Answer:
The value is [tex]R_f = \frac{4}{5} R[/tex]
Explanation:
From the question we are told that
The initial velocity of the proton is [tex]v_o[/tex]
At a distance R from the nucleus the velocity is [tex]v_1 = \frac{1}{2} v_o[/tex]
The velocity considered is [tex]v_2 = \frac{1}{4} v_o[/tex]
Generally considering from initial position to a position of distance R from the nucleus
Generally from the law of energy conservation we have that
[tex]\Delta K = \Delta P[/tex]
Here [tex]\Delta K[/tex] is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
[tex]\Delta K = K__{R}} - K_i[/tex]
=> [tex]\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2[/tex]
=> [tex]\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2[/tex]
=> [tex]\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2[/tex]
And [tex]\Delta P[/tex] is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
[tex]\Delta P = P_f - P_i[/tex]
Here [tex]P_i[/tex] is zero because the electric potential energy at the initial stage is zero so
[tex]\Delta P = k * \frac{q_1 * q_2 }{R} - 0[/tex]
So
[tex]\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0[/tex]
=> [tex]\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}[/tex]
=> [tex]- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )[/tex]
Generally considering from initial position to a position of distance [tex]R_f[/tex] from the nucleus
Here [tex]R_f[/tex] represented the distance of the proton from the nucleus where the velocity is [tex]\frac{1}{4} v_o[/tex]
Generally from the law of energy conservation we have that
[tex]\Delta K_f = \Delta P_f[/tex]
Here [tex]\Delta K[/tex] is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
[tex]\Delta K_f = K_f - K_i[/tex]
=> [tex]\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2[/tex]
=> [tex]\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2[/tex]
=> [tex]\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2[/tex]
And [tex]\Delta P[/tex] is the change in electric potential energy from initial position to a position of distance [tex]R_f[/tex] from the nucleus , this is mathematically represented as
[tex]\Delta P_f = P_f - P_i[/tex]
Here [tex]P_i[/tex] is zero because the electric potential energy at the initial stage is zero so
[tex]\Delta P_f = k * \frac{q_1 * q_2 }{R_f } - 0[/tex]
So
[tex]\frac{1}{2} * m * \frac{1}{8} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R_f }[/tex]
=> [tex]\frac{1}{2} * m *v_o^2 [-\frac{15}{16} ] = k * \frac{q_1 * q_2 }{R_f }[/tex]
=> [tex]- \frac{15}{32} * m *v_o^2 = k * \frac{q_1 * q_2 }{R_f } ---(2)[/tex]
Divide equation 2 by equation 1
[tex]\frac{- \frac{15}{32} * m *v_o^2 }{- \frac{3}{8} * m *v_0^2 } } = \frac{k * \frac{q_1 * q_2 }{R_f } }{k * \frac{q_1 * q_2 }{R } }}[/tex]
=> [tex]-\frac{15}{32 } * -\frac{8}{3} = \frac{R}{R_f}[/tex]
=> [tex]\frac{5}{4} = \frac{R}{R_f}[/tex]
=> [tex]R_f = \frac{4}{5} R[/tex]
18
AX
A bicycle rider travels 15 km in 1.25 hours. What is the rider's average speed? v=
At
rage speed [v-ante
10.5 km/h
13.75 km/h
12 km/h
22.5 km/h
The required average speed of bicycle rider is 12 km/h.
Given data:
The distance travelled by the bicycle rider is, d = 15 km.
Time taken to cover the distance is, t = 1.25 h.
Here, we need to use the simple relation between the speed, time and distance. The distance covered covered by any object per unit of time is known as average speed of object. And the mathematical expression for the average speed is given as,
[tex]v = \dfrac{d}{t}[/tex]
Here, v is the average speed.
Solving as,
[tex]v = \dfrac{15}{1.25}\\\\v = 12\;\rm km/h[/tex]
Thus, we can conclude that the required average speed of bicycle rider is 12 km/h.
Learn more about the average speed here:
https://brainly.com/question/12322912
oo hi granger ru online here i have a doubt in physics .
Answer:
.
Explanation:
In order to model the motion of an extinct ape, scientists measure its hand and arm bones. From shoulder to wrist, the arm bones are 0.60 m long and their mass is 4.0 kg. From wrist to the tip of the fingers, the hand bones are 0.10 m long and their mass is 1.0 kg. In the model above, each bone is assumed to have a uniform density. When the arm and hand hang straight down, the distance from the shoulder to the center of mass of the arm-hand system is most nearly
Answer:
0.37 m
Explanation:
Let the shoulder be the origin.
The center of mass of the arm bones is 0.60 m/2 = 0.30 m and the center of mass of the hand bones is 0.10 m/2 = 0.05 m since they are modeled as straight rods with uniform density and the center of mass of a rod is x = L/2 where L is the length of the rod.
The center of mass y = (m₁y₁ + m₂y₂)/(m₁ + m₂) where m₁ = mass of arm bones = 4.0 kg, y₁ = distance center of mass of arm bones from shoulder = 0.30 m, m₂ = mass of hand bones = 1.0 kg and y₂ = distance of center of mass hand bones from shoulder = x₁ + distance of center of hand bones from wrist = 0.60 m + 0.05 m = 0.65 m
Substituting these into the equation for the center of mass, we have
y = (m₁y₁ + m₂y₂)/(m₁ + m₂)
y = (4.0 kg × 0.30 m + 1.0 kg × 0.65 m)/(4.0 kg + 1.0 kg)
y = (1.20 kgm + 0.65 kgm)/5.0 kg
y = 1.85 kgm/5.0 kg
y = 0.37 m
The distance of the center of mass from the shoulder is thus y = 0.37 m
Two workers are sliding 500 kg crate across the floor. One worker pushes forward on the crate with a force of 440 N while the other pulls in the same direction with a force of 340 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor
Answer:
0.159
Explanation:
Given that
F1 = 440 N
F2 = 340 N
m = 500 kg
acceleration, a = 0 m/s²
Remember that F = m.a
If F - f(f) = 0, then F = f(f)
N = mg
f(f) = μN, substituting for N, we have
f(f) = μmg
The total force, F = F1 + F2
F = 440 + 340
F = 780 N
Recall that we'd already proven that
F = f(f), so F = 780 N = f(f)
And again, f(f) = μmg, if we substitute for all the values, we have
780 = μ * 500 * 9.81
780 = μ * 4905
μ = 780 / 4905
μ = 0.159
Therefore, the coefficient of static friction of the crate on the floor is 0.159
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be when it is very far from the Earth
Answer:
The value is [tex]v = 2.3359 *10^{4} \ m/s[/tex]
Explanation:
From the question we are told that
The initial speed is [tex]u = 2.05 *10^{4} \ m/s[/tex]
Generally the total energy possessed by the space probe when on earth is mathematically represented as
[tex]T__{E}} = KE__{i}} + KE__{e}}[/tex]
Here [tex]KE_i[/tex] is the kinetic energy of the space probe due to its initial speed which is mathematically represented as
[tex]KE_i = \frac{1}{2} * m * u^2[/tex]
=> [tex]KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2[/tex]
=> [tex]KE_i = 2.101 *10^{8} \ \ m \ \ J[/tex]
And [tex]KE_e[/tex] is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as
[tex]KE_e = \frac{1}{2} * m * v_e^2[/tex]
Here [tex]v_e[/tex] is the escape velocity from earth which has a value [tex]v_e = 11.2 *10^{3} \ m/s[/tex]
=> [tex]KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2[/tex]
=> [tex]KE_e = 6.272 *10^{7} \ \ m \ \ J[/tex]
Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as
[tex]KE_p = \frac{1}{2} * m * v^2[/tex]
Generally from the law energy conservation we have that
[tex]T__{E}} = KE_p[/tex]
So
[tex]2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2[/tex]
=> [tex]5.4564 *10^{8} = v^2[/tex]
=> [tex]v = \sqrt{5.4564 *10^{8}}[/tex]
=> [tex]v = 2.3359 *10^{4} \ m/s[/tex]
What happens between particles of different charges?
Answer:
Oppositely charged objects will exert an attractive influence upon each other. In contrast to the attractive force between two objects with opposite charges, two objects that are of like charge will repel each other.
Explanation:
A 3 kg exercise ball is held 2m above the ground. What is the gravitational potential energy?
Answer:
58.56 J
Explanation:
Since the formula for gravitational potential energy is:
GE = m x g x h - where m = mass (kg), g = acceleration due to gravity (9.81 m/s²), h = height (m)
GE= 3 x 9.81 x 2
GE = 58.86 J
Hope this helps
During an isothermal process one mole of a monoatomic gas did 3000 J of work on its surrounding. The final volume and pressure of the gas are 25 L and 1 atm, respectively. What was the initial volume of the gas
Answer: The initial volume of the gas is 7.72 L
Explanation:
For an isothermal process the temperature is constant.
[tex]PV=nRT[/tex]
as P = pressure = 1 atm ,
V = Volume = 25 L
n = moles
R= gas constant
T = temperature
[tex]PV=1atm\times 25L[/tex]
[tex]nRT=25Latm=25\times 101.3J=2532.5J[/tex] (1Latm=101.3 J)
For isothermal reaction :
[tex]w=-2.303nRT\log\frac{V_2}{V_1}[/tex]
where , w = work done by system = -ve
n = moles = 1
[tex]V_2[/tex] = final volume = 25 L
[tex]V_1[/tex] = initial volume = ?
[tex]-3000J=-2.303\times 2532.5\log \frac{25}{V_1}[/tex]
[tex]V_1=7.72L[/tex]
Thus initial volume of the gas is 7.72 L
Define each of the three heat transfer methods:
Conduction
Convection
Radiation
What determines the type of air mass that forms in an area?
Question 11 options:
The amount of oxygen present
The amount of air present
The direction of air flow
The location where it forms
Answer: no because you have left the number
Explanation:
a box is 30 cm wide, 40 cm long and 25 cm high calculate: what is the area of its base.
Answer:
1200cm²
Explanation:
Width= 30cm
Length= 40cm
Area of base(A) = Width×Length
= 30cm×40cm
= 1200cm²
Taking into account the definition of reactanguar prism and area of rectangle, the area of its base is 1200 cm².
A rectangular prism is a polyhedron whose surface is formed by two equal and parallel rectangles called bases and by four lateral faces that are also parallel and equal rectangles two by two.
So, the base being a rectangle, its area is calculated as the multiplication between the base and the height. In this case, these values correspond to the width and length of the figure.
Then, in this case, you know:
Width= 30cm Length= 40cm
Being:
Area of base (A) = Width×Length
Then, the are of base (A) is calculated as:
Area of base (A)= 30cm×40cm
Solving:
Area of base (A)= 1200cm²
Finally, the area of its base is 1200 cm².
Learn more:
https://brainly.com/question/20360303The temperature of the surface of the Sun is 5500°C.
a. What is the average kinetic energy, in joules, of hydrogen atoms on the surface of the Sun?
b. What is the average kinetic energy, in joules, of helium atoms in a region of the solar corona where the temperature is 6.00 times 10^5 K?
Answer:
a. the average kinetic energy of hydrogen atoms is 1.20 × 10^-19J
b. the average kinetic energy of helium atoms is 1.24 × 10^-17J
Explanation:
The computation is shown below;
As we know that
Kinetic energy = 3 ÷ 2 kT
where,
K = Boltzmann constant
And, T = Temperature
a. Now the temperature in kelvin is
T = (5,500 × (°C ÷ K) + 273.15 K)
= 5773.15 K
As
Kinetic energy = 3 ÷ 2 kT
So now 1.38 × 10^-23 J/K for K would be substituted and 5773.15 K for Temperature T
Now Kinetic energy is
= 3 ÷ 2 (1.38 × 10^-23 J/K) ( 5773.15 K)
= 1.20 × 10^-19J
hence, the average kinetic energy of hydrogen atoms is 1.20 × 10^-19J
b. As
Kinetic energy = 3 ÷ 2 kT
now 1.38 × 10^-23 J/K for K would be substituted and 6 × 10^5K for Temperature T
Now Kinetic energy is
= 3 ÷ 2 (1.38 × 10^-23 J/K) (6 × 10^5K )
= 1.24 × 10^-17J
hence, the average kinetic energy of helium atoms is 1.24 × 10^-17J
It is difficult to lift a bigger stone than the smaller
stone. why?
plzzz give me short and brillient answer
A spring with a spring constant of 1730 N/m is compressed 0.136 m from its equilibrium position with an object having a mass of 1.72 kg. The spring is released and the object moves along a frictionless surface when it reaches a small embankment. If the speed of the object is 2.45 m/s at location A, what is the embankment height h
Given :
A spring with a spring constant of 1730 N/m is compressed 0.136 m from its equilibrium position with an object having a mass of 1.72 kg.
To Find :
The embankment in the height.
Solution :
Since no external force is acting in the system, therefore total energy will be conserved.
Initial kinetic energy of the object = Energy stored in spring
[tex]K.E _i = \dfrac{kx^2}{2}\\\\K.E_i = \dfrac{1730\times 0.136^2}{2}\\\\K.E_i = 16\ J[/tex]
Also, initial potential energy is 0.
Now,
[tex]K.E_i + P.E_i = K.E_f + P.E_f\\\\16 + 0 = \dfrac{1.72\times 2.45^2}{2}+ mgh\\\\mgh =16 - 5.16\\\\h = \dfrac{16 - 5.16}{1.72 \times 9.8}\\\\h = 0.64\ m[/tex]
Therefore, the embankment height is 0.64 m.
Compare the weight of a 60 kg person on the earth with the weight of the same person on
the moon. Then, describe a quick (but very costly) way for dieters
at NASA to lose weight.
Answer:
Explanation:
The formula for weight is
W = mg, where
W = the weight of the object or person
m = mass of the object or person
g = acceleration due to gravity
Now, we're given the mass of the person to be 60 kd, and thus, the weight of that person would be
W = 60 * 9.81
W = 588.6 N
On the surface of the moon, the weight of the person would be
W = 60 * 1.625
W = 97.5 N
Therefore, the weight of the person on both surfaces are 588.6 and 97.5 respectively
How would taping a coin to the balloon affect the overall motion of the balloon?
Answer:
the motion of the coin taping the balloon is the balloon squshing down
the tighter the threads of the screw the greater
A rotating space station simulates artificial gravity by means of centripetal acceleration at the rim. The radius of the station is 1156 m and the apparent acceleration of gravity at the rim is 6.13 m/s2. What is the rotation rate of the station in rpm (revolutions per minute).
a= 1156× 6.13= 7086.28
Explanation:
a = m× m/ s^2
Ist Law: a object continues in a state of
rest or motion unless an external force
applied to it
This image shows an example of
the first law because:
Answer:
Explanation:
This image is the example of Newton's first law of motion because:
As per Newton's first law, football will remain in the state of rest until a player applies an external force by kicking the ball.
And the ball will keep on moving until unless the net of a goal post exerts the external force to stop the ball.
Car 1 drives 20 mph to the south, and car 2 drives 30 mph to the north. From
the frame of reference of car 1, what is the velocity of car 2?
Answer:
10 mph faster than car 1 is going
Explanation:
g Estimate the number of photons emitted by the Sun in a second. The power output from the Sun is 4 × 1026 W and assume that the average wavelength of each photon is 550 nm.
Answer:
The value is [tex]N = 1.107 *10^{45 } \ photons[/tex]
Explanation:
From the question we are told that
The power output from the sun is [tex]P_o = 4 * 10^{26} \ W[/tex]
The average wavelength of each photon is [tex]\lambda = 550 \ nm = 550 *10^{-9} \ m[/tex]
Generally the energy of each photon emitted is mathematically represented as
[tex]E_c = \frac{h * c }{ \lambda }[/tex]
Here h is the Plank's constant with value [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
So
[tex]E_c = \frac{6.62607015 * 10^{-34} * 3.0 *10^{8} }{ 550 *10^{-9} }[/tex]
=> [tex]E_c = 3.614 *10^{-19} \ J[/tex]
Generally the number of photons emitted by the Sun in a second is mathematically represented as
[tex]N = \frac{P }{E_c}[/tex]
=> [tex]N = \frac{4 * 10^{26} }{3.614 *10^{-19}}[/tex]
=> [tex]N = 1.107 *10^{45 } \ photons[/tex]
Please help
4. What are the lowest points on a transverse wave called?
a. Crests
b. troughs
C. compressions
d. rarefractions
5. Any substance that a wave moves through is called a
a. medium
b. vibrate
C. crest
d. frequency
Answer:
Explanation:
the lowest points of a transverse wave are called the troughs what is the wavelength of a wave traveling through a rope if the distance from one crest to the next is 1 meter a. 2 m
4. Low points are called troughs
5. Medium: substance through which a wave can travel
A toy car weighing 3.2 N gets pushed a distance of 10 m in 1.3 s. What is the momentum of the car?
Answer:
2.51 kg * m/s
Explanation:
In order to find momentum, use the equation below:
momentum = mass * velocity.
Since neither mass nor velocity was given, you must solve for both variables.
In order to solve for mass, use the force equation for its weight / gravitational force.
Fg (gravitational force) = 3.2 N = ma = 9.8m
mass = 3.2 N / 9.8 m/s^2 = 0.326531 kg
In order to solve for velocity, use the equation:
velocity = displacement / time
velocity = 10m / 1.3 s = 7.69231 m/s
Momentum = mass * velocity = 0.326531 kg * 7.69231 m/s = 2.51177 kg * m/s = 2.51 kg * m/s
what belongs in the center section
Answer:
The second one I think
Explanation:
B
diffine thermodynamics
Answer:
The science of the conversions between heat and other forms of energy is thermodynamics.
Hope it helps :)
Answer:
thermodynamic is the branch of physics that deals with the relationship between heat and other forms of energy .it describes how thermal energy is converted to other forms and how it will affect matter.
Explanation:
To celebrate a victory, a pitcher throws her glove straight upward with an initial speed of 5.3 m/s. How long does it take for the glove to reach its maximum height
Hello!!
For the maximum height the final velocity is zero, because can't up more.
Then, use the formula:
V = Vi + gt
Replacing, we have:
0 m/s = 5,3 m/s + (-9,8 m/s² * t)
0 m/s - 5,3 m/s = -9,8 m/s² * t
(-5,3 m/s) / -9,8 m/s² = t
t = 0,54 s
The time it will take to reach the maximum height is 0,54 seconds.
A person who weighs 614 N is riding a 85-N mountain bike. Suppose the entire weight of the rider plus bike is supported equally by the two tires. If the gauge pressure in each tire is 9.00 x 10^5 Pa. What is the area of contact between each tire and the ground?
Answer:
3.88 × 10^-4 m
Explanation:
Given that a person who weighs 614 N is riding a 85-N mountain bike. Suppose the entire weight of the rider plus bike is supported equally by the two tires. If the gauge pressure in each tire is 9.00 x 10^5 Pa. What is the area of contact between each tire and the ground?
The total weight = 614 + 85
The total weight = 699N
Let the total area of contact = A
Pressure = Force / A
Substitute all the parameters into the formula
900000 = 699 /A
A = 699 / 900000
A = 7.77 × 10^-4 m
The area of contact between each tire and the ground will be = A/2
That is, 7.77 / 2 = 3.88 × 10^-4 m
Therefore, the area of contact between each tire and the ground is 3.88 × 10^-4 m approximately.
A ball is thrown with a speed of 100 ft/s in a direction of 30 degrees above the horizontal. determine the horizontal distance in m, the time of flight, and the height to which it rises..Note: you have to label your answer and final answer must be two decimal place
Answer:
Horizontal distance covered is 270.63ft
Time of flight is 3.125secs
Height to which it rises is 39.06feet
Explanation:
The horizontal distance covered is the range;
Range = U²sin2theta/g
Range = 100²sin2(30)/32
Range = 10000sin60/32
Range = 10000(0.8660)/32
Range = 8660/32
Range = 270.625ft
Hence the horizontal distance covered is 270.63ft
The time of flight is expressed using the formula;
T = 2Usin theta/g
g is the acceleration due to gravity
T = 2(100)sin30/32
T = 200(0.5)/32
T = 100/32
T = 3.125secs
Hence the time of flight is 3.125secs
Maximum height H = u²sin²theta/2g
H = 100²(sin30)²/2(32)
H = 10000(0.5²)/64
H = 10000(0.25)/64
H = 2500/64
H = 39.0625
Hence the height to which it rises is 39.06feet
How original were Newton’s contributions to science? (In what ways did Newton depend on the mechanical view?)
A child pulls a sled up a snow covered hill. If the child does 504J of work on the sled while pulling the sled 23m up the hill then how much force did they exert?
Explanation:
ans is equal to 504j* 23 m* 10 ms