Answer:
Work done by the bullet is 612.26 J.
Explanation:
mass of bullet, m = 0.5 kg
initial velocity of bullet, u = 50 m/s
coefficient of friction = 0.2
mass of block, M = 3 kg
let the final speed of the bullet block system is v.
use conservation of momentum
Momentum of bullet + momentum of block = momentum of bullet block system
0.5 x 50 + 3 x 0 = (3 + 0.5) v
v = 7.14 m/s
let the stopping distance is
The work done is given by change in kinetic energy of bullet
initial kinetic energy of bullet, K = 0.5 x 0.5 x 50 x 50 = 625 J
Final kinetic energy of bullet, K' = 0.5 x 0.5 x 7.14 x 7.14 = 12.74 J
So, the work done by the bullet
W = 625 - 12.74 = 612.26 J
don't answer for points you will be reported
Explanation:
Glasses or Contacts. You might not realize it, but if you wear glasses or contact lenses, this is light refraction at play. ...
Human Eyes. Human eyes have a lens. ...
Prism. ...
Pickle Jar. ...
Ice Crystals. ...
Glass. ...
Twinkling Stars. ...
Microscope or Telescope.
A metallic sphere of radius 5 cm is charged such that the potential of its surface is 100 V (relative to infinity). Draw the plots that correctly shows the potential as a function of distance from the center of the sphere.
Answer:
Attached below
Explanation:
The plot showing the potential as a function of distance from the center of the sphere is attached below
note : The Potential inside the sphere will remain constant as potential remains constant on surface and it start to decrease as 1/r
____ are the foundation of psychoanalytic theory.
Answer:
an unconscious needs rooted in childhood are the foundation of psychoanalytic theory.
Explanation:
Three forces are pulling on the same object such that the system is in equilibrium. Their magnitudes are F1 = 2.83 N.F= 3.35 N. and F3 = 3.64 N, and they make angles of 0, = 45.0°, 02 = -63.43 and 03 =164.05° with respect to the x-axis, respectively.
Required:
a. What is the x-component of the force vector F1?
b. What is the y-component of the force vector F1?
(a) 2.001N
(b) 2.001N
Explanation:A sketch of the scenario has been attached to this response.
Since only the force vector F₁ is required, the only force shown in the sketch is F₁.
As shown in the sketch;
The x-component of the force vector F₁ = [tex]F_{x}[/tex]
The y-component of the force vector F₁ = [tex]F_{y}[/tex]
The magnitude of F₁ as given in the question = 2.83N
The angle that the force makes with respect to the x-axis = 45.0°
Using the trigonometric ratio, we see that;
(a) cos 45.0° = [tex]\frac{F_x}{F_1}[/tex]
=> [tex]F_{x}[/tex] = F₁ cos 45.0°
=> [tex]F_{x}[/tex] = 2.83 cos 45.0°
=> [tex]F_{x}[/tex] = 2.83 x 0.7071
=> [tex]F_{x}[/tex] = 2.001N
(b) Also;
sin 45.0° = [tex]\frac{F_y}{F_1}[/tex]
=> [tex]F_{y}[/tex] = F₁ sin 45.0°
=> [tex]F_{y}[/tex] = 2.83 sin 45.0°
=> [tex]F_{y}[/tex] = 2.83 x 0.7071
=> [tex]F_{y}[/tex] = 2.001N
Therefore, the x-component and y-component of the force vector F₁ is 2.001N
The x and y component of vector F1 is mathematically given as
F_x = 2.001N
F_y= 2.001N
What is the x and y component of vector F1?Question Parameters:
Generally, the equation for the x-component is mathematically given as
x=Fsin\theta
Therefore
F_x = F₁ cos 45.0°
F_x = 2.83 x 0.7071
F_x = 2.001N
For y component
x=Fcos\theta
F_y = F₁ sin 45.0
F_y = 2.83 x 0.7071
F_y= 2.001N
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Một xe hơi nặng 1000kg đang kéo một toa mo1oc 300kg. Cả hai cùng tiến về phía trước với gia tốc 2.15m/s2. Bỏ qua lực cản không khí xác định: Tổng lực tác dụng lên xe hơi
Answer:
Một ô tô có khối lượng m=1000kg đang chạy với vận tốc 18km/h thì hãm phanh.Biết lực hãm là 2000N. Tính quãng đường xe còn chạy thêm trước khi dừng ... Chiếu phương trình của định luật II Newtơn mà →F=m. ... chuyển động ta có F=ma, suy ra gia tốc chuyển động của xe ( với F=2000N) ...
Explanation:
An +9.7 C charge moving at 0.75 m/s makes an angle of 45∘ with a uniform, 1.5 T magnetic field. What is the magnitude of the magnetic force F that the charge experiences?
Answer:
F = 7.72 N
Explanation:
The magnetic force on the charge can be given by the following formula:
[tex]F = qvB Sin\theta[/tex]
where,
F = magnetic force = ?
q = magnitude of charge = 9.7 C
v = speed of charge = 0.75 m/s
B = magnetic field = 1.5 T
θ = angle = 45°
Therefore,
[tex]F = (9.7\ C)(0.75\ m/s)(1.5\ T)Sin45^{o}[/tex]
F = 7.72 N
voltage is 50v and capacitance is 30c what is the charge
Explanation:
voltage is 50v and capacitance is 30c what is the charge
q=1500c
Bonnie and Clyde are sliding a 289 kg bank safe across the floor to their getaway car. The safe slides with a constant speed if Clyde pushes from behind with 387 N of force while Bonnie pulls forward on a rope with 341 N of force.
Required:
What is the safe's coefficient of kinetic friction on the bank floor?
Answer:
= 0.257
Explanation:
Applied force on the safe by Bonnie and Clyde is F= 387 N + 341 N
= 728 N
Given safe slides with constant speed .So, force of friction =applied force
= 728 N
μ*normal force = 728 N
from this the safes coefficient of kinetic friction on the bank floor is
μ = 728 N / normal force
= 728 N / Mg
= 728 N / ( 289 * 9.8 )
= 0.257
____________is obtained from the fleece of animals.
Answer:
wool and fibers
Explanation:
I HAVE A PHYSICS LOCKDOWN EXAM TODAY, THEY ARE 25 QUESTIONS AND I HAVE ABOUT AN HOUR TO SOLVE IT, I NEED HELP WITH THEM ASAP. PLEASE IF YOU'RE GOOD AT PHYSICS LET ME KNOW ILL BE SO GRATEFUL.
Answer:
I’ll try my best!
The figure shows three displacement vectors, which are
labeled a, 5, and c. What is the magnitude and direction
of the resultant vector found by adding 5 and ć?
N
4
S
15 m west
āt
2.0 m east
7.0 m east
Ĉ
A. 5.0 m east
B. 5.0 m west
ОО
C. 9.0 m west
0
D. 9.0 m east
Answer:
the correct answer is D
Explanation:
In this exercise, the vectors are in the same west-east direction, so we can assume that the positive direction is east and perform the algebraic sum.
R = δ + ε
where
δ = 2.0 m
ε = 7.0 m
the positive sign indicates that it is heading east
R = 2.0 + 7.0
R = 9.0 m
the direction is east
the correct answer is D
a particle of mass m=375 g is launched with velocity of A =4 moves with a length AB=2.5m under the action of tractive force F=3.46 N making an angle 30. during its motion between A and B which is subjected to a frictional force f=1.5 N
calculate V of B by applying the kinetic energy theorem
Answer:
The final speed is 5.78 m/s.
Explanation:
mass, m = 375 g = 0.375 kg
initial velocity, u = 4 m/s
Distance, s = 2.5 m
Angle, A = 30 degree
Force, F = 1.5 N
let the final velocity is v.
Use the work energy theorem
Work done = change in kinetic energy
[tex]W= 0.5 m(v^2 - u^2)\\\\F s cos A= 0.5 m (v^2 - u^2)\\\\1.5\times 2.5\times cos30= 0.5\times 0.375\times (v^2 - 16)\\\\v = 5.78 m/s[/tex]
The wavelength of visible light range of 400 to 750mm .what is the corresponding range of photon energies for visible light
Answer:
The range of the photon energies is between:
2.652 x 10⁻²⁵ J to 4.973 x 10⁻²⁵ J
Explanation:
The energy of a photon is calculated using the following equation;
E = hf
where;
h is Planck's constant = 6.63 x 10⁻³⁴ Js
f is frequency of the photon
[tex]E = h \frac{c}{\lambda} \\\\where;\\\\\lambda \ is \ the \ wavelength\\\\c \ is \ the \ speed \ of \ light \ = 3\times 10^8 \ m/s\\\\When \ \lambda = 400 \ mm = 400 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{400 \times 10^{-3}} \\\\E = 4.973 \times 10^{-25} \ J[/tex]
[tex]When \ \lambda = 750 \ mm = 750 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{750 \times 10^{-3}} \\\\E = 2.652 \times 10^{-25} \ J[/tex]
The range of the photon energies is between:
2.652 x 10⁻²⁵ J to 4.973 x 10⁻²⁵ J
A woman stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.64 of her regular weight. Calculate the magnitude of the acceleration of the elevator.
Answer:
The downwards acceleration is 3.53 m/s2.
Explanation:
Let the true weight is m g.
The reading of the balance, R = 0.64 mg
Let the acceleration is a.
As the apparent weight is less than the true weight so the elevator goes down wards with some acceleration.
Use Newton's second law
m g - R = m a
m g - 0.64 m g = m a
0.36 g = a
a = 3.53 m/s2
When a single high-resistance (long) bulb is connected to a 1.5 V battery, the current through the battery is about 80 mA. If you add another high-resistance (long) bulb in parallel, the battery current of course increases to 160 mA. Select all of the true statements given this situation.
a. The battery is ohmic.
b. The battery is not ohmic.
c. Current through the battery is proportional to ΔV across the battery.
d. Current through the battery is not proportional to ΔV across the battery.
e. The battery always puts out the same current.
Answer:
a) True. The battery obeys ohm's law, it is formed by an ideal source with a fixed internal resistance
c) True. Ohm's law is V = iR therefore voltage and current are proportional
Explanation:
In this problem let's analyze the load of the system, when a resistance is placed the current is 80 mA, if we place two resistors in parallel the voltage remains the same, but the current is divided between each resistance (bulb), therefore the current in the battery it must be 160 mA
Let's analyze the answers
a) True. The battery obeys ohm's law, it is formed by an ideal source with a fixed internal resistance
b) false
c) True. Ohm's law is V = iR therefore voltage and current are proportional
d) False
e) False. The current coming out of the battery is proportional to the load placed
An object starts to rotate about an axis from rest wih a, uniform angular acceleration of 2pi rads-2 what is the no.of rotations it can complete in 5s
Answer:
θ = 12.5 rotations
Explanation:
The number of rotations can be found by using the second equation of motion:
[tex]\theta = \omega_i t + \frac{1}{2}\alpha t^2\\\\[/tex]
where,
[tex]\theta[/tex] = angular displacement = ?
ωi = initial angular speed = 0 rad/s
t = time = 5 s
α = angular acceleration = 2π rad/s²
Therefore,
[tex]\theta = (0\ rad/s)(5\ s)+\frac{1}{2}(2\pi\ rad/s^2)(5\ s)^2\\\\\theta = 78.54\ rad[/tex]
converting it to no. or rotations:
[tex]\theta = (78.54\ rad)(\frac{1\ rotation}{2\pi\ rad})[/tex]
θ = 12.5 rotations
What is needed to Run A Brushless DC motor
Two connection methods are used for brushless DC motors. One method is to connect the coils in a loop as we compared it with the rotor winding of DC motors in Fig. 2.27. This method is called a Δ (delta) connection.
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The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to Ip, the moment of inertia of the same object about a parallel axis passing through point p. The mathematical statement of the theorem is Ip=Icm+Md2, where d is the perpendicular distance from the center of mass to the axis that passes through point p, and M is the mass of the object. Part A Suppose a uniform slender rod has length L and mass m. The moment of inertia of the rod about about an axis that is perpendicular to the rod and that passes through its center of mass is given by Icm=112mL2.
Required:
Find Iend, the moment of inertia of the rod with respect to a parallel axis through one end of the rod.
Answer:
I = ⅓ m L²
Explanation:
They tell us to use the parallel axes theorem
I = [tex]I_{cm}[/tex] + M d²
The moment of inertia of a rod with respect to the center of mass is
I_{cm} = [tex]\frac{1}{12}[/tex] m L²
the distance from the center of mass that coincides with its geometric center to the ends of the rod is
d = L / 2
we substitute
I =[tex]\frac{1}{12}[/tex] m L² + m (L/2)²
I = m L² ( [tex]\frac{1}{12} + \frac{1}{4}[/tex] )
I = m L² ( [tex]\frac{1+3}{12}[/tex] )
I = ⅓ m L²
A dipole placed in an electric field will... Try to align along the direction of the field. Try to align antiparallel to the direction of the field. try to align perpendicular to the field. none of these choices.
Answer:
The dipole always tries to align in the direction of electric field so that the net torque is zero.
Explanation:
When a dipole placed in an electric field at any arbitrary angle, it experiences a torque.
Let the dipole moment is p, electric field is E and the angle is A.
The torque is given by
Torque = p E sin A
where, A is the angle between the electric field and dipole moment.
So, the dipole always tries to align in the direction of electric field so that the net torque is zero.
A student on a new planet wants to determine the value of gravity on that planet. Luckily for them they brought equipment that they can use to set up an oscillating spring or an oscillating pendulum. Which procedure would allow the student to determine the value of gravity on the new planet
Answer:
By measure the effective length and the time period of the pendulum.
Explanation:
Let the student take the oscillating pendulum at the planet.
He measure the time period of the pendulum by using the stop watch or the ordinary watch.
Then measure the effective length of the pendulum which is the distance between the center of gravity of the bob and the point of suspension of the pendulum.
Now, use the formula of the time period of the pendulum,
[tex]T =2\pi\sqrt\frac{L}{g}[/tex]
Here, L is the effective length of the pendulum, g is the acceleration due to gravity at the planet and T is time period of the pendulum.
By rearranging the terms, we get
[tex]T =2\pi\sqrt\frac{L}{g}\\\\T^{2}=4\pi^2\times\frac{L}{g}\\\\g =\frac{4\pi^2L}{T^2}[/tex]
Here, by substituting the values of L and T, the student get the value of acceleration due to gravity at that planet.
The process of finding the age of fossil is called which answer carbondating,urinumdating ,both of them,none
Answer:
the process of finding the age of fossils is called carbon dating because they use samples of carbon and compare it's characteristics to the samples of carbon from each set time
explain how a lever can act as a force multiplier
Answer:
Example:Opening of a bottle cap by tool
when we hold a tool and open the bottle cap this is because , force x tool force .
The load arm is usually shorter than the effort arm in second order levers. Moving a large weight hence requires less effort. A force multiplier lever or effort multiplier lever is the name for this kind of lever. A boat's oars, for instance, can increase the force.
What is second order levers?Second-order levers are devices with the input force farthest from the fulcrum and the output force on the same side of the fulcrum. A wheelbarrow is an excellent illustration of a second-order lever.
A second-order lever will have an output force greater than an input force, similar to first-order levers. The output journey, however, will be shorter than the input length. Both the input and output forces in this situation will move in the same direction.
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1. It is an object's tendency to resist a change in motion.
2. The study of celestial object such as moon, planets stars and galaxies.
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Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of 33 pC experiences an electric force in the earth's electric field, which is typically 100 N/C, directed downward.
1. What is the ratio of the electric force on the bee to the bee's weight?
2. What electric field strength would allow the bee to hang suspended in the air?
3. What electric field direction would allow the bee to hang suspended in the air?
Answer:
A) 3.367 × 10^(-6)
B) 2.97 × 10^(7) N/C
C) Upwards
Explanation:
We are given;
Mass of bee; m = 100 mg = 100 × 10^(-6) kg
Charge on bee;q=33 pC = 33 × 10^(-12)C
Electric field strength; E = 100 N/C
A) Formula for weight of bee; W = mg = 100 × 10^(-6) × 9.8 = 9.8 × 10^(-4) N
Electric force on Bee; F = qE = 33 × 10^(-12) × 100 = 33 × 10^(-10) N
ratio of the electric force on the bee to the bee's weight; F/W = (33 × 10^(-10))/(9.8 × 10^(-4)) = 3.367 × 10^(-6)
B) For the bee to be suspended in the air, it means the weight of the bee must be equal to the electric force. Thus;
mg = qE
100 × 10^(-6) × 9.8 = 33 × 10^(-12) × E
E = (100 × 10^(-6) × 9.8)/(33 × 10^(-12))
E = 2.97 × 10^(7) N/C
C) From Newton's law, sum of forces = 0.
Thus;
F_n + F + W = 0
Where F is the normal force.
Thus;
F_n = -(F + W)
F_n = - ((33 × 10^(-10)) + (9.8 × 10^(-4)))
F_n = -9.8 × 10^(-4) N
Thus, applied electric field is;
E_a = F_n/q = (-9.8 × 10^(-4))/(33 × 10^(-12)) = -2.97 × 10^(7) N/C
This is negative and so it means the direction will be opposite the Earth's electric filed which is upwards.
A 3-kg ball is thrown with a speed of 10 m/s at an unknown angle above the horizontal. The ball attains a maximum height of 1.7 m before striking the ground. If air resistance is negligible, what is the value of the kinetic energy of the ball at its highest point?
Answer:
Explanation:
Kinetic energy is at an absolute max when the potential energy is 0. At the ball's highest point, at its most absolute highest point, the velocity of the ball is 0, making KE = 0 and PE the only energy the ball has. So if this isn't a trick question, the wording is off.
A particle moves along a straight line. Its position at any instant is given by x = 32t− 38t^3/3 where x is in metre and t in second. Find the acceleration of the particle at the instant when particle is at rest.
Answer:
The acceleration of the object is -69.78 m/s²
Explanation:
Given;
postion of the particle:
[tex]x = 32t - 38\frac{t^3}{3} \\\\[/tex]
The velocity of the particle is calculated as the change in the position of the particle with time;
[tex]v = \frac{dx}{dt} = 32 - 38t^2\\\\when \ the \ particle \ is \ at \ rest, \ v = 0\\\\32-38t^2 = 0\\\\38t^2 = 32\\\\t^2 = \frac{32}{38} \\\\t = \sqrt{\frac{32}{38} } \\\\t = 0.918 \ s[/tex]
Acceleration is the change in velocity with time;
[tex]a = \frac{dv}{dt} = -76t\\\\recall , \ t = 0.918 \ s\\\\a = -76(0.918)\\\\a = -69.78 \ m/s^2[/tex]
What is the acceleration of a 0.30 kilogram ball that is hit
with a force of 27 N?
Answer: 90
Explanation: a=Fnet/M
=27/0.30
=90
the acceleration is 90
Answer:
The acceleration of the ball is 83.333ms2 [forward].
Explanation:
i hope it helps :)
A box with mass 25.14 kg is sliding at rest from the top of the slope with height 13.30 m
and slope angle 30 degree, suppose the coefficient of friction of the slope surface is
0.25, find (neglect air resistance,take g=10 m/s^2)
The friction force experienced by the box.
00) The acceleration of the box along the slope.
(1) The time T required for the object to reach the bottom of the slope from the slope top.
Answer:
Explanation:
The first thing we are asked to find is the Force experienced by the box. That is found in the formula:
F - f = ma where F is the force exerted by the box, f is the friction opposing the box, m is the mass, and a is the acceleration (NOT the same as the pull of gravity). But F can be rewritten in terms of the angle of inclination also:
[tex]wsin\theta-f=ma[/tex] where w is the weight of the box. We will use this version of the formula because it will help us answer the second question, which is to solve for a. Filling in:
First we need the weight of the box. Having the mass, we find the weight:
w = mg so
w = 25.14(10) so
w = 251.4 N (I am not paying any attention at all to the sig fig's here, since I noticed no one on this site does!) Now we have the weight. Filling that in:
251.4sin(30) - f = ma Before we go on to fill in for f, let's answer the first question. F = 251.4sin(30) so
F = 125.7 And in order to answer what a is equal to, we find f:
f = μ[tex]F_n[/tex] where Fn is the weight of the object.
f = .25(251.4) so
f = 62.85. Filling everything in now altogether to solve for a, the only missing value:
125.7 - 62.85 = 25.14a and
62.85 = 25.14a so
a = 2.5 m/s/s
Now we have to move on to another set of equations to answer the last part. The last part involves the y-dimension. In this dimension, what we know is that
a = -10 m/s/s
v₀ = 0 (it starts from rest)
Δx = -13.30 m (negative because the box falls this fr below the point fro which it started). Putting all that together in the equation for displacement:
Δx = v₀t + [tex]\frac{1}{2}at^2[/tex] and we are solving for time:
[tex]-13.30=0t+\frac{1}{2}(-10)t^2[/tex] and
[tex]t=\sqrt{\frac{2(-13.30)}{-10} }[/tex] so
t = 1.6 seconds to reach the bottom of the slope from 13.30 m high.
Henry, whose mass is 95 kg, stands on a bathroom scale in an elevator. The scale reads 830 N for the first 3.6 s after the elevator starts moving, then 930 N for the next 3.6 s.
What is the elevator's velocity 4.0 s after starting?
Answer: [tex]3.816\ m/s[/tex]
Explanation:
Given
Mass of Henry is 95 kg
Normal weight of Henry is [tex]mg=95\times 9.8=931\ N[/tex]
The scale reads the weight as 830 N for first 3.6 s i.e. less than the normal weight i.e. Elevator is moving downwards
Apparent weight is given by
[tex]\Rightarrow 830=m(g-a)\quad [a=\text{acceleration of elevator}]\\\Rightarrow 830=95(9.8-a)\\\Rightarrow 8.736=9.8-a\\\Rightarrow a=1.06\ m/s^2[/tex]
After 3.6 s weight becomes 930 N which is approximately equal to normal weight. It implies elevator starts moving with constant velocity i.e. no acceleration.
If elevator starts from rest, it velocity after 3.6 s is
[tex]v=u+at\\\Rightarrow v=0+1.06(3.6)\\\Rightarrow v=3.816\ m/s[/tex]
This velocity will remain continues as after 3.6 s, elevator starts moving with constant velocity.
A hole is made in a square tile of uniform thickness. The diagram shows the tile hanging loosely on a nail. Where is the centre of gravity of the tile?
Answer:
The center of gravity of the tile is at the center of the tile:
Examining 1/2 of the tile, each mass point will have an equal mass point
on the other side of the tile (considering a line thru the center of the tile)
with coordinates opposite and equal of the point being considered.
Each mass point will have an identical mass point on the opposite side of the tile, with coordinates opposite and equal to the point being analyzed, while looking at only half of the tile.
What is center of gravity ?
A hypothetical location where the gravitational force is said to appear to act is the human body's centre of gravity (COG). That is the location where the entire body's bulk appears to be concentrated[2]. The COG need not be located inside the actual boundaries of an object or a person because it is a hypothetical point. Imagine the thing balanced on one finger to get an idea of its COG (there are objective measurements).
The COG is positioned anatomically so that it is about anterior to the second sacral vertebra. The specific placement of the COG, however, continually shifts with each new posture of the torso and limbs since humans do not remain stable in the anatomical position. The dimensions of the body
The center of the gravity is at half of the length of tile.
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