rm s)[/tex]. The velocity of this particle is continuously changing.
Explanation:
Differentiate the expression for position [tex]x[/tex] with respect to time [tex]t[/tex] to find an expression for velocity.
[tex]\begin{aligned}v(t) &= \frac{d}{dt}[x(t)] \\ &= \frac{d}{dt} \left[ 7.4 + 9.2\, t - 2.1\, t^{3}\right]\\ &= 9.2 - 6.3\, t^{2}\end{aligned}[/tex].
Hence, at [tex]t = 3.5\; \rm s[/tex], velocity would be [tex]v(3.5) = 9.2 - 6.3 \times (3.5)^{2} = -67.975\; \rm m[/tex].
Since velocity [tex]v(t)[/tex] changes with time [tex]t[/tex], the velocity of this particle is continuously changing.
(This is for other people with this Question i hope you find this when you need help) Need answer for this Help!!!
Question 7 of 20 You plan to use a slingshot to launch a ball that has a mass of 0.025 kg. You want the ball to accelerate straight toward your target at 19 m/s2. How much force do you need to apply to the ball? O A. 19.03 N OB. 0.48 N O C. 4.51 N D. 760.00 N
Answer:
0.48N
Explanation:
according to the second law of motion
force=mass×acceleration
the mass in this question is 0.025,the acceleration 19
therefore f=0.025×19
=0.48N
I hope this helps
