Find the ratio of speeds of a proton and an alpha particle accelerated through the same voltage, assuming nonrelativistic final speeds. Take the mass of the alpha particle to be 6.64 ✕ 10−27 kg.
Answer:
The required ratio is 1.99.
Explanation:
We need to find the atio of speeds of a proton and an alpha particle accelerated through the same voltage.
We know that,
[tex]eV=\dfrac{1}{2}mv^2[/tex]
The LHS for both proton and an alpha particle is the same.
So,
[tex]\dfrac{v_p}{v_a}=\sqrt{\dfrac{m_a}{m_p}} \\\\\dfrac{v_p}{v_a}=\sqrt{\dfrac{6.64\times 10^{-27}}{1.67\times 10^{-27}}} \\\\=1.99[/tex]
So, the ratio of the speeds of a proton and an alpha particle is equal to 1.99.
You throw a football straight up. Air resistance can be neglected. When the football is 4.00 mm above where it left your hand, it is moving upward at 0.500 m/sm/s. What was the speed of the football when it left your hand
Answer:
u=8.868 m/s
Explanation:
The displacement of the ball is 4 meters
The final speed of the ball is 0.5 m/s
The initial speed of the ball is to be calculated. Using the equation of the rectilinear motion,
[tex]v^{2} =u^{2} +2as[/tex]
Plugging the values in the above expression,
[tex]\\0.5^{2} =u^{2} +2*(-9.8)*4\\\\u^{2} =78.4+0.25\\\\u^{2} =78.65\\\\u=8.868 m/s[/tex]
A kangaroo kicks downward with a 1000N force. According to Newton's Law the kangaroo is propelled into the air by:
A) gravitational force
B) his muscles
C) The earth
D) wallabies
Explanation:
Specifically his leg muscles. As the leg muscles expand, they push down on the ground. Newton's 3rd law says that for any action, there's an opposite and equal reaction. That means a downward push into the ground will have the ground push back, more or less, and that's why the kangaroo will jump. The ground (and the earth entirely) being much more massive compared to the animal means that the ground doesn't move while the kangaroo does move. Perhaps on a very microscopic tiny level the ground/earth does move but it's so small that we practically consider it 0.
This experiment can be done with a wall as well. Go up to a wall and lean against it with your hands. Then do a pushup to move further away from the wall, but you don't necessarily need to lose contact with the wall's surface. As you push against the wall, the wall pushes back, and that causes you to move backward. If the wall was something flimsy like cardboard, then you could easily push the wall over and you wouldn't move back very much. It all depends how much mass is in the object you're pushing on.
26.
Which one of the following is not a vector quantity?
(2)
A) acceleration
C) displacement
E) instantaneous velocity
B) average speed
D) average velocity
Answer:
Answer: Speed is not a vector quantity. It has only magnitude and no direction and hence it is a scalar quantity.
A flat (unbanked) curve on a highway has a radius of 260 mm . A car successfully rounds the curve at a speed of 32 m/sm/s but is on the verge of skidding out.
Required:
a. If the coefficient of static friction between the car’s tires and the road surface were reduced by a factor of 2, with what maximum speed could the car round the curve?
b. Suppose the coefficient of friction were increased by a factor of 2; what would be the maximum speed?
I suppose you meant to say the radius of the curve is 260 m, not mm?
There are 3 forces acting on the car as it makes the turn,
• its weight mg pulling it downward;
• the normal force exerted by the road pointing upward, also with magnitude mg since the car is in equilibrium in the vertical direction; and
• static friction keeping the car from skidding with magnitude µmg (since it's proportional to the normal force), pointing horizontally toward the center of the curve.
By Newton's second law, the net force on the car acting in the horizontal direction is
F = ma => µmg = ma => a = µg
where a is the car's radial acceleration given by
a = v ^2 / R
with v = the car's tangential speed and R = radius of the curve. At the start, the car's radial acceleration is
a = (32 m/s)^2 / (260 m) ≈ 3.94 m/s^2
(a) If µ were reduced by a factor of 2, then the radial acceleration would also be halved:
1/2 a = 1/2 µg
Then the car can have a maximum speed v of
1/2 a = v ^2 / R => v = √(aR/2) = √((3.94 m/s^2) (260 m) / 2) ≈ 22.6 m/s
(b) If µ were increased by a factor of 2, then the acceleration would also get doubled. Then the maximum speed v would be
2a = v ^2 / R => v = √(2aR) = √(2 (3.94 m/s^2) (260 m)) ≈ 45.3 m/s
50 POINTS‼️‼️‼️‼️‼️
A 4.88 x 10-6 C charge moves 265 m/s
perpendicular (at 90°) to a magnetic
field of 0.0579 T. What is the magnetic
force on the charge?
Answer: 0
Explanation: Trust
The magnetic force on the charge is approximately 6.47 x 10^(-4) Newtons.
The magnetic force on a charged particle moving through a magnetic field can be calculated using the formula:
F = q * v * B * sin(θ)
Where:
F is the magnetic force,
q is the charge of the particle (in this case, 4.88 x 10^(-6) C),
v is the velocity of the particle (in this case, 265 m/s),
B is the magnetic field strength (in this case, 0.0579 T),
θ is the angle between the velocity vector and the magnetic field vector (in this case, 90 degrees).
Plugging in the values:
F = (4.88 x 10^(-6) C) * (265 m/s) * (0.0579 T) * sin(90°)
Since sin(90°) is equal to 1, the equation simplifies to:
F = (4.88 x 10^(-6) C) * (265 m/s) * (0.0579 T) * 1
Calculating the value:
F = 6.47 x 10^(-4) N
Therefore, the magnetic force on the charge is approximately 6.47 x 10^(-4) Newtons.
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PLEASE HELP ME WITH THIS ONE QUESTION
The color orange has a wavelength of 590 nm. What is the energy of an orange photon? (h = 6.626 x 10^-19, 1 eV = 1.6 x 10^-19 J)
A) 2.81 eV
B) 3.89 eV
C) 2.10 eV
D) 2.78 eV
k = [tex] \dfrac{ (\dfrac{h}{ \lambda} )^{2} }{2m} [/tex]
k = (6.626×10-¹⁹/590 × 10-⁹ )^{2} /2 × 1.673 × 10-²⁷
k = (1.12 × 10-³⁰)^2/3.346×10-²⁷
k = 1.25 × 10-⁶⁰ /3.346×10-²⁷
k = 0
ldk why, my answer is coming this :(
A 12 kg hanging sculpture is suspended by a 80-cm-long, 6.0 g steel wire. When the wind blows hard, the wire hums at its fundamental frequency. What is the frequency of the hum
Answer:
[tex]F=78.3hz[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=12[/tex]
Length [tex]l=80cm=0.8m[/tex]
Linear density [tex]\mu= 6.0g[/tex]
Generally the equation for Frequency is mathematically given by
[tex]F=\frac{1}{2l}\sqrt{\frac{T}{K}}[/tex]
[tex]F=\frac{1}{2(0.8)}\sqrt{\frac{12*9.8*0.8}{6*10^{-3}}}[/tex]
[tex]F=78.3hz[/tex]
Erica (37 kg ) and Danny (45 kg ) are bouncing on a trampoline. Just as Erica reaches the high point of her bounce, Danny is moving upward past her at 4.7 m/s . At that instant he grabs hold of her. What is their speed just after he grabs her?
Answer:
V = 2.58 m/s
Explanation:
Below is the calculation:
Given the weight of Erica = 37 kg
The weight of Danny = 45 kg
Danny's speed to move upward = 4.7 m/s
Use below formula to find the answer.
m1 * u1 = (m1+m2) * V
V = m1*u1 / (m1+m2)
Here, m1 = 45
u1 = 4.7
m1 = 45
m2 = 37
Now plug the values in formula:
V = m1*u1 / (m1+m2)
V = (45*4.7)/(45+37)
V = 2.58 m/s
A mass of 4 kg is traveling over a quarter circular ramp with a radius of 10 meters. At the bottom of the incline the mass is moving at 21.3 m/s and at the top of the incline the mass is moving at 2.8 m/s. What is the work done by all non-conservative force in Joules?
Answer:
499.7 J
Explanation:
Since total mechanical energy is conserved,
U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at bottom of incline = mgh₁, K₁ = kinetic energy at bottom of incline = 1/2mv₁² and W₁ = work done by friction at bottom of incline, and U₂ = potential energy at top of incline = mgh₂, K₁ = kinetic energy at top of incline = 1/2mv₂² and W₂ = work done by friction at top of incline. m = mass = 4 kg, h₁ = 0 m, v₁ = 21.3 m/s, W₁ = 0 J, h₂ = radius of circular ramp = 10 m, v₂ = 2.8 m/s, W₂ = unknown.
So, U₁ + K₁ + W₁ = U₂ + K₂ + W₂
mgh₁ + 1/2mv₁² + W₁ = mgh₂ + 1/2mv₂² + W₂
Substituting the values of the variables into the equation, we have
mgh₁ + 1/2mv₁² + W₁ = mgh₂ + 1/2mv₂² + W₂
4 kg × 9.8 m/s²(0) + 1/2 × 4 kg × (21.3 m/s)² + 0 = 4 kg × 9.8 m/s² × 10 m + 1/2 × 4 kg × (2.8 m/s)² + W₂
0 + 2 kg × 453.69 m²/s² = 392 kgm²/s² + 2 kg × 7.84 m²/s² + W₂
907.38 kgm²/s² = 392 kgm²/s² + 15.68 kgm²/s² + W₂
907.38 kgm²/s² = 407.68 kgm²/s² + W₂
W₂ = 907.38 kgm²/s² - 407.68 kgm²/s²
W₂ = 499.7 kgm²/s²
W₂ = 499.7 J
Since friction is a non-conservative force, the work done by all the non-conservative forces is thus W₂ = 499.7 J
PLEASE HELP ME WITH THIS ONE QUESTION
The color orange has a wavelength of 590 nm. What is the energy of an orange photon? (h = 6.626 x 10^-19, 1 eV = 1.6 x 10^-19 J)
A) 2.81 eV
B) 3.89 eV
C) 2.10 eV
D) 2.78 eV
The color orange has a wavelength of 590 nm. The energy of an orange photon is approximately 0.337 eV.
The correct answer is option E.
To calculate the energy of a photon, we can use the equation:
E = (hc) / λ
where E is the energy of the photon, h is the Planck's constant (6.626 x [tex]10^-^3^4[/tex]J·s or 6.626 x[tex]10^-^1^9^[/tex] eV·s), c is the speed of light (3.00 x [tex]10^8[/tex] m/s), and λ is the wavelength of the light.
Given that the wavelength of orange light is 590 nm (or 590 x [tex]10^-^9[/tex]m), we can substitute the values into the equation:
E = [(6.626 x[tex]10^-^1^9^[/tex] eV·s) x (3.00 x [tex]10^8[/tex] m/s)] / (590 x[tex]10^-^9[/tex]m)
E = (1.9878 x [tex]10^-^1^0[/tex]eV·m) / (590 x [tex]10^-^9[/tex] m)
E = 3.3695 x [tex]10^-^1[/tex] eV
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The question probable may be:
The color orange has a wavelength of 590 nm. What is the energy of an orange photon? (h = 6.626 x [tex]10^-^1^9^[/tex], 1 eV = 1.6 x[tex]10^-^1^9^[/tex]J)
A) 2.81 eV
B) 3.89 eV
C) 2.10 eV
D) 2.78 eV
E) 0.337 eV
A house has well-insulated walls. It contains a volume of 105 m3 of air at 305 K.
Consider heating it at constant pressure. Calculate the energy required to increase the temperature of this diatomic ideal gas by 0.7
Answer: [tex]85.46\ kJ[/tex]
Explanation:
Given
Volume of air [tex]V=105\ m^3[/tex]
Temperature of air [tex]T=305\ K[/tex]
Increase in temperature [tex]\Delta T=0.7^{\circ}C[/tex]
Specific heat for diatomic gas is [tex]C_p=\dfrac{7R}{2}[/tex]
Energy required to increase the temperature is
[tex]\Rightarrow Q=nC_pdT\\\\\Rightarrow Q=n\times \dfrac{7R}{2}\times \Delta T\\\\\Rightarrow Q=\dfrac{7}{2}nR\Delta T\\\\\Rightarrow Q=\dfrac{7}{2}\times \dfrac{PV}{T}\times \Delta T\quad [\text{using PV=nRT}][/tex]
Insert the values
[tex]\Rightarrow Q=\dfrac{7}{2}\times \dfrac{1.01325\times 10^5\times 105}{305}\times 0.7\\ \text{Assuming air pressure to be atmospheric P=}1.01325\times 10^5\ N/m^2\\\\\Rightarrow Q=0.8546\times 10^5\\\Rightarrow Q=85.46\ kJ[/tex]
A solenoid used to produce magnetic fields for research purposes is 2.1 m long, with an inner radius of 28 cm and 1000 turns of wire. When running, the solenoid produced a field of 1.3 T in the center. Given this, how large a current does it carry?
Answer:
I = 2172.46 A
Explanation:
Given that,
The length of a solenoid, l = 2.1 m
The inner radius of the solenoid, r = 28 cm = 0.28 m
The number of turns in the wire, N = 1000
The magnetic field in the solenoid, B = 1.3 T
We need to find the current carried by it. We know that, the magnetic field in a solenoid is given by :
[tex]B=\mu_o nI\\\\or\\\\B=\mu_o \dfrac{N}{L}I\\\\I=\dfrac{BL}{\mu_o N}[/tex]
Put all the values,
[tex]I=\dfrac{1.3\times 2.1}{4\pi \times 10^{-7}\times 1000}\\\\I=2172.46\ A[/tex]
So, it carry current of 2172.46 A.
Using pascals principle, F1/A1=F2/A2, solve this like a proportion.
A force of 50 N is applied to an area of 200 sq feet, how much force will be applied of the area to be covered is 50 sq feet?
Answer:
NO CLUE
Explanation:
GOOD LUCK THOUGH
If a magnifying glass has a power of 10.0 D, what is the magnification it produces when held 6.55 cm from an object?
Answer: The magnification of the magnifying glass is -2.9
Explanation:
The equation for power is given as:
[tex]P=\frac{1}{f}[/tex]
where,
P = Power = 10 D
f = focal length
Putting values in above equation, we get:
[tex]f=\frac{1}{10}=0.1m=10 cm[/tex] (Conversion factor: 1 m = 100 cm)
The equation for lens formula follows:
[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}[/tex]
where,
v = image distance
u = Object distance = 6.55 cm
Putting values in above equation, we get:
[tex]\frac{1}{v}=\frac{1}{10}-\frac{1}{(6.55)}\\\\\frac{1}{v}=\frac{6.55-10}{10\times 6.55}\\\\v=\frac{65.5}{-3.45}=-18.98cm[/tex]
Magnification (m) can be written as:
[tex]m=\frac{-v}{u}[/tex]
Putting values in above equation, we get:
[tex]m=\frac{-18.98}{6.55}\\\\m=-2.9[/tex]
Hence, the magnification of the magnifying glass is -2.9
With neat circuit diagrams where possible show any two (2) ways of direct current motor
excitations,
How much power does it take to lift 70.0 N to 5.0 m high in 5.00 s?
Answer:
Power = 70 W
Explanation:
Given that,
Force, F = 70 N
Height, h = 5 m
Time, t = 5 s
We need to find the power of the object. We know that,
Power = work done/time
Put all the values,
[tex]P=\dfrac{Fd}{t}\\\\P=\dfrac{70\times 5}{5}\\\\P=70\ W[/tex]
So, the required power is 70 W.
A 45-kg skydiver jumps out of an airplane and falls 450 m, reaching a maximum speed of 51 m/s before opening her parachute. How much work, in joules, did air resistance do on the skydiver before she opened her parachute
Answer:
The work done by the friction force is - 139927.5 J.
Explanation:
mass of diver, m = 45 kg
distance falls, h = 450 m
initial speed, u = 0 m/s
final speed = 51 m/s
According to the work energy theorem,
Work done by the gravity + work done by the friction force = change in kinetic energy
[tex]m g h + W' = 0.5 m ()v^2 - u^2)\\\\45\times 9.8\times 450 + W' = 0.5\times 45\times (51^2 - 0)\\\\198450 + W' = 58522.5\\\\W' = - 139927.5 J[/tex]
In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0342 s, during which time it experiences an acceleration of 186 m/s2. The ball is launched at an angle of 45.9 ° above the ground. Determine the (a) horizontal and (b) vertical components of the launch velocity.
Answer:
b) v_y = 4.57 m / s
a) vₓ = 4.43 m / s
Explanation:
This is an exercise in kinematics, where we assume that the acceleration is in the direction of the force and the initial body with zero velocity
v = v₀ + a t
v = 0 + a t
v = 186 0.0342
v = 6.36 m / s
let's use trigonometry to decompose this velocity
sin 45.9 = v_y / v
cos 45.9 = vₓ / v
v_y = v sin 45.9
vₓ = v cos 45.9
v_y = 6.36 sin 45.9
vₓ = 6.36 cos 45.9
v_y = 4.57 m / s
vₓ = 4.43 m / s
ₓ
Hydrogen carried in light phase
Answer:
because it is helpful to human beings I think
The correct formula for finding the relative velocity of an object is:
WILL MARK BRAINLIEST TO THE CORRECT ANSWER!!
Answer:
[tex]V_{a/c} = V_{a/b} + V_{b/c}[/tex]
Explanation:
The relative velocity of an object is the velocity of the object relative to the observer or frame of reference.
The velocity of particle "A" with respect to particle "B" is written as [tex]V_{A/B} = V_{A} - V_{B}[/tex]
From the given options, the second option is the correct answer.
[tex]V_{a/c} = V_{a/b} + V_{b/c}\\\\Re-arrange \ the \ above \ equation;\\\\V_{a/c} - V_{b/c}= V_{a/b}\\\\or\\\\V_{a/b}= V_{a/c} - V_{b/c}[/tex]
The drag force Fd, imposed by the surrounding air on a vehicle moving with velocity V is given by:
Fd =C dA 2 pV2
where Cd is a constant called the drag coefficient, A is the projected frontal area of the vehicle, and \rhorho is the air density.
Determine the power, in hp, required to overcome aerodynamic drag for an automobile moving at
(a) 25 miles per hour,
(b) 70 miles per hour.
Assume Cd=0.28,
A= 25ft2
and p=0.075Ib/ft2
Answer:
Explanation:
a)
Given that:
V = 25 mi/hr
To ft/sec, we have:
[tex]V = 25 \times \dfrac{5280}{3600} ft/s[/tex]
[tex]V = \dfrac{110}{3} ft/s[/tex]
[tex]\rho = 0.075 \ lb/ft^3[/tex]
[tex]\rho = 0.075 \times \dfrac{1 \ lbf s^2/ft}{32.174 \ lbm}[/tex]
[tex]\rho = \dfrac{0.075}{32.174 } lbf.s^2/ft^4[/tex]
[tex]C_d = 0.28[/tex]
A = 25ft²
Recall that:
The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]
[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{110}{3})^2[/tex]
[tex]F_d =10.967 \ lbf[/tex]
[tex]P = F_dV \\ \\ P = 10.97 \times (\dfrac{110}{3}} \\ \\ P = 402.3 \ hp[/tex]
For 70 miles per hour, we have:
[tex]V = 70 \times \dfrac{5280}{3600} ft/s[/tex]
[tex]V = \dfrac{308}{3} ft/s[/tex]
The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]
[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{308}{3})^2[/tex]
[tex]F_d =85.99 \ lbf[/tex]
[tex]P = F_dV \\ \\ P = 85.99 \times (\dfrac{308}{3}}) \\ \\ P = 8828.2 \ hp[/tex]
. A car increases velocity from 20 m/s to 60 m/s in a time of 10 seconds. What was the acceleration of the car?
Answer:
0.3333
Explanation:
Acceleration = change in velocity/time
a = 20 m/s / 60 m/s
a = 0.3333 m/s^2
an artificial satellite is moving in a circular orbit of radius 36000 kilometre calculate its speed if it takes 24 hours to revolve around the earth
Explanation:
9420 km/hr is the correct answer
Hope this helps...☺
Is the following chemical reaction balanced?
2H202-H2O + O2
yes
no
scripture union was founded by who in what year
Answer:
Josiah Spiers in 1867 was when scripture union was founded
The steps to determine the sum are shown. (6.74x104)+(8.95 x 104) Step 1. Rearrange the expression: (6.74+8.95) 104 Step 2. Add the coefficients: (15.69) 104 Step 3. Write in scientific notation: 1.569x 10 What is the value of k in Step 3? =
Answer:
We want to solve the sum:
6.74*10⁴ + 8.95*10⁴
first, we take the common factor 10⁴ out, so we get:
(6.74 + 8.95)*10⁴
Now we solve the sum:
(15.66)*10⁴
Now we want to rewrite it in exponential form, wo we can rewrite it as:
(15.66)*10⁴ = (1.566*10)*10⁴ = (1.566)*10*10⁴ = (1.566)*10⁴⁺¹ = 1.566*10⁵
k = 5.
A reservoir located in the mountain 250 m above sea level flows through a pipe to a hydroelectric plant in a town at sea level. Assuming the pressure in both locations are the same and the density of water is 1000 kg/m3. How fast will the water flow into the plant?
Answer:
v₂ = 70 m / s
Explanation:
For this exercise let's use Bernoulli's equation
where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂
indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute
ρ g y₁ = ½ ρ v₂²
v₂ = [tex]\sqrt {2g \ y_1}[/tex]
let's calculate
v₂ = √( 2 9.8 250)
v₂ = 70 m / s
Which statement describes an action-reaction pair?
O A. You push on a car, and the car pushes back on you.
B. A book pushes down on a table, and the table pushes down on the
Earth.
C. The Moon pulls on Earth, and Earth pulls on the Sun.
D. You push down on your shoe, and Earth's gravity pulls down on the
shoe.
Answer:
A
Explanation:
a pex
Electromagnetic radiation from a 8.25 mW laser is concentrated on a 1.23 mm2 area. Suppose a 1.12 nC static charge is in the beam, and moves at 314 m/s. What is the maximum magnetic force it can feel
Answer:
The maximum magnetic force is 2.637 x 10⁻¹² N
Explanation:
Given;
Power, P = 8.25 m W = 8.25 x 10⁻³ W
charge of the radiation, Q = 1.12 nC = 1.12 x 10⁻⁹ C
speed of the charge, v = 314 m/s
area of the conecntration, A = 1.23 mm² = 1.23 x 10⁻⁶ m²
The intensity of the radiation is calculated as;
[tex]I = \frac{P}{A} \\\\I = \frac{8.25 \times 10^{-3} \ W}{1.23 \ \times 10^{-6} \ m^2} \\\\I = 6,707.32 \ W/m^2[/tex]
The maximum magnetic field is calculated using the following intensity formula;
[tex]I = \frac{cB_0^2}{2\mu_0} \\\\B_0 = \sqrt{\frac{2\mu_0 I}{c} } \\\\where;\\\\c \ is \ speed \ of \ light\\\\\mu_0 \ is \ permeability \ of \ free \ space\\\\B_0 \ is \ the \ maximum \ magnetic \ field\\\\B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times 6,707.32 }{3\times 10^8} } \\\\B_0 = 7.497 \times 10^{-6} \ T[/tex]
The maximum magnetic force is calculated as;
F₀ = qvB₀
F₀ = (1.12 x 10⁻⁹) x (314) x (7.497 x 10⁻⁶)
F₀ = 2.637 x 10⁻¹² N
