Answer:
a
Explanation:
you take 32,000kg ÷2.0m
A proton has been accelerated from rest through a potential difference of -1350 V. What is the proton's kinetic energy, in electron volts? What is the proton's kinetic energy, in joules? What is the proton's speed?
Answer:
1 eV = 1.60 * 10^-19 J work done in accelerating electron thru 1 V
KE (total energy) = 1350 ^ 1 eV (note proton goes from + to -)
KE = 1.60 * 10^-19 * 1350 = 2.16 * 10^-16 Joules
1/2 m v^2 = KE = 2.16 * 10^-16 J
v^2 = 4.32 * 10E-16 / 1.67 * 10-27 = 2.59 * 10^11
v = 5.09 * 10^5 m/s
The proton's kinetic energy, in joules is 2.16 *[tex]10^{-16}[/tex] J. The proton's velocity is 5.09 * [tex]10^{5}[/tex]m/s.
What is velocity?
When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time.
Uniform motion an object is said to have uniform motion when object cover equal distance in equal interval of time within exact fixed direction. For a body in uniform motion, the magnitude of its velocity remains constant over time.
1 eV = 1.60 * [tex]10^{-19} J[/tex] work done in accelerating electron throw 1 V
K.E (total energy) = 1350 ^ 1 eV (note proton goes from + to -)
K.E = 1.60 * [tex]10^{-19}[/tex]J * 1350 = 2.16 * [tex]10^{-16}[/tex] Joules
1/2 m v² = KE = 2.16 *[tex]10^{-16}[/tex] J
Velocity of proton is,
v² = 4.32 * 10[tex]e^{-16}[/tex] / 1.67 * [tex]10{-27}[/tex] = 2.59 * [tex]10^{11}[/tex]
v = 5.09 * [tex]10^{5}[/tex]m/s
The proton's kinetic energy, in joules is 2.16 *[tex]10^{-16}[/tex] J. The proton's velocity is 5.09 * [tex]10^{5}[/tex]m/s.
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the spring was compressed three times farther and then the block is released, the work done on the block by the spring as it accelerates the block is
Answer:
The work done on the block by the spring as it accelerates the block is 4kx².
Explanation:
Let initial distance is x.
It was compressed three times farther and then the block is released, new distance is 3x.
The work done in compressing the spring is given by :
[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)[/tex]
[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\W=\dfrac{1}{2}k((3x)^2-x^2)\\\\W=\dfrac{1}{2}k((9x^2-x^2)\\\\W=\dfrac{1}{2}k\times 8x^2\\\\W=4kx^2[/tex]
So, the work done on the block by the spring as it accelerates the block is 4kx².
Explain why the flow from the battery increases when the switch is closed. Give the label of the concept(s) that you use from the model of electricity. [
Answer:
Due to the applied filed the electrons move in a particular direction.
Explanation:
Initially when the switch is off, the free electrons move here and there in any random directions in the conductor with the random speeds called thermal velocity. So, tat the net flow is almost zero.
When the battery is connected is switch is ON, the random motion of the electrons aligned in a particular direction due to the force applied by the electric filed, so the net flow is not zero it increases and thus the current flow.
Light energy is part of a larger form of energy known as __________.
Light energy is part of a larger form of energy known as electromagnetic energy. Details about electromagnetic energy can be found below.
What is electromagnetic radiation?Electromagnetic spectrum is the entire range of wavelengths of all known electromagnetic radiations extending from gamma rays through visible light, infrared, and radio waves, to X-rays.
Visible light is the part of the electromagnetic spectrum, between infrared and ultraviolet, that is visible to the human eye.
Therefore, Light energy is part of a larger form of energy known as electromagnetic energy.
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What is the work done if a Boulder of mass 100 kilogram is rolled 40 meter up slope an angle of 20 degrees assuming the force of friction is negligible
Answer:
The work done is 13680.8 J.
Explanation:
The work done can be calculated as follows:
[tex] W = F*d [/tex]
Where:
F: is the force
d: is the displacement = 40 m
The force acting on the boulder is given by:
[tex] F = mgsin(\theta) [/tex]
Where:
m: is the mass = 100 kg
g: is the acceleration due to gravity = 10 m/s²
θ: is the angle = 20°
Then, the work is:
[tex] W = mgsin(\theta)d = 100 kg*10 m/s^{2}*sin(20)*40 m = 13680.8 J [/tex]
Therefore, the work done is 13680.8 J.
I hope it helps you!
Diffuse reflection occurs when parallel light waves strike which surface? a mirror a rippling fountain a polished silver plate a still pond
Answer: a rippling fountain
Explanation: diffuse reflection happens on rough surfaces, so using the process of elimination, that leaves us with b, a rippling fountain (I also just took this test I'm pretty sure I'm right)
describe the cause of earth's magnetism ?
A 1.0 ball moving at 2.0 / perpendicular to a wall rebounds from the wall at 1.5 /. If the ball was in contact with the wall for 0.1 , what force did the wall impart onto the ball?
Answer:
5N
Explanation:
We have a simple problem of momentum here.
ΔMomentum= mΔv= FΔt
Solve for F
mΔv/Δt=F
Plug in givens
1*(2-1.5)/0.1=F
F=5N
The amount of force that the wall imparts on the ball is 5.0N
According to Newton's second law, the formula for calculating the force applied is expressed as:
[tex]F=ma[/tex]
m is the mass of the object
a is the acceleration of the object
Since acceleration is the change in velocity of an object, hence [tex]a=\frac{\triangle v}{t}[/tex]
The applied force formula becomes [tex]F=\frac{m\triangle v}{t}[/tex]
Given the following parameters
m = 1.0kg
[tex]\triangle v=2.0-1.5\\\triangle v=0.5m/s[/tex]
t = 0.1sec
Substitute the given parameter into the formula
[tex]F=\frac{1.0\times 0.5}{0.1}\\F=\frac{0.5}{0.1}\\F=5N[/tex]
Hence the amount of force that the wall imparts on the ball is 5.0N
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A block slides down a frictionless plane that makes an angle of 24.0° with the horizontal. What is the
acceleration of the block?
Answer:
F = m g sin theta force accelerating block
m a = m g sin theta
a = 9.8 sin 24 = 3.99 m/sec^2
I need help with this physics question.
The acceleration will increase by 61.3%.
Explanation:
The centripetal acceleration [tex]a_c[/tex] is given by
[tex]a_c = \dfrac{v^2}{r}[/tex]
If the velocity of the object increases by 27.0%, then its new velocity v' becomes
[tex]v' = 1.270v[/tex]
The new centripetal acceleration [tex]a'_c[/tex] becomes
[tex]a'_c = \dfrac{(1.270v)^2}{r} = 1.613 \left(\dfrac{v^2}{r} \right)[/tex]
[tex]\:\:\:\:\:\:\:\:\:= 1.613a_c[/tex]
The drawings show (in cross section) two solid spheres and two spherical shells. Each object is made from copper and has a net charge, as the plus and minus signs indicate. Which drawing correctly shows where the charges reside when they are in equilibrium?
a) shows a lot of negative signs in the interior of circle
b) shows a lot of positive signs in the interior of circle
c) shows a hollowed out "hole" in the interior of the circle, with negative signs surrounding the opening.
d) shows a hollowed out "hole" in the interior of the circle, with positive signs surrounding the exterior edge
Answer:
d
Explanation:
The minimum energy configuration in electrostatics states that Charges always reside on the surface of a conductor. If anyhow they were inside, an electric field would exist inside and would act to move them to the surface,
Therefore, the drawing that shows where the charges reside when they are in equilibrium is a hollowed-out "hole" in the interior of the circle, with positive signs surrounding the exterior edge. This means that the d part is the correct answer.
The weight of a hydraulic barber's chair with a client is 2100 N. When the barber steps on the input piston with a force of 44 N, the output plunger of a hydraulic system begins to lift the chair. Determine the ratio of the radius of the output plunger to the radius of the input piston.
Answer:
[tex]\frac{r_1}{r_2}=6.9[/tex]
Explanation:
According to Pascal's Law, the pressure transmitted from input pedal to the output plunger must be same:
[tex]P_1 = P_2\\\\\frac{F_1}{A_1}=\frac{F_2}{A_2}\\\\\frac{F_1}{F_2}=\frac{A_1}{A_2}\\\\\frac{F_1}{F_2}=\frac{\pi r_1^2}{\pi r_2^2}\\\\\frac{F_1}{F_2}=\frac{r_1^2}{r_2^2}[/tex]
where,
F₁ = Load lifted by output plunger = 2100 N
F₂ = Force applied on input piston = 44 N
r₁ = radius of output plunger
r₂ = radius of input piston
Therefore,
[tex]\frac{r_1^2}{r_2^2}=\frac{2100\ N}{44\ N}\\\\\frac{r_1}{r_2}=\sqrt{\frac{2100\ N}{44\ N}} \\\\\frac{r_1}{r_2}=6.9[/tex]
What is the total number of moles of products involved in the following reaction?
CaCO3 (s) + 2HCl (aq) - CaCl2 (aq) + CO2 (g) + H20 (g)
O 6
2.
3
5
Answer:
3
Explanation:
You must first make sure the equation is balanced. This one is. Then, you simply add up the coefficients of each compound on the products side of the equation. When the coefficient is not specified, you can assume it is 1 mole. So, in this equation, there is 1 mole of CaCl₂, 1 mole of CO₂, and 1 mole of H₂O = 3 moles.
The reactant side of the equation also has three moles:
1 mole of CaCO₃ and 2 moles of HCl.
~~~~~NEED HELP ASAP~~~~~
A point on a rotating wheel (thin loop) having a constant angular velocityy of 300 rev/min, the wheel has a radius of 1.5m and a mass of 30kg. (I = mr^2)
a.) Determine the linear regression
b.) At this given angular velocity, what is the rotational kinetic energy?
Answer:
Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J
Explanation:
a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)
a= ω^2*r
ω= 300rev/min
convert into rev/s
300/60= 5rev/s
a= 18.75m/s^2
b) use Krot= 1/2 Iω^2
plug in gives
1/2(30*2.25)(25)= 843.75 J
A 2.90 m segment of wire supplying current to the motor of a submerged submarine carries 1400 A and feels a 2.00 N repulsive force from a parallel wire 4.50 cm away. What is the direction and magnitude (in A) of the current in the other wire? magnitude A direction
Answer:
[tex]I_2=30.9A[/tex]
Explanation:
From the question we are told that:
Wire segment [tex]l_s=2.9m[/tex]
Initial Current [tex]I_1=1400A[/tex]
Force [tex]F=2.00N[/tex]
Distance of Wire [tex]d=4.50cm=>0.0450m[/tex]
Generally the equation for Force is mathematically given by
[tex]F=\frac{\mu_0 * I_1*I_2*l_s}{2 \pi *r}[/tex]
[tex]F=\frac{4 \pi*10^{-7} *1400 I*I_2*2.9}{2 \pi *0.0450}[/tex]
[tex]I_2=\frac{22.5*10^-2}{2*10^{-7}*1400*2.6}[/tex]
[tex]I_2=30.9A[/tex]
how did kepler discoveries contribute to astronomy
Answer:
They established the laws of planetary motion. They explained how the Sun rises and sets. They made astronomy accessible to people who spoke Italian.
Explanation:
Which one of the following statements concerning resistors in "parallel" is true? Question 7 options: The voltage across each resistor is the same. The current through each resistor is the same. The total current through the resistors is the sum of the current through each resistor. The power dissipated by each resistor is the same.
Answer: The correct statement is:
--> The voltage across each resistor is the same.
Explanation:
RESISTORS are defined as the components of an electric circuit which are capable of creating resistance to the file of electric current in the circuit. They work by converting electrical energy into heat, which is dissipated into the air. These resistors can be divided into two according to their arrangements in the electric cell. It include:
--> Resistors in parallel and
--> Resistors in series
RESISTORS are said to be in parallel when two or more resistance or conductors are connected to common terminals so that the potential difference ( voltage) across each conductor IS THE SAME but with different current flow through each of them. Also, Individual resistances diminish to equal a smaller total resistance rather than add to make the total.
The mass of the moon is 7.2 × 10^22 kg and its radius is 1.7×10^6 m.What will be the gravity of the moon to a body of the mass 1 kg on the surface of the moon.
Answer:
1.66 N
Explanation:
The force of gravity of the moon on the body is given by
F = GMm/R² where G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of moon = 7.2 × 10²² kg, m = mass of body = 1 kg and R = radius of moon = 1.7 × 10⁶ m
Substituting the values of the variables into the equation, we have
F = GMm/R²
F = 6.67 × 10⁻¹¹ Nm²/kg² × 7.2 × 10²² kg × 1 kg/(1.7 × 10⁶ m)²
F = 48.024 × 10¹¹ Nm²/2.89 × 10¹² m²
F = 16.62 × 10⁻¹ N
F = 1.662 N
F ≅ 1.66 N
So, the gravity on the moon is 1.66 N
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
A 10 kg box is at static equilibrium and the downward pull of gravity acting on the box is 98 Newton’s what is the minimum force that would require to just pick up the box
Explanation:
static equilibrium means its on the floor or something
so slightly greater than 98 newtons in the upward direction
A farmhand pushes a 26-kg bale of hay 3.9 m across the floor of a barn. If she exerts a horizontal force of 88 N on the hay, how much work has she done
Answer:
W = 343.2 J
Explanation:
Given that,
Mass of bale of hay = 26 kg
Horizontal force exerted = 88 N
Distance moved, d = 3.9 m
Work done, W = Fd
Put all the values,
W = 88 N × 3.9 m
= 343.2 J
So, the work done is 343.2 J.
In order to keep a leaking ship from sinking, it is necessary to pump 12.0 lb of water each second from below deck up a height of 2.00 m and over the side. What is the minimum horse-
power motor that can be used to save the ship?
Answer:
P = 0.14 hp
Explanation:
The power required by the ship is given as:
[tex]P = \frac{Work}{Time} = \frac{Potential\ Eenrgy}{t}\\\\P = \frac{mgh}{t}[/tex]
where,
P = Power = ?
m = mass to pump = (12 lb)(1 kg/2.20 lb) = 5.44 kg
g = acceleration due to gravity = 9.81 m/s²
h = height = 2 m
t = time = 1 s
Therefore,
[tex]P = \frac{(5.44\ kg)(9.81\ m/s^2)(2\ m)}{1\ s}\\\\P = 106.8\ W[/tex]
Converting to horsepower (hp):
[tex]P = (106.8\ W)(\frac{1\ hp}{746\ W})[/tex]
P = 0.14 hp
A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?
Answer:
The correct answer is "[tex]4.49\times 10^{10} \ joules[/tex]".
Explanation:
According to the question,
The work will be:
⇒ [tex]Work=-\frac{kQq}{R}[/tex]
[tex]=-\frac{1}{4 \pi \varepsilon \times (18-30)\times 3\times 0.2}[/tex]
[tex]=-\frac{1}{4 \pi \varepsilon \times (-12)\times 3\times 0.2}[/tex]
[tex]=\frac{0.3978}{\varepsilon }[/tex]
[tex]=4.49\times 10^{10} \ joules[/tex]
Thus the above is the correct answer.
We have that the workdone is mathematically given as
W=4.49*10e10 J
From the question we are told
A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?WorkdoneGenerally the equation for the workdone is mathematically given as
W=-kQq/R
Therefore
0.3978/ε0 =-1/(4πε0*(18-30)*3*0.2
Hence
W=4.49*10e10 JFor more information on Charge visit
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The period of a simple pendulum is 3.5 s. The length of the pendulum is doubled. What is the period T of the longer pendulum?
Explanation:
The period T of a simple pendulum is given by
[tex]T = 2 \pi \sqrt{\dfrac{l}{g}}[/tex]
Doubling the length of the pendulum gives us a new period T'
[tex]T' = 2 \pi \sqrt{\dfrac{l'}{g}} = 2 \pi \sqrt{\dfrac{2l}{g}}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{2} \left(2 \pi \sqrt{\dfrac{l}{g}} \right)[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{2}\:T = \sqrt{2}(3.5\:\text{s})= 4.95\:\text{s}[/tex]
A proton enters a region of constant magnetic field, perpendicular to the field and after being accelerated from rest by an electric field through an electrical potential difference of 330 V. Determine the magnitude of the magnetic field, if the proton travels in a circular path with a radius of 23 cm.
Answer:
B = 1.1413 10⁻² T
Explanation:
We use energy concepts to calculate the proton velocity
starting point. When entering the electric field
Em₀ = U = q V
final point. Right out of the electric field
em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
q V = ½ m v²
v = [tex]\sqrt{2qV/m}[/tex]
we calculate
v = [tex]\sqrt{\frac{ 2 \ 1.6 \ 10^{-19} \ 300}{1.67 \ 0^{-27}} }[/tex]
v = [tex]\sqrt{632.3353 \ 10^8}[/tex]
v = 25.15 10⁴ m / s
now enters the region with magnetic field, so it is subjected to a magnetic force
F = m a
the force is
F = q v x B
as the velocity is perpendicular to the magnetic field
F = q v B
acceleration is centripetal
a = v² / r
we substitute
qvB =1/2 m v² / r
B = v[tex]\frac{m v}{2 q r}[/tex]
we calculate
B = [tex]\frac{1.67 \ 10^{-27} 25.15 \ 10^4 }{1.6 \ 10^{-19} 0.23}[/tex]
B = 1.1413 10⁻² T
(a) If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve on wet concrete where the coefficient of kinetic friction is 0.5 and the coefficient of static friction is 0.7.
(b) Will a metal cabinet lying on the wooden bed of the truck slip if it accelerates at this rate where the coefficient of kinetic friction is 0.3 and the coefficient of static friction is 0.55?
(c) If the truck has four-wheel drive, and the cabinet is wooden, what is it's maximum acceleration (in m/s2)?
Answer:
a) a = 27.44 m / s², b) a = 5.39 m / s², c) a = 156.8 m / s², cabinet maximum acceleration does not change
Explanation:
a) In this exercise the wheels of the truck rotate to provide acceleration, but the contact point between the ground and the 2 wheels remains fixed, therefore the coefficient of friction for this point is static.
Let's apply Newton's second law
we set a regency hiss where the x axis is in the direction of movement of the truck
Y axis y
N- W = 0
N = W = m g
X axis
2fr = m a
the expression for the friction force is
fr = μ N
fr = μ m g
we substitute
2 μ m g = m /2 a
a = 4 μ g
a = 4 0.7 9.8
a = 27.44 m / s²
b) let's look for the maximum acceleration that can be applied to the cabinet
fr = m a
μ N = ma
μ m g = m a
a = μ g
a = 0.55 9.8
a = 5.39 m / s²
as the acceleration of the platform is greater than this acceleration the cabinet must slip
c) the friction force is in the four wheels as well
With when the truck had two-wheel Thracian the weight of distributed evenly between the wheels, in this case with 4-wheel Thracian the weight must be distributed among all
applying Newton's second law
4 fr = (m/4) a
16 mg = (m) a
a = 16 g
a = 16 9.8
a = 156.8 m / s²
cabinet maximum acceleration does not change
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
A 0.500-kg block slides up a plane inclined at a 30° angle. If it slides 1.50 m before coming to rest while encountering a frictional force of 2 N, find (a) its acceleration, and (b) its initial velocity.
A point charge of -3.0 x 10-5C is placed at the origin of coordinates. Find the electric field at the point 3. r= 50 m on the x-axis
Answer: -5×10-3
Explanation:
E=kq/r
A seesaw made of a plank of mass 10.0 kg and length 3.00 m is balanced on a fulcrum 1.00 m from one end of the plank. A 20.0-kg mass rests on the end of the plank nearest the fulcrum. What mass must be on the other end if the plank remains balanced?
Answer:
7.5 kg
Explanation:
We are given that
[tex]m_1=10 kg[/tex]
Length of plank, l=3 m
Distance of fulcrum from one end of the plank=1 m
[tex]m_2=20 kg[/tex]
We have to find the mass must be on the other end if the plank remains balanced.
Let m be the mass must be on the other end if the plank remains balanced.
In balance condition
[tex]20\times 1=10\times (1.5-1)+m\times (1.5+0.5)[/tex]
[tex]20=10(0.5)+2m[/tex]
[tex]20=5+2m[/tex]
[tex]2m=20-5=15[/tex]
[tex]\implies m=\frac{15}{2}[/tex]
[tex]m=7.5 kg[/tex]
Hence, mass 7.5 kg must be on the other end if the plank remains balanced.
Answer:
The mass at the other end is 7.5 kg.
Explanation:
Let the mass is m.
Take the moments about the fulcrum.
20 x 1 = 10 x 0.5 + m x 2
20 = 5 + 2 m
2 m = 15
m = 7.5 kg
