Answers
Answer:
B . energy cannot be created or destroyed
Related Questions
A proton is held at rest in a uniform electric field. When it is released, the proton will gain:_________
a) electrical potential energy.
b) kinetic energy.
c) both kinetic energy and electric potential energy.
d) either kinetic energy or electric potential energy.
Answers
A horizontal, mass spring system undergoes simple harmonic motion. which of the following statements is correct reguarding the mass in the system when it is located at its maximum distance from the equilibrium position?
a. The acceleration of the mass is zero.
b. The potential energy of the spring attached to the mass is at a minimum.
c. The total mechanical energy of the mass is zero.
d. The kinetic energy of the mass is at a maximum.
e. The speed of the mass is zero.
Answers
Answer:
Option (e)
Explanation:
A body executing SHM moves to and fro or back and forth about its mean position.
When the particle is at mean position, its velocity is maximum and when it is at extreme position its velocity is zero.
So, when it is at maximum distance:
a.
The acceleration is maximum.
b.
The potential energy is maximum.
c.
The total mechanical energy is non zero.
d.
The kinetic energy is zero.
e. The speed is zero. Correct
While an object near the earths surface is in free fall, its
A) velocity increases
B) acceleration increases
Answers
Answer:
a
Explanation:
The rate of change of an object's location with relation to a reference point is its velocity, which is dependent on time. when an object is dropped from space at rest (t = 0) under the influence of gravity, the velocity of the object changes and increases with time while the acceleration decreases.
The Lamborghini Huracan has an initial acceleration of 0.85g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode along, what would the car's acceleration be?
Answers
Answer:
7.9 [tex]\frac{m}{s^{2} }[/tex]
Explanation:
Take the fact that mass is inversely proportional to accelertation:
m ∝ a
Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:
[tex]\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\[/tex]
Using algebra, we can rearrange our equation to find the final acceleration, [tex]a_{2}[/tex]:
[tex]a_{2} = \frac{a_{1}*m_{1} }{m_{2} } \\[/tex]
Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:
0.85g * 9.8 [tex]\frac{m }{s^{2} }[/tex] = 8.33 [tex]\frac{m }{s^{2} }[/tex]
Plug everything in:
7.9 [tex]\frac{m }{s^{2} }[/tex] = [tex]\frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}[/tex]
(1590kg the initial weight plus the weight of the added passenger)
Which one is the better material to use for an inexpensive compass? hard iron, soft iron, or any conductor
Answers
Answer:
Soft iron
Explanation:
The free-fall acceleration at the surface of planet 1 is 22 m/s^2. The radius and the mass of planet 2 are twice those of planet 1. What is the free-fall acceleration on planet 2?
Answers
Answer:
g₂ = 11 m/s²
Explanation:
The value of free-fall acceleration on the surface of a planet is given by the following formula:
[tex]g = \frac{Gm}{r^2}[/tex]
where,
g = free-fall acceleration
G = Universal Gravitational Constant
m = mass of the planet
r = radius of planet
FOR PLANET 1:
[tex]g_1 = \frac{Gm_1}{r_1^2}\\\\\frac{Gm_1}{r_1^2} = 22 m/s^2[/tex] --------------------- equation (1)
FOR PLANET 2:
[tex]g_2 = \frac{Gm_2}{r_2^2}\\\\g_2 = \frac{G(2m_1)}{(2r_1)^2}\\\\g_2 = \frac{1}{2}\frac{Gm_1}{r_1^2}\\\\[/tex]
using equation (1):
[tex]g_2 = \frac{g_1}{2}\\\\g_2 = \frac{22\ m/s^2}{2}[/tex]
g₂ = 11 m/s²
Q 26.12: Assume current flows in a cylindrical conductor in such a way that the current density increases linearly with radius, from zero at the center to 1.0 A/m2 at the surface of the conductor. If the conductor has a cross sectional area of 1.0 m2, what can you say about the current in this conductor
Answers
Answer:
The current is 0.67 A.
Explanation:
Density, J = 1 A/m^2
Area, A = 1 m^2
Let the radius is r. And outer is R.
Use the formula of current density
[tex]I = \int J dA = \int J 2\pi r dr\\\\I = \int_{0}^{R}\frac{2\pi r^2}{R} dr\\\\I = \frac{2 \pi R^2}{3}.... (1)Now A = \pi R^2\\\\1 =\pi R^2\\\\R^2 = \frac{1}{\pi}\\\\So, \\\\I = \frac{2\pi}{3}\times \frac{1}{\pi}\\\\I = 0.67 A[/tex]
A cylindrical tank with radius 7 m is being filled with water at a rate of 2 m3/min. How fast is the height of the water increasing (in m/min)?
Answers
Answer:
0.013 m/min
Explanation:
Applying,
dV/dt = (dh/dt)(dV/dh)............. Equation 1
Where
V = πr²h................ Equation 2
Where V = volume of the tank, r = radius, h = height.
dV/dh = πr²............ Equation 3
Substitute equation 3 into equation 1
dV/dt = πr²(dh/dt)
From the question,
Given: dV/dt = 2 m³/min, r = 7 m, π = 3.14
Substitute these values into equation 3
2 = (3.14)(7²)(dh/dt)
dh/dt = 2/(3.14×7²)
dh/dt = 0.013 m/min
A water-balloon launcher with mass 2 kg fires a 0.75 kg balloon with a
velocity of 14 m/s to the west. What is the recoil velocity of the launcher?
What is the answer
Answers
Answer:
5.25 m/s to the east
Explanation:
Applying,
MV = mv.............. Equation 1
Where M = mass of the launcher, V = recoil velocity of the launcher, m = mass of the balloon, v = velocity of the balloon
make V the subject of the equation
V = mv/M............ Equation 2
From the question,
M = 2 kg, m = 0.75 kg, v = 14 m/s
Substitute these values into equation 2
V = (0.75×14)/2
V = 5.25 m/s to the east
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.
Answers
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.
TIME REMAINING
45:43
What are possible units for impulse? Check all that apply.
kg • m
kg • meters per second
N • s
N • m
StartFraction Newtons times meters per second EndFraction
Answers
Answer:
n.m maby
Explanation:
i think or its kg m/s
Answer:
answer (B) & (C)
Explanation:
kg • /N • s
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at 10 rev/s; 60 revolutions later, its angular speed is 15 rev/s. Calculate
(a) the angular acceleration,
(b) the time required to complete the 60 revolutions,
(c) the time required to reach the 10 rev/s angular speed, and
(d) the number of revolutions from rest until the time the disk reaches the 10 rev/s angular speed.
Answers
Explanation:
Given:
[tex]\omega_0[/tex] = 10 rev/s = [tex]20\pi\:\text{rad/s}[/tex]
[tex]\omega[/tex] = 15 rev/s = [tex]30\pi\:\text{rad/s}[/tex]
[tex]\theta[/tex] = 60 rev = [tex]120\pi\:\text{rads}[/tex]
a) the angular acceleration [tex]\alpha[/tex] is given by
[tex]\alpha = \dfrac{\omega^2 - \omega_0^2}{2\theta}[/tex]
[tex]\:\:\:\:\:\:\:=\dfrac{(30\pi)^2 - (20\pi)^2}{240\pi} = 6.5\:\text{rad/s}^2[/tex]
b) [tex]t = \dfrac{\omega - \omega_0}{\alpha} = \dfrac{30\pi - 20\pi}{6.5} = 4.8\:\text{s}[/tex]
c) [tex]t = \dfrac{\omega - \omega_0}{\alpha}[/tex]
[tex]=\dfrac{20\pi - 0}{6.5} = 9.7\:\text{s}[/tex]
d)[tex]\theta = \frac{1}{2}\alpha t^2[/tex]
[tex]\:\:\:\:\:\:\:=\frac{1}{2}(6.5\:\text{rad/s}^2)(9.7\:\text{s})^2 = 305.8\:\text{rad}[/tex]
[tex]\:\:\:\:\:\:\:= 48.7\:\text{revs}[/tex]
Find the force on a negative charge that is placed midway between two equal positive charges. All charges have the same magnitude.
Answers
Answer: The force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.
Explanation:
Let us assume that
[tex]q_{1} = q_{2} = +q[/tex]
[tex]q_{3} = -q[/tex]
As [tex]q_{3}[/tex] is the negative charge and placed midway between two equal positive charges ([tex]q_{1}[/tex] and [tex]q_{2}[/tex]).
Total distance between [tex]q_{1}[/tex] and [tex]q_{2}[/tex] is 2r. This means that the distance between [tex]q_{1}[/tex] and [tex]q_{3}[/tex], [tex]q_{2}[/tex] and [tex]q_{3}[/tex] = d = r
Now, force action on charge [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.
[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})[/tex]
where,
k = electrostatic constant = [tex]9 \times 10^{9} Nm^{2}/C^{2}[/tex]
Substitute the values into above formula as follows.
[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})\\= 9 \times 10^{9} (\frac{q \times (-q)}{r^{2}})\\= - 9 \times 10^{9} (\frac{q^{2}}{r^{2}})[/tex] ... (1)
Similarly, force acting on [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.
[tex]F_{32} = k \frac{q_{2}q_{3}}{d^{2}}\\= -9 \times 10^{9} \frac{q^{2}}{r^{2}}\\[/tex] ... (2)
As both the forces represented in equation (1) and (2) are same and equal in magnitude. This means that the net force acting on charge [tex]q_{3}[/tex] is zero.
Thus, we can conclude that the force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.
The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 10.6 ft/s at point A and 15.6 ft/s at point C. The cart takes 4.00 s to go from point A to point C, and the cart takes 1.80 s to go from point B to point C. What is the cart's speed at point B
Answers
Answer:
a) [tex]a_{avg}=1.25ft/s^2[/tex]
b) [tex]v_b=13.35ft/s[/tex]
Explanation:
From the question we are told that:
Speed at point A [tex]v_A=10.6ft/s[/tex]
Speed at point C [tex]v_C=15.6ft/s[/tex]
Time from Point A to C [tex]T_{ac}=4.00s[/tex]
Time from Point B to C [tex]T_{bc}=1.80s[/tex]
Generally the equation for acceleration From A to B is mathematically given by
[tex]a_{avg}=\frac{v_c-v_a}{\triangle t}[/tex]
[tex]a_{avg}=\frac{15.6-10.6}{4.0 }[/tex]
[tex]a_{avg}=1.25ft/s^2[/tex]
Generally the equation for cart speed at B is mathematically given by
[tex]a_{avg}=\frac{v_c-v_a}{T_{bc}}[/tex]
[tex]v_b=v_c-a_{avg}*T_{bc}[/tex]
[tex]v_b=15.6ft/s-(1.25ft/s^2)(1.80)[/tex]
[tex]v_b=13.35ft/s[/tex]
You are driving to the grocery store at 20 m/s. You are 150 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.
Required:
a. How far are you from the intersection when you begin to apply the brakes?
b. What acceleration will bring you to rest right at the intersection?
c. How long does it take you to stop?
Answers
Hi there!
a.
Use the formula d = st to solve:
d = 20 × 0.5 = 10m
150 - 10 = 140m away when brakes are applied
b.
Use the following kinematic equation to solve:
vf² = vi² + 2ad
Plug in known values:
0 = 20² + 2(150)(a)
Solve:
0 = 400 + 300a
-300a = 400
a = -4/3 (≈ -1.33) m/s² required
c.
Use the following kinematic equation to solve:
vf = vi + at
0 = 20 - 4/3t
Solve:
4/3t = 20
Multiply both sides by 3/4 for ease of solving:
t = 15 sec
1. On each of your equipotential maps, draw some electric field lines with arrow heads indicating the direction of the field. (Hint: At what angle do field lines intersect equipotential lines?) Draw sufficient field lines that you can "see" the electric field.
Answers
Answer:
The angle between the electric field lines and the equipotential surface is 90 degree.
Explanation:
The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.
The electric field lines are always perpendicular to the equipotential surface.
As
[tex]dV = \overrightarrow{E} . d\overrightarrow{r}\\\\[/tex]
For equipotential surface, dV = 0 so
[tex]0 = \overrightarrow{E} . d\overrightarrow{r}\\\\[/tex]
The dot product of two non zero vectors is zero, if they are perpendicular to each other.
I need help with this physics question.
Answers
Answer:
5.04 m
Explanation:
You are told that the homeowner wants to increase their fences by 34 percent meaning Original+ 34 percent. If the original is 100 percent, then the new fence size will be 134 % of the original. You are given the original which is 3.76 meters, to find new fence size 1.34 * 3.76m to get 5.0384 meters, rounded to 5.04 m.
Answer:
5.0384m
Explanation:
% increase = 100 x (Final - Initial / | initial | )
( |~~| Bars indicate absolute value since you can't have a negative height)
A 56.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 11.2 s, what is the spring constant of the bungee cord, assuming it has negligible mass compared to that of the jumper in N/m
Answers
Answer:
2.80N/m
Explanation:
Given data
mass m= 56kg
perios T= 11.2s
The expression for the period is given as
T=2π√m/k
Substitute
11.2= 2*3.142*√56/k
square both sides
11.2^2= 2*3.142*56/k
125.44= 351.904/k
k=351.904/125.44
k= 2.80N/m
Hence the spring constant is 2.80N/m
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
Answers
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 box-shaped metal can has dimensions 8 in. by 4 in. by 10 in. high. All of the air inside the can is removed with a vacuum pump. Assuming normal atmospheric pressure outside the can, find the total force on one of the 8-by-10-in. sides
Answers
Answer:
The force on the side is 5252 N.
Explanation:
Area, A = 8 in x 10 in = 80 in^2 = 0.052 m^2
height, h = 10 in
The force on the area is
F = P x A
where, P is the atmospheric pressure and A is the area.
P = 1.01 x 10^5 Pa
Force = 1.01 x10^5 x 0.052 = 5252 N
the Period T of oscillation of a Single Pendulum depends on the length l, and acceleration g. Determine the exact form of the dependence.
Answers
Answer:
[tex]{ \tt{check \: in \: the \: pic}}[/tex]
One hazard of space travel is the debris left by previous missions. There are several thousand objects orbiting Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. Calculate the force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of 4.00×10^3 m/s, given the collision lasts 6.00×10^8s.
Answers
Answer:
F = 6666.7 N
Explanation:
Given that,
Mass of a chip, m = 0.1 mg
Initial speed, u = 0
Final speed,[tex]v=4\times 10^{3}\ m/s[/tex]
Time of collision,[tex]t=6\times 10^{-8}\ s[/tex]
We know that,
Force, F = ma
Put all the values,
[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.1\times 10^{-6}\times (4\times 10^3-0)}{6\times 10^{-8}}\\\\F=6666.7\ N[/tex]
So, the required force is 6666.7 N.
0. The temperature of source is 500K with source energy 2003, what is the temperature of sink with sink energy 100 J? a. 500 K b. 300 K c. 250 K d. 125 K
Answers
Answer:
c. 250k
Explanation:
The temperature of the sink is approximately 250 K.
To find the temperature of the sink, we can use the formula for the efficiency of a heat engine:
Efficiency = 1 - (Temperature of Sink / Temperature of Source)
Given that the temperature of the source (T_source) is 500 K and the source energy (Q_source) is 2003 J, and the sink energy (Q_sink) is 100 J, we can rearrange the formula to solve for the temperature of the sink (T_sink):
Efficiency = (Q_source - Q_sink) / Q_source
Efficiency = (2003 J - 100 J) / 2003 J
Efficiency = 1903 J / 2003 J
Efficiency = 0.9497
Now, plug the efficiency back into the first equation to solve for T_sink:
0.9497 = 1 - (T_sink / 500 K)
T_sink / 500 K = 1 - 0.9497
T_sink / 500 K = 0.0503
Now, isolate T_sink:
T_sink = 0.0503 * 500 K
T_sink = 25.15 K
Since the temperature should be in Kelvin, we round down to the nearest whole number, which is 25 K. Thus, the temperature of the sink is approximately 250 K.
To learn more about sink energy, here
https://brainly.com/question/10483137
#SPJ2
An object is moving from north to south what is the direction of the force of friction of the object
Answers
Answer:
North
Explanation:
Friction is a reaction force against the direction of movement. So since the direction of movement is south the friction would be opposite and move north.
Answer:
South To North
Explanation:
Frictional force acts in the direction opposite to the direction of motion of a body. Because the object is moving from north to south, the direction of frictional force is from south to north
Pure water is an example of alan
A. Insulator
B. Metalloid
C. Conductor
D. Nonmetal
Answers
Answer: I think its A or C I'm not sure though sorry.
The total resistance of a parallel circuit is 25 ohms. If the total current is 100mA, how much current is through a 220 ohm resistor that makes up part of the parallel circuit?
Answers
Answer:
The current across the resistance is 0.011 A.
Explanation:
Total resistance, R = 25 ohms
Total current, I = 100 mA = 0.1 A
Let the voltage is V.
By the Ohm's law
V = I R
V = 0.1 x 25 = 2.5 V
Now the resistance is R' = 220 ohm
As they are in parallel so the voltage is same. Let the current is I'.
V = I' x R'
2.5 = I' x 220
I' = 0.011 A
What are the differences among elements, compounds, and mixtures?
Answers
Answer:
Elements have a characteristic number of electrons and protons.Both Hydrogen(H) and oxygen(O) are two different elements.
••••••••••••••••
Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of .Hydrogen(H) and Oxygen(O) both qr4 naturally gases,but they react to form water(H2O),which is liquid compound.
•••••••••••••••
A mixture is made of atleast two parts》 solid,liquid or gas.The difference is that it's not a chemical substance that's bonded by other elements.
------------------------------
Hope it helps...
Have a great day!!!
Answer: Elements have a characteristic number of electrons and protons. Both Hydrogen(H) and oxygen(O) are two different elements. Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of.Hydrogen(H) and Oxygen(O) both qr4 naturally gases, but they react to form water(H2O), which is liquid compound. A mixture is made of at least two parts solid, liquid, or gas. The difference is that it's not a chemical substance that's bonded by other elements.
the magnitude of the magnetic field at point p for a certain electromagnetic wave is 2.21. What is the magnitude of the elctic field for that wave at P
Answers
Answer:
[tex]6.63\times 10^8\ N/C[/tex]
Explanation:
Given that,
The magnitude of magnetic field, B = 2.21
We need to find the magnitude of the electric field. Let it is E. So,
[tex]\dfrac{E}{B}=c\\\\E=Bc[/tex]
Put all the values,
[tex]E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C[/tex]
So, the magnitude of the electric field is equal to [tex]6.63\times 10^8\ N/C[/tex].
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
Answers
Answer: -5×10-3
Explanation:
E=kq/r
* A ball is projected horizontally from the top of
a building 19.6m high.
a, How long when the ball take to hit the ground?
b, If the line joining the point of projection to
the point where it hits the ground is 45
with the horizontal. What must be the
initial velocity of the ball?
c,with what vertical verocity does the ball strike
the grounds? (9= 9.8 M152)
Answers
Explanation:
Given
Ball is projected horizontally from a building of height [tex]h=19.6\ m[/tex]
time taken to reach ground is given by
[tex]\text{Cosidering vertical motion}\\\Rightarrow h=ut+0.5at^2\\\Rightarrow 19.6=0+0.5\times 9.8t^2\\\Rightarrow t^2=4\\\Rightarrow t=2\ s[/tex]
(b) Line joining the point of projection and the point where it hits the ground makes an angle of [tex]45^{\circ}[/tex]
From the figure, it can be written
[tex]\Rightarrow \tan 45^{\circ}=\dfrac{h}{x}\\\\\Rightarrow x=h\cdot 1\\\Rightarrow x=19.6[/tex]
Considering horizontal motion
[tex]\Rightarrow x=u_xt\\\Rightarrow 19.6=u_x\times 4\\\Rightarrow u_x=4.9\ m/s[/tex]
(c) The vertical velocity with which it strikes the ground is given by
[tex]\Rightarrow v^2-u_y^2=2as\\\Rightarrow v^2-0=2\times 9.8\times 19.6\\\Rightarrow v=\sqrt{384.16}\\\Rightarrow v=19.6\ m/s[/tex]
Thus, the ball strikes with a vertical velocity of [tex]19.6\ m/s[/tex]
Explanation:
Given
Ball is projected horizontally from a building of height
time taken to reach ground is given by
(b) Line joining the point of projection and the point where it hits the ground makes an angle of
From the figure, it can be written
Considering horizontal motion
(c) The vertical velocity with which it strikes the ground is given by
Thus, the ball strikes with a vertical velocity of