Answer:
wareffctgggyyggghhhh
two point charges with charge q are initially separated by a distance d. if you double the charge on both charges, how far should the charges be separated for the potential energy between them to remain the same
Answer:
r ’= 4 r
Explanation:
Electric potential energy is
U = [tex]k \frac{q_1q_2}{r_{12}}[/tex]k q1q2 / r12
in this exercise
q₁ = q₂ = q
U = k q² / r
for when the charge change
U ’= k q’² / r’
indicate that
q ’= 2q
U ’= U
we substitute
U = k (2q) ² / r ’
U = 4 k q² / r ’
we substitute
[tex]k \ \frac{q^2}{r} = 4 k \ \frac{q^2}{r'}[/tex]k q² / r = 4 k q² / r ’
r ’= 4 r
A double-slit experiment is performed with light of wavelength 550 nm. The bright interference fringes are spaced 2.3 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 360 nm?
Answer:
[tex]d_2=1.5*10^-3m[/tex]
Explanation:
From the question we are told that:
Initial Wavelength [tex]\lambda_1=550nm=550*10^{-9}[/tex]
Space 1 [tex]d_1=2.3*10^{-3}[/tex]
Final wavelength [tex]\lambda_2=360*10^{-9}[/tex]
Generally the equation for Fringe space at [tex]\lambda _2[/tex] is mathematically given by
[tex]d_2=\frac{d_1}{\lambdaI_1}*\lambda_2[/tex]
[tex]d_2=\frac{2.3*10^{-3}}{550*10^{-9}}*360*10^{-9}[/tex]
[tex]d_2=1.5*10^-3m[/tex]
A uniform metre rule of mass 10g is balanced on a knife edge placed at 45cm mark. Calculate the distance of a mass 25g from the pivot
Answer:
The distance of a mass 25g from the pivot is 18cm
Explanation:
Given
[tex]m_1 = 10g[/tex]
[tex]d_1 = 45cm[/tex]
[tex]m_2 = 25g[/tex]
Required
Distance of m2 from the pivot
To do this, we make use of:
[tex]m_1 * d_1 = m_2 * d_2[/tex] --- moments of the masses
So, we have:
[tex]10 * 45= 25* d_2[/tex]
[tex]450= 25* d_2[/tex]
Divide both sides by 25
[tex]18= d_2[/tex]
Hence:
[tex]d_2 = 18[/tex]
collisions may take place between
Answer:
Collisions may take place between the reactants.
Explanation:
The collision frequency must be greater than the frequency factor for the reaction. A collision between the reactants must occur.
High-speed stroboscopic photographs show that the head of a -g golf club is traveling at m/s just before it strikes a -g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at m/s. Find the speed of the golf ball just after impact.
The question is incomplete. The complete question is :
High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 60 m/s just before it strikes a 50 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40 m/s. Find the speed of the golf ball just after impact.
Solution :
We know that momentum = mass x velocity
The momentum of the golf club before impact = 0.200 x 60
= 12 kg m/s
The momentum of the ball before impact is zero. So the total momentum before he impact is 12 kg m/s. Therefore, due to the conservation of momentum of the two bodies after the impact is 12 kg m/s.
Now the momentum of the club after the impact is = 0.2 x 40
= 8 kg m/s
Therefore the momentum of the ball is = 12 - 8
= 4 kg m/s
We know momentum of the ball, p = mass x velocity
4 = 0.050 x velocity
∴ Velocity = [tex]$\frac{4}{0.050}$[/tex]
= 80 m/s
Hence the speed of the golf ball after the impact is 80 m/s.
In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed from its unstretched length
Answer: hello below is the missing part of your question
A mass m = 10 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5029 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.49. The mass leaves the spring at a speed v = 3.4 m/s.
answer
x = 0.0962 m
Explanation:
First step :
Determine the length of the rough patch/spot
F = Uₓ (mg)
and w = F.d = Uₓ (mg) * d
hence;
d( length of rough patch) = w / Uₓ (mg) = 46.55 / (0.49 * 10 * 9.8) = 0.9694 m
next :
work done on unstretched spring length
Given that block travels halfway i.e. d = 0.9694 / 2 = 0.4847 m
w' = Uₓ (mg) * d
= 0.49 * 10 * 9.81 * 0.4847 = 23.27 J
also given that the Elastic energy of spring = work done ( w')
1/2 * kx^2 = 23.27 J
x = [tex]\sqrt{\frac{2*23.27}{5029} }[/tex] = 0.0962 m
write down the following units in the ascending of their value A) mm nm cm um B) 1m 1cm 1km 1mm. convert the following units into SI without changing their values? A)3500g B)2.5km C)2h
Answer:
A) nm, um, mm, cm
B) 1mm, 1cm, 1m, 1km
A) 3500g, B) 2500m, C) 7200 seconds
recognizing forms of energy
Answer:
hi the question isn't obvious and need a photo I guess
A particle of mass 1.2 mg is projected vertically upward from the ground with a velocity of 1.62 x 10 cm/h. Use the above information to answer the following four questions: 7. The kinetic energy of the particle at time t = 0 s is A. 1.215 x 10-3 J B. 2.430 J C. 1215 J D. 9.72 x 106 J E. OJ (2)
Answer:
K = 0 J
Explanation:
Given that,
The mass of the particle, m = 1.2 mg
The speed of the particle, [tex]v=1.62\times 10\ cm/h[/tex]
We need to find the kinetic energy of the particle at time t = 0 s.
At t = 0 s, the particle is at rest, v = 0
So,
[tex]K=\dfrac{1}{2}mv^2[/tex]
If v = 0,
[tex]K=0\ J[/tex]
So, the kinetic energy of the particle at time t = 0 s is 0 J.
There are two beakers of water on the table. We can compare the average kinetic energy of the water molecules in the two beakers by measuring their
A temperatures.
B volumes.
C densities.
D masses.
Answer: masses
Explanation:
Trust me
When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not noticeable?
Answer:
Yes, when an apple falls towards the earth, the apple gets accelerated and comes down due to the gravitational force of attraction used by the earth. The apple also exerts an equal and opposite force on the earth but the earth does not move because the mass of the apple is very small, due to which the gravitational force produces a large acceleration in it (a = F/m) but the mass of the earth is very large, the same gravitational force produces very small acceleration in the earth and we don't see the earth rising towards the apple.
If a 1.3 kg mass stretches a spring 4 cm, how much will a 5.8 kg mass stretch the
spring? Show MATH, answer and unit.
Answer:
17.8cm
Explanation:
1.3kg --> 4cm
1kg --> 3, 1/13cm
5.8kg --> 18.8cm
IS ANYONE THERE..??!
Answer:
hmmmmmmmmmmmmmmmmmmmmmmmm y
At which point is the kinetic energy of the pendulum the greatest?
C
A
D
B
Answer:
Point C
Explanation:
Greatest Kinetic Energy means lowest potential energy since energy is conserved. Lowest potential energy means lowest height which is at Point C.
What quantity of heat is transferred when a 150.0g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of the heat flow?
Answer:
Heat is flowing into the metal.
Explanation:
From the question given above, the following data were obtained:
Mass (M) of iron = 150 g
Initial temperature (T₁) = 25.0°C
Final temperature (T₂) = 73.3°C
Direction of heat flow =?
Next, we shall determine the change in the temperature of iron. This can be obtained as follow:
Initial temperature (T₁) = 25.0 °C
Final temperature (T₂) = 73.3 °C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 73.3 – 25
ΔT = 48.3 °C
Next, we shall determine the heat transfered. This can be obtained as follow:
Mass (M) of iron = 150 g
Change in temperature (ΔT) = 48.3 °C
Specific heat capacity (C) of iron = 0.450 J/gºC
Heat (Q) transfered =?
Q = MCΔT
Q = 150 × 0.450 × 48.3
Q = 3260.25 J
Since the heat transferred is positive, it means the iron metal is absorbing the heat. Thus, heat is flowing into the metal.
In 2009 Usain Bolt set the world record time by running 100 meters in 9.58 s. Assume that during this race he ran in a straight line with constant acceleration a. What would be the required constant acceleration a
In 2009 Usain Bolt set the world record time by running 100 meters in 9.58 seconds, assuming that he ran this race with constant acceleration, then the required constant acceleration would have been
What are the three equations of motion?There are three equations of motion given by Newton
v = u + at
S = ut + 1/2×a×t²
v² - u² = 2×a×s
By using the second equation of motion given by Newton,
S = ut + 1/2at²
100= 0 + 0.5*a*9.58²
a = 2.17 meters / second²
Thus,the required constant acceleration of Usain Bolt would have been 2.17 meters / second².
To learn more about equations of motion here, refer to the link;
brainly.com/question/5955789
#SPJ2
Typhoon signal number 2 is raised. What is the speed of the expected typhoon?
the simple answer is from 61kmph to 120kmph
Explanation:
no explanation is needed
A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg. How far will block m drop in the first seconds after the system is released?
How long will block M move during above time?
At the time, calculate the velocity of block M
Find out the deceleration of the block M, if the connected string is
removal by cutting after the first second. Then, calculate the time
taken to contact block M and pulley.
Answer:
a) y = 0.98 t², t=1s y= 0.98 m,
b) he two blocks must move the same distance
c) v = 1.96 m / s, d) a = -1.96 m / s², e) x = 0.98 m
Explanation:
For this exercise we can use Newton's second law
Big Block
Y axis
N-W = 0
N = M g
X axis
T- fr = Ma
the friction force has the expression
fr = μ N
fr = μ Mg
small block
w- T = m a
we write the system of equations
T - fr = M a
mg - T = m a
we add and resolved
mg- μ Mg = (M + m) a
a = [tex]g \ \frac{m - \mu M}{m+M}[/tex]
a = [tex]9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}[/tex]
a = 9.8 (6/30)
a = 1.96 m / s²
a) now we can use the kinematic relations
y = v₀ t + ½ a t²
the blocks come out of rest so their initial velocity is zero
y = ½ a t²
y = ½ 1.96 t²
y = 0.98 t²
for t = 1s y = 0.98 m
t = 2s y = 1.96 m
b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.
As the curda is in tension the two blocks must move the same distance
c) the velocity of the block M
v = vo + a t
v = 0 + 1.96 t
for t = 1 s v = 1.96 m / s
t = 2 s v = 3.92 m / s
d) the deceleration if the chain is cut
when removing the chain the tension becomes zero
-fr = M a
- μ M g = M a
a = - μ g
a = - 0.2 9.8
a = -1.96 m / s²
e) the distance to stop the block is
v² = vo² - 2 a x
0 = vo² - 2a x
x = vo² / 2a
x = 1.96² / 2 1.96
x = 0.98 m
the time to travel this distance is
v = vo - a t
t = vo / a
t = 1.96 /1.96
t = 1 s
Explain what a circuit breaker is and how it helps protect your house?
Explanation:
A circuit breaker is an electrical switch designed to protect an electrical circuit from damage caused by overcurrent/overload or short circuit. Its basic function is to interrupt current flow after protective relays detect a fault.
Circuit breakers have been designed to detect when there is a fault in the electricity, so it will “trip” and shut down electrical flow. ... This detection is key to preventing surges of electricity that travel to appliances or other outlets, which can cause them to break down
The Cleveland City Cable Railway had a 14-foot-diameter pulley to drive the cable. In order to keep the cable cars moving at a linear velocity of 14 miles per hour, how fast would the pulley need to turn (in revolutions per minute)
Answer:
13.94 rpm
Explanation:
Given that,
The diameter of the pulley, d = 14 foot
Radius, r = 7 foot
The linear velocity of the pulley, v = 14 mph = 20.53 ft/s
We need to find the angular velocity in rpm.
We know that, the relation between the linear velocity and the angular velocity is as follows :
[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\\omega=\dfrac{20.53}{14}\\\\\omega=1.46\ rad/s[/tex]
or
[tex]\omega=13.94\ rpm[/tex]
So, the angular velocity of the pulley is 13.94 rpm.
What word chemical equation describes this chemical reaction?
Answer : sodium + chlorine → sodium chloride
what is the dimensional formula of force and torque
Answer:
Units. Torque has the dimension of force times distance, symbolically T−2L2M. Although those fundamental dimensions are the same as that for energy or work, official SI literature suggests using the unit newton metre (N⋅m) and never the joule. The unit newton metre is properly denoted N⋅m.
Dimension: M L2T−2
In SI base units: kg⋅m2⋅s−2
Other units: pound-force-feet, lbf⋅inch, ozf⋅in
Answer:
hope it is helpful to you
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Explain the following observations:
a) A balloon filled with hydrogen gas floats in air;
B) A ship made of steel floats on water.
Answer and Explanation:
a. An oxygen-filled balloon is not able to float in the air, because the oxygen inside the balloon is of the same density, that is, the same "weight" as the oxygen outside the balloon and present in the atmosphere. The balloon can only float if the gas inside it is less dense than atmospheric oxygen. Helium gas is less dense than atmospheric gas, so if a balloon is filled with helium gas, that balloon will be able to float because of the difference in density.
b. The ship is able to float in the water because its steel construction is hollow and full of air. This makes the average density of this ship less than the density of water, which makes the ship lighter than water and for this reason, this ship is able to float. In addition, the ship is partially immersed, allowing the weight of the ship on the water to counteract the buoyant force that the water promotes on the ship. Weight and buoyant are two opposing forces that keep the ship afloat.
A hot air balloon is a sphere of volume 2210 m3. The density of the hot air inside is 1.13 kg/m3, while the air outside has a density of 1.29 kg/m3. The balloon itself has a mass of 240 kg. What is the TOTAL NET force acting on the balloon?
[?]N
The total net force acting on the balloon will be 24498 Newtons
Given that
Volume of the balloon = 2210 cubic meter
Density of the air inside the balloon = 1.13 kg/m3
What will be the net force exerted on the balloon ?Here force on the balloon will be equal to the weight of the air displaced by balloon
[tex]F= mass of air displaced\times gravity[/tex]
[tex]F= Density \times volume \times gravity[/tex]
[tex]F=1.13 \times 2210 \times 9.81[/tex]
[tex]F=24498 N[/tex]
The total net force acting on the balloon will be 24498 Newtons
To know more about buoyancy force follow
https://brainly.com/question/117714
Every object around you is attracted to you. In fact, every object in the galaxy is attracted to every other object in the galaxy.
a. True
b. False
Answer:
True
Explanation:
With the gravitational pull that our planets have, we are able to remain in orbit. This demonstrates how every object in the galaxy is attracted to every other object. Every object in the universe that has mass exerts a gravitational pull on every other mass. We as humans do it too, but since our force isn't strong, we don't have much of an effect. I hope this helped and please don't hesitate to reach out with more questions!
A car is moving with a velocity of45m/sis brought to rest in 5s.the distance travelled by car before it comes to rest is
Answer:
The car travels the distance of 225m before it comes to rest.
Explanation:
Given,
v = 45m/s
t = 5s
Therefore,
d = v × t
= 45 × 5
= 225m
Question 8 of 10
What was the name of the book that Ibn al-Haytham wrote?
A. Weather and Air Flow
B. Book of Optics
C. Light and Vision
D. Book of Sound
Answer:b
Explanation:
Two charged particles exert an electric force of 27 N on each other. What will the magnitude of the force be if the distance between the particles is reduced to one-third of the original separation
Answer:
243 N
Explanation:
The formula for electromagnetic force is F= Kq1q2/r^2
where r is the distance between the charges, if the distance between the charges is reduced by 1/3 then F will increase by 9 [(1/3r)^2 becomes 1/9r which is 9F] so 27*9 is 243N
A 0.495-kg hockey puck, moving east with a speed of 4.50 m/s , has a head-on collision with a 0.720-kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed (magnitude of the velocity) of each object after the collision?
Answer:
a) [tex]v_1=-0.833m/s[/tex]
b) [tex]V_2=12.5m/s[/tex]
Explanation:
From the question we are told that:
Hockey puck Mass [tex]m_1=0.495kg[/tex]
Hockey puck Speed [tex]u_1=4.50m/s[/tex]
Puck Mass [tex]m_2=0.720kg[/tex]
Assuming
Initial speed of Puck [tex]u_2=0[/tex]
Generally the equation for Speed of First Puck is mathematically given by
[tex]v_1=(\frac{m_1-m_2}{m_1+m_2})*v_1+(\frac{2m_2}{m_1+m_2})u_2[/tex]
[tex]v_1=(\frac{0.495-0.720}{0.495+0.720})*4.50+0[/tex]
[tex]v_1=-0.833m/s[/tex]
Generally the equation for Speed of Second Puck is mathematically given by
[tex]V_2=(\frac{2m_1}{m_1+m_2})u_2-(\frac{m_1-m_2}{m_1+m_2})v[/tex]
[tex]V_2=(\frac{2*0.495}{0.495*0.720})*4.50-0[/tex]
[tex]V_2=12.5m/s[/tex]
Three 30 g metal balls, one of aluminum, copper and lead, are placed in a large beaker of hot water for a few minutes. [The specific heats of aluminum, copper, and lead are 903, 385, and 130 J / (kg ° C), respectively].
to. Which of the balls, if any, will reach the highest temperature? Explain.
b. Which of the balls, if any, will have the most heat energy? Explain.
Answer:
The answer is below
Explanation:
Specific heat capacity is an intensive property of a material. The specific heat of a material is the amount of energy required to raise the temperature of one unit mass m of material by one unit of temperature.
a) Temperature is inversely proportional to specific heat capacity. If the same amount of heat is applied to all three balls, the ball that will reach the highest temperature is the ball with the least specific heat capacity.
Hence lead will have the highest temperature since it has the least specific heat capacity.
b) The quantity of heat is directly proportional to the specific heat capacity. Hence if all balls experience the same temperature change, the ball that have the most energy will be that with the highest specific heat capacity.
Hence aluminum will have the most heat since it has the highest specific heat capacity.
