Answer:
[tex]v_{0,new} = v0\sqrt{}2[/tex]
Explanation:
Initial work done on the proton is given by, [tex]\DeltaW0 = q V_o[/tex]
we know that, [tex]\DeltaW = \DeltaK.E[/tex]
[tex]qV0 = (1/2) m v_0^2[/tex]
[tex]v_0 = \sqrt{}2 q V_0 / m[/tex] { eq.1 }
If it were accelerated instead through a potential difference of 2V0, then it would gain a speed will be given as :
using the above formula, we have
[tex]v_{0,new} = \sqrt{}2 q (2V0) / m[/tex]
[tex]v_{0,new} = \sqrt{}4 q V0 / m[/tex]
[tex]v_{0,new} = v0\sqrt{}2[/tex]
A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 6600 m. You can ignore any effects of air resistance.
Required:
a. What was the rocket's acceleration during the first 16s?
b. What is the rocket's speed as it passes through acloud 5100 m above the ground?
Answer:
a) a = 34.375 m / s², b) v_f = 550 m / s
Explanation:
This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.
a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed
v_f = [tex]\frac{x-x_1}{t}[/tex]
we substitute the values
v_f = [tex]\frac{ 6600 -x_1}{4}[/tex]
The initial part of the movement is carried out with acceleration
v_f = v₀ + a t
x₁ = x₀ + v₀ t + ½ a t²
the rocket starts from rest v₀ = 0 with an initial height x₀ = 0
x₁ = ½ a t²
v_f = a t
we substitute the values
x₁ = 1/2 a 16²
x₁ = 128 a
v_f = 16 a
let's write our system of equations
v_f = [tex]\frac{6600 - x_1}{4}[/tex]
x₁ = 128 a
v_f = 16 a
we substitute in the first equation
16 a = [tex]\frac{6600 -128 a}{4}[/tex]
16 4 a = 6600 - 128 a
a (64 + 128) = 6600
a = 6600/192
a = 34.375 m / s²
b) let's find the time to reach this height
x = ½ to t²
t² = 2y / a
t² = 2 5100 / 34.375
t² = 296.72
t = 17.2 s
We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part
v_f = 16 a
v_f = 16 34.375
v_f = 550 m / s
I need help with this assignment
If the light energy passes through material in straight lines they are __?___ to light.
Answer:
transparent
Explanation:
When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal speed. Once he has reached terminal speed Group of answer choices his acceleration is equal to g. the force of air drag on him is equal to zero. the force of air drag on him is equal to g. his speed is equal to g. None of the above choices are correct the force of air drag on him is equal to his weight.
Answer:
None of the above forces on air drag on him is equal to his weight
Explanation:
In the velocity-time graph,the gradient of the curve where it is flatten shows the parachutist reaches the terminal velocity when it reaches terminal velocity which means the parachutist reaches constant velocity or speed,indicating that the acceleration of free fall(g) is zero.And according to the resultant force formula weight - air drag= mass*acceleration. so when accelerate is zero,resultant force is zero. And hence the equation will be like this: weight= air drag
A good soccer player can kick the ball up to 25 m/sec. A soccer ball has a mass of 800 grams (0.8 kg). What force must a goalie exert on the ball to bring it to rest in 0.1 sec?
Answer:
200 N
Explanation:
Applying,
The force a golie must exert on the ball is,
F = ma...................... Equation 1
Where m = mass of the ball, a = acceleration of the ball.
But,
a = Δv/t............... Equation 2
Where Δv = change in velocity, t = time.
Substitute equation 2 into equation 1
F = m(Δv/t)............... Equation 3
From the question,
Given: m = 0.8 g, t = 0.1 s, Δv = 25 m/s
Substitute these values into equation 3
F = 0.8×25/0.1
F = 200 N
In a double-slit arrangement, the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits. (a) Calculate the angular separation, !, in radians between the central maximum and the 1st order maximum
Solution :
The conditions for the maximum in the Young's experiment is :
d sin θ = m λ, where m = 0, 1, 2, 3, .....
The angle between the central maximum and the 1st order maximum can be determined by setting the m = 1. So,
d sin θ = λ
[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{d}\right)$[/tex]
Given : d = 100 λ
[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{100 \lambda}\right)$[/tex]
[tex]$\theta = \sin^{-1}\left(\frac{1}{100}\right)$[/tex]
[tex]$=0.573^\circ$[/tex]
= 0.01 rad
True or false, wrrect the false
statement:
• The magnetic field created by a flat coil is
uniform.
• Inside a solenoid, the lines of field are
oriented from the north face to the south
face.
• The magnetic field outside Helmholtz
coils is uniform.
• Le champ B à l'intérieur d'un solénoïde
est uniforme.
• The magnitude of B, created by a flat coil
of radius R, at any point in its plane is
B= 2m x 10-NI
R
• The designation of the faces of a wil
depend the sense of the current
traversing it.
Answer:
false
Explanation:
Explain how muscles are effected by space travel
After turning on the power source connected to your two electrodes, we expect to see the microbeads moving through the solution. What forces are acting on the microbeads as they move (ignore vertical forces)
Answer:
the pearls have an electrical charge induced by contact with the ions of the solution and these charges are attracted by the electrode by a force electric
Explanation:
The pearls are suspended in a solution, when connecting the power source, it is subjected to an electric shock, the pearls have an electrical charge induced by contact with the ions of the solution and these charges are attracted by the electrode by a force electric
F = q E
What kind of model is shown below?
O A. A mathematical model
B. An experimental model
O C. A computer model
D. A physical Model
Answer:
B is excellent answer..............
The model of the brain that is shown here is the experimental model that is present in Option B, as it is used to study the brain's parts and its function, which is helpful for a better understanding of the brain.
What is an experimental model of the brain?There are various experimental models of the brain that have been developed to better understand its functions and mechanisms, such as Animal models, such as mice, rats, and primates, have been widely used to study the brain due to their similarity to the human brain in terms of structure and function. Computer models can simulate brain function and behavior at various levels of detail, from individual neurons to large-scale brain networks. These models are useful for testing hypotheses and predicting outcomes, as well as for designing new experiments.
Hence, the model of the brain that is shown here is the experimental model that is present in Option B.
Learn more about the experimental model of the brain here.
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Calculate the spring constant of the spring in a child's pogo stick if the child has a mass of 32 kg and bounces once every 2.7 seconds
Answer:
The spring constant is 173.12 N/m.
Explanation:
mass, m = 32 kg
Period, T = 2.7 s
let the spring constant is K.
Use the formula of period,
[tex]T = 2\pi\times \sqrt \frac{m}{K}\\\\2.7 =2\times 3.14\sqrt\frac{32}{K}\\\\K = 173.12 N/m[/tex]
An electric field has a positive test charge of 5.00 C placed in it. The force on the test charge is
6.000 N. The magnitude of the electric field at the location of the test charge is
o 30.0 NVC
0 1.20 N/C
0 120, NVC
O 3.00 N/C
01.02 N/C
Answer:
yes
Explanation:
this means the answer is yes
In an experiment, a student brings up the rotational speed of a piece of laboratory apparatus to 24 rpm. She then allows the apparatus to slow down uniformly on its own, and counts 236 revolutions before the apparatus comes to a stop. The moment of inertia of the apparatus is known to be 0.076 kg m2. What is the magnitude of the torque on the apparatus
Answer:
T = 6.43 x 10⁻⁵ N.m
Explanation:
First, we will calculate the deceleration of the apparatus by using the third equation of motion:
[tex]2\alpha \theta = \omega_f^2-\omega_i^2[/tex]
where,
α = angular decelration = ?
θ = angular displacement = (236 rev)(2π rad/rev) = 1482.83 rad
ωi = initial angular speed = (24 rpm)(2π rad/1 rev)(1 min/ 60 s) = 2.51 rad/s
ωf = final angular speed = 0 rad/s
Therefore,
[tex]2\alpha(1482.83\ rad) = (0\ rad/s)^2-(2.51\ rad/s)^2\\\\\alpha = -\frac{(2.51\ rad/s)^2}{2965.66\ rad} \\\\\alpha = - 8.46\ x\ 10^{-4}\ rad/s^2[/tex]
negative sign shows deceleration
Now, for torque:
T = Iα
where,
T = Torque = ?
I = moment of inertia = 0.076 kg.m²
Therefore,
T = (0.076 kg.m²)(8.46 x 10⁻⁴ N.m)
T = 6.43 x 10⁻⁵ N.m
The Heat Force
이
18
1 point
-
If two objects are the same temperature and are physically touching which of the following would be true?
The objects would be in thermodynamic equilibrium and would transfer energy through conduction.
ОООО
1
The objects would not be in thermodynamic equilibrium and heat would transfer through conduction.
The objects would not be in thermodynamic equilibrium and as a result there would be no heat transfer
The objects would be in thermodynamic equilibrium and as a result there would be no heat transfer.
2
If two objects are the SAME temperature and are physically touching,
then
. . .
. . .
. . .
The objects would be in thermodynamic equilibrium and as a result there would be no heat transfer.
A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains air with a volume of 0.175 m^3 at a pressure of 0.350 atm. The piston is slowly pulled out until the volume of the gas is increased to 0.365 m^3.
Required:
If the temperature remains constant, what is the final value of the pressure?
Answer:
the value of the final pressure is 0.168 atm
Explanation:
Given the data in the question;
Let p₁ be initial pressure, v₁ be initial volume.
After expansion, p₂ is final pressure and v₂ is final volume.
So using the following equations;
p₁v₁ = nRT
p₂v₂ = nRT
hence, p₁v₁ = p₂v₂
we find p₂
p₂ = p₁v₁ / v₂
given that; initial volume v₁ = 0.175 m³, Initial pressure p₁ = 0.350 atm,
final volume v₂ = 0.365 m³
we substitute
p₂ = ( 0.350 atm × 0.175 m³ ) / 0.365 m³
p₂ = 0.06125 atm-m³ / 0.365 m³
p₂ = 0.168 atm
Therefore, the value of the final pressure is 0.168 atm
A space probe is launched from Earth headed for deep space. At a distance of 10,000 miles from Earth's center, the gravitational force on it is 435 lb. What is the size of the force when it is at 20,000, 30,000, and 100,000 miles from the earth's center?
Answer:
2
Explanation:
3. How do you think a hot air balloon works?
Answer:hot air
Explanation:Hot air goes up, so when you put it in a hot air balloon fire is placed at the oppening to creat hot air and lift the balloon! :)
HELP ME PLEASE !!!!!!!!!!!!
Answer:
D
Explanation:
Because the y axis is meter. If it is straight line at time and meter graph then it velocity and speed is 0
Coherent monochromatic light falls perpendicularly on two slits (each of width 0.10 mm) separated by 0.50 mm. In the resulting interference pattern on a screen 2.80 m away, adjacent bright fringes are separated by 2.80 mm. (a)What is the wavelength of the light that falls on the slits
Answer:
The correct answer is "[tex]0.5\times 10^{-6} \ m[/tex]".
Explanation:
Given:
[tex]\frac{\lambda D}{d} =2.8\times 10^{-3}[/tex]
[tex]d = 0.5\times 10^{-3}[/tex]
[tex]D = 2.80[/tex]
Now,
The wavelength will be:
⇒ [tex]\lambda = 2.8\times 10^{-3}\times \frac{d}{D}[/tex]
By putting the values, we get
⇒ [tex]=\frac{2.8\times 10^{-3}\times 0.5\times 10^{-3}}{2.8}[/tex]
⇒ [tex]=\frac{1.4\times 10^{-6}}{2.8}[/tex]
⇒ [tex]=0.5\times 10^{-6} \ m[/tex]
while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8,32 meters per second take the speed of sound as 340 meters per second calculate frequency
Complete question:
while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8.32 meters per second take the speed of sound as 340 meters per second. calculate the frequency reflected off the wall to the bat?
Answer:
The frequency reflected by the stationary wall to the bat is 41 kHz
Explanation:
Given;
frequency emitted by the bat, f = 39 kHz
velocity of the bat, [tex]v_b[/tex] = 8.32 m/s
speed of sound in air, v = 340 m/s
Apply the doppler-effect principle to solve this problem.
The apparent frequency of sound striking the wall is calculated as;
[tex]f' = f(\frac{v}{v- v_b} )\\\\f' = 39,000(\frac{340}{340 -8.32} )\\\\f' = 39978.29 \ Hz[/tex]
The frequency reflected by the stationary wall to the bat is calculated as;
[tex]f_s = f'(\frac{v + v_b}{v} )\\\\f_s = 39978.29(\frac{340 + 8.32}{340} )\\\\f_s = 40,956.56 \ Hz[/tex]
[tex]f_s\approx 41 \ kHz[/tex]
The function s(t)s(t) describes the position of a particle moving along a coordinate line, where ss is in feet and tt is in seconds. (a) Find the velocity and acceleration functions. (b) Find the position, velocity, specd, and acceleration at time t
Answer:
Explanation:
From the given information:
Let's assume that the missing function is:
s(t) = t³ - 6t², t ≥ 0
From part (b), we are to find the given required terms when time t = 2
So; from the function s(t) = t³ - 6t², t ≥ 0
[tex]velocity \ v(t) \ = \dfrac{d}{dt}s(t)[/tex]
[tex]velocity \ v(t) \ = \dfrac{d}{dt}(t^3 - 6t^2)[/tex]
[tex]velocity \ v(t) \ = 3t^2 - 12t[/tex]
[tex]acceleration a(t) = \dfrac{d}{dt}*v(t)[/tex]
[tex]acceleration a(t) = \dfrac{d}{dt}(3t^2 - 12 t)[/tex]
[tex]acceleration\ a(t) = 6t - 12[/tex]
At time t = 2
The position; S(2) = (2)² - 6(2)²
S(2) = 8 - 6(4)
S(2) = 8 - 24
S(2) = - 16 ft
v(2) = 3(2)² - 12 (2)
v(2) = 3(4) - 24
v(2) = 12 - 24
v(2) = - 12 ft/s
speed = |v(2)|
|v(2)| = |(-12)|
|v(2)| = 12 ft/s
acceleration = 6t - 12
acceleration = 6(2) - 12
acceleration = 12 - 12
acceleration = 0 ft/s²
100 POINTS AND BRAINLIEST!!! How does the egg sucked into a glass bottle trick work?
Answer: The egg will get sucked into the bottle. To get the egg out of the bottle, turn the bottle upside down and blow into it, so that the egg acts as a one-way valve. The increased air pressure in the bottle will cause the egg to pop back out.
Explanation:
Quickly place the egg over the mouth of the bottle. The egg will get sucked into the bottle. To get the egg out of the bottle, turn the bottle upside down and blow into it, so that the egg acts as a one-way valve. The increased air pressure in the bottle will cause the egg to pop back out.
What is measured by the change in velocity of a moving object?
Answer:
acceleration is measured
Q5: An ice skater moving at 12 m/s coasts
to a halt in 95m on an ice surface. What is the coefficient
of (kinetic) friction between ice and skates?
u = 0.077
Explanation:
Work done by friction is
Wf = ∆KE + ∆PE
-umgx = ∆KE,. ∆PE =0 (level ice surface)
-umgx = KEf - KEi = -(1/2)mv^2
Solving for u,
u = v^2/2gx
= (12 m/s)^2/2(9.8 m/s^2)(95 m)
= 0.077
Kinetic friction is the ratio of the friction force to the normal force experienced by a body in moving state.The coefficient of kinetic friction between the ice and skates is 0.077.
Given-
velocity of the ice skater is 12 m/ sec.
Work done by the friction is the sum of the change of the kinetic energy and the change in potential energy.
[tex]W_{f}=\bigtriangleup KE +\bigtriangleup PE[/tex]
The value for the potential energy will be equal to Zero in this case. Therefore the work done by the friction is,
[tex]W_{f}=\bigtriangleup KE +0[/tex]
Kinetic energy is directly proportional to the mass of the object and to the square of its velocity and work done can be given as,
[tex]W_{f} =u_{f} mgx[/tex]
Here, [tex]u_{f}[/tex] is friction force, [tex]m[/tex] is mass, [tex]g[/tex] is gravity and x is the distance .
Equate the value of kinetic energy and work done of friction for further result, we get,
[tex]u_{f} mgx=\dfrac{1}{2} \times mv^2[/tex]
[tex]u_{f} =\dfrac{1}{2gx} \times v^2[/tex]
[tex]u_{f} =\dfrac{1}{9.8\times 95} \times 12^2[/tex]
[tex]u_{f} =0.077[/tex]
Hence, the coefficient of kinetic friction between the ice and skates is 0.077.
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How much work can a motor with a power output of 0.70 hp do in 2 s?
Answer:
the work done by the motor is 1,044 J.
Explanation:
Given;
the output power of the motor, P = 0.7 hp
duration of the work, t = 2 s
The relationship between horse-power and watt is given as;
1 hp = 745.7 W
0.7 hp = ?
0.7 hp = 522 W = 522 J/s
The work done by the motor is calculated as;
W = Power x time
W = 522 J/s x 2 s
W = 1,044 J
Therefore, the work done by the motor is 1,044 J.
A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the battery supplies energy at a rate of W, how large is the resistance
Answer:
Resistance is as large as 2.8 ohm
Explanation:
Complete question
A 3.0 A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the battery supplies energy at a rate of 25 W, how large is the resistance?
Solution -
The relation between Power and current is as follows
P = I^2*R
R = P/I^2
Were P = Power
R = resistance and
I = current
Given-
P = 25 W
I = 3.0 A
Substituting the given values, in above equation, we get -
R = 25/3.0^2
R = 2.8 ohm
What is the resistance a circult when 9V batter is connected the circult to generate 0.002A
Answer:
v= IR then you can R= v/I 9v÷0.002 = 450 Ohms
Given this relationship, if you and your twin sibling (assuming you have the same mass) were to be separated by three times your original distance, what is the new gravitational force between you?
Answer:
The new force becomes (1/9)th of the original force.
Explanation:
The gravitational force between two masses is given by :
[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]
Where
r is the distance between masses,
If the new distance is, r' = 3r
The new force is given by :
[tex]F'=G\dfrac{m_1m_2}{r'^2}\\\\F'=G\dfrac{m_1m_2}{(3r)^2}\\\\F'=\dfrac{1}{9}\times G\dfrac{m_1m_2}{r^2}\\\\F'=\dfrac{F}{9}[/tex]
So, the new force becomes (1/9)th of the original force.
If the loop is removed from the field region in a time interval of 2.8 ms , find the average emf that will be induced in the wire loop during the extraction process. Express your answer in volts.
The question is incomplete. The complete question is :
A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magnetic field. A field of 1.2 T is directed along the positive z-direction, which is upward. (a)If the loop is removed from the field region in a time interval of 2.8 ms, find the average emf that will be induced in the wire loop during the extraction process.
Solution :
Let us consider a [tex]$\text{circular loo}p \text{ of wire}$[/tex] which has a [tex]\text{radius}[/tex] of r = [tex]15[/tex] cm.
It is oriented horizontally along the xy-plane and is located in the region of an [tex]$\text{uniform magnetic field}$[/tex], such that it points in the positive z direction and having a magnitude of B = 1.2 T.
Now if the loop [tex]$\text{is removed from the field region}$[/tex] in a time interval of Δt = 2.8 ms. Initially the magnetic field and the area points is in the same direction, so that the angle between them is Ф = 0°, thus the initial and the final fluxes are :
[tex]$\phi_{B,i}=BA \cos (\phi) = BA $[/tex] and [tex]$\phi_{B,f} = 0$[/tex]
Area A = [tex]$\pi r^2.$[/tex] The induced emf equals to the change in the flux, and is divided by the time that it takes to go from the initial flux, Δt and multiplied by the number of turns N = 1, i.e. ,
[tex]$\epsilon = -\frac{\Delta \phi_{B}}{\Delta t}$[/tex]
[tex]$=-\frac{0-(1.2 T)\pi(0.15^2)}{2.8 \times 10^{-3}}$[/tex]
= 30.27 V
Therefore, the emf generated is 30.27 V.
A large, metallic, spherical shell has no net charge. It is supported on an insulating stand and has a small hole at the top. A small tack with charge Q is lowered on a silk thread through the hole into the interior of the shell.
Required:
a. What is the charge on the inner surface of the shell?
b. What is the charge on the outer surface of the shell?
Answer:
(a) Negative Q
(b) Positive Q
Explanation:
Charge is the inherent property of matter due to the transference of electrons.
There are three methods of charging a body.
(i) Charging by friction: When two uncharged bodies rubbed together, then one body gets positive charged and the other is negatively charges it is due to the transference of electrons form one body to another.
(ii) Conduction: when a charged body comes in contact with the another uncharged body, the uncharged body gets the same charge and the charge is distributed equally.
(iii) Induction: When a uncharged body keep near the charged body, the uncharged body gets the same amount of charge but opposite in sign.
(a) When a small tack of charge Q is lowered into the hole, then due to the process of induction, the charge on the inner surface of the shell is - Q.
(b) Due to the process of conduction, the charge on the outer surface of the shell is Q.
The charge on the inner surface of the shell is negative whereas the charge on the outer surface of the shell is positive.
Reasons for change of charge on a body
Due to the process of induction the inner surface of the shell creates negative charge because when a uncharged body bring near to the charged body, the uncharged body gets the same amount of charge but opposite in sign.
While on the other hand, there is no charge interaction with the outer surface so it remains positively charge so we can conclude that the charge on the inner surface of the shell is negative whereas the charge on the outer surface of the shell is positive.
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