Answer:
the component of light reflected by that object
Explanation:
which of the following is not an equation of motion:- v=u+at; s=ut+at; v^2-u^2=2as; s=ut+1/2 at^2
Answer:
s=ut+at
its not the equation of motion..
hope it helps stay safe healthy and happy.....The electric potential energy of an electric dipole is ______ when the dipole is aligned with an electric field. The electric potential energy of an electric dipole is ______ when the dipole is aligned with an electric field. most positive most negative zero infinite undefined
Answer:
Most negative
Explanation:
The electric potential energy of a dipole U = -pEcosθ where p = dipole moment, E = electric field and θ = angle between p and E.
When the electric dipole is aligned with the electric field, the angle between p and E is 0°. That is θ = 0°.
So, U = -pEcosθ
U = -pEcos0°
U = -pE
which is the most negative value it can have.
So, the electric potential energy of an electric dipole is most negative when the dipole is aligned with an electric field.
So, most negative is the answer.
Internal energy of a diatomic gas consists of:
OA. kinetic energy due to vibration and rotation.
B. kinetic energy due to translation, vibration, and rotation.
C. potential energy due to intermolecular forces.
D. kinetic energy due to translation only.
Answer:
C) Potential energy due to intermolecular forces.
Which describes an atom that has fewer neutrons than protons and more electrons than protons?
negative compound
positive compound
positive ion
negative ion
Answer:
Negative ion
.................
A student sects a leaf of length 7.2 cm to draw. Her drawing is 28.8 cm in length. What is the magnification of the drawing?
a) x4
b) x207.36
c) x0.25
d) x36
Answer:
A) x4
Explanation:
Magnification is equal to image size divided by the actual size, or M = I/A.
The image size is the student's drawing, which is 28.8 cm, and the actual size is 7.2 cm. Divide them, and cancel out the units, and you should get:
28.8 cm/7.2 cm = 4
An electrostatic paint sprayer has a 0.17 m-diameter metal sphere at a potential of 25.0 kV that repels charged paint droplets onto the object to be painted, which is grounded (electric potential of 0).
Answer:
[tex]q=0.236uC[/tex]
Explanation:
From the question we are told that:
Diameter [tex]d=0.17m[/tex]
Radius [tex]r=0.17/2=>0.085[/tex]
Potential [tex]E=25.0kV[/tex]
Generally the equation for Potential on spere is mathematically given by
[tex]E=\frac{1}{4 \pi e_0}*\frac{q}{r}[/tex]
Therefore
[tex]q=\frac{25*10^3*0.085}{\frac{1}{4 \pi e_0}}[/tex]
Where
[tex]\frac{1}{4 \pi e_0}=9*10^9[/tex]
Therefore
[tex]q=\frac{25*10^3*0.085}{(9*10^9}}[/tex]
[tex]q=0.236uC[/tex]
A thin rod of length 1.4 m and mass 180 g is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed 1.80 rad/s. Neglecting friction and air resistance, find (a) the rod's kinetic energy at its lowest position and (b) how far above that position the center of mass rises.
Answer:
[tex]K.E = 0.1905 J[/tex]
Explanation:
From the question we are told that:
Length [tex]L=1.4m[/tex]
Mass [tex]m=180g[/tex]
Angular Velocity [tex]\omega=1.80rads/s[/tex]
Generally the equation for Kinetic energy K.E is mathematically given by
[tex]K.E =0.5 (1/3 ML^2 )w^2[/tex]
[tex]K.E =0.5 ( 1/3 * 0.18 * 1.4^2 ) 1.8^2[/tex]
[tex]K.E = 0.1905 J[/tex]
Imagine you see Mars rising in the east at 6:30 pm. Six hours later what direction would you face (look) to see Mars when it is highest in the sky
Answer:
The Mars appears in the direction of South.
Explanation:
Mars is rising in the east at 6: 30 PM. The period of rotation of earth is 24 hours.
So, 6 hours is the one fourth of the period of rotation of earth. Earth rotates counter clockwise on its axis, so after 6 hours, we see the Mars in the direction of South.
A system has a pressure of 5 N/m2
If a force of 2000N is applied, what is the area that the force is applied to?
Give the units.
400 m2
Explanation:
Pressure = Force ÷ Area
5 N/m2 = 2000 N ÷ A
A = 2000 N ÷ 5
= 400 m2
If a force of 2000N is applied, the area that the force is applied to is 400 m²
What is force?The word "force" has a specific meaning in science. At this level, calling a force a push or a pull is entirely appropriate. A force is not something an object "has in it" or that it "contains."
One thing experiences a force from another. There are both living things and non-living objects in the concept of a force.
The amount of force applied to a certain region is referred to as pressure. The force per unit area is called pressure. F in this condensed version of the equation stands in for the force, which is expressed in newtons.
Given that the pressure of 5 N/m²
Force is 2000N
Pressure = Force ÷ Area
5 N/m² = 2000 N ÷ A
A = 2000 N ÷ 5 = 400 m²
Therefore, the area that the force is applied to is 400 m².
To learn more about force, refer to the link:
https://brainly.com/question/19529052
#SPJ2
If the radius of curvature of a mirror is 15m and the distance of the object from the mirror is 10m. Find the distance of the image from the mirror and the magnification of the object in meter
Answer:
Data given.
focal length (f)=15m÷2=7.5m
Distance of the object(U)=10m
Image distance (v)=?
Magnification (M)=?
Solution:
From:
1/f=1/u+1/v
1/7.5=1/10+1/v=75
then v=75m
Magnification, M=u/v
=75/10=7.5
Then magnification=7.5
Answer:
v = 30 m and m = 3
Explanation:
Given that,
The radius of curvature of the mirror, R = 15 m
Focal length, f = 7.5 m
Object distance, u = -10 m
We need to find the image distance and the magnification of the object.
Using mirror's formula,
[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(7.5)}+\dfrac{1}{(-10)}\\\\v=30\ m[/tex]
The magnification of the object in mirror is given by :
[tex]m=\dfrac{-v}{u}\\\\m=\dfrac{-30}{-10}\\\\m=3[/tex]
So, the distance of the image from the mirror and the magnification of the object are 30 m and 3 respectively.
A gas is enclosed in a confainer fitted with a piston of cross sectional area 0.10 the pressureof the gas is maintained in 8000 when hat is slowlh transferred the piston is pushed up through a distance of 4.0cm If 42j of heat is transferred to the system during expansion wht is the change im internal energy of th system
Answer:
10 Joule
Explanation:
The solution and answer are well written in the Pic above.
Please help me. Please help me.
Answer:
Make a nice powerpoint and think about the question
Explanation:
Its easier than you think :)
Find the sum. Express the answer in scientific notation. (1.54 x 10^6)+(6.15 x 10^6)
Answer:
[tex] { \tt{(1.54 \times {10}^{6}) + (6.15 \times {10}^{6}) }} \\ = { \tt{(1.54 + 6.15) \times {10}^{6} }} \\ = { \tt{7.69 \times {10}^{6} }}[/tex]
A ball of mass 0.3 kg is released from rest at a height of 8 m. How fast is it going when it hits the ground? (Gravity being equal to 9.8)
Answer:
Explanation:
Mass doesn't matter here because when something is falling, gravity plays fairly; an elephant falls at the same rate of acceleration as does a feather. What DOES matter is everything pertinent to the y-dimension of free-fall:
a = -9.8 m/s/s
v₀ = 0 (since the ball was held before it was dropped)
v = ??
Δx = -8 m (negative because the ball drops this far below the point from which it was released).
Putting all this together in one equation:
v² = v₀² + 2aΔx and filling in this equation:
v² = (0)² + 2(-9.8)(-8) and
v² = 156.8 so
v = 12.5 which rounds to 13 if you're using 2 sig figs, and rounds to 10 if you're only using 1 (which you should be, according to the way the numbers have been given in this problem)
A student walks 350 m [S], then 400 m [E20°N], and finally 550 m [N10°W]. Using the component method, find the resultant (total) displacement). Round your answer to the appropriate significant figures. Round your angle to the nearest degree.
In component form, the displacement vectors become
• 350 m [S] ==> (0, -350) m
• 400 m [E 20° N] ==> (400 cos(20°), 400 sin(20°)) m
(which I interpret to mean 20° north of east]
• 550 m [N 10° W] ==> (550 cos(100°), 550 sin(100°)) m
Then the student's total displacement is the sum of these:
(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m
≈ (280.371, 328.452) m
which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].
A sample from of water is heated with 5000 J of energy and its temperature goes up by 6 K. What is the mass of the sample?
Answer:
mass= 0.1993 kg
Explanation:
Using the formula c = Q / (mΔT)
1. 20kg of water is ejected horizontally in 10s; the speed of the water leaving the nozzle is 30m/s. Calculate the force experienced by a fire-fighter holding the hose.
Answer 60 NEWTON
Explanation:
FORCE = MASS * acceleration
acceleration= VELOCITY / TIME
acceleration= 30 / 10 = 3 M/S2
FORCE = MASS * acceleration
FORCE = 20 *3 = 60 NEWTON
The work function for silver is 4.73 eV. (a) Convert the value of the work function from electron volts to joules.
Answer:
[tex]W=7.56\times 10^{-19}\ J[/tex]
Explanation:
Given that,
The work function for silver is 4.73 eV.
We need to find the value of the work function from electron volts to joules.
We know that,
[tex]1\ eV=1.6\times 10^{-19}\ J[/tex]
For 4.73 eV,
[tex]4.73\ eV=1.6\times 10^{-19}\times 4.73\\\\=7.56\times 10^{-19}\ J[/tex]
So, the work function for silver is [tex]7.56\times 10^{-19}\ J[/tex].
A. A
B. C
C. D
D. B
HELLLP GUYS!!!
Answer:
B
Explanation:
why does a desert cooler cool better than a hot dry day
On a hot dry day, the amount of water vapour present in atmosphere is less. Thus, water present inside the desert cooler evaporates more, thereby cooling the surroundings more. Hence, a desert cooler cools better on a hot dry day.
what are MA and VR of a lever?
Explanation:
Mechanical advantage (MA) = Load/Effort. Velocity ratio (VR) = distance effort moves/ distance load moves in the same time
Question 5 of 10
A 15 N force is applied to an object, which then accelerates at 5.0 m/s2. What
is the mass of the object?
A. 10 kg
B. 20 kg
C. 75 kg
D. 3.0 kg
Hi there!
[tex]\large\boxed{D. \text{ } 3.0 kg}}[/tex]
Use the following equation to solve:
F = ma, where:
F = Force (N)
m = mass (kg)
a = acceleration (m/s²)
Thus:
15 = 5m
15/5 = m
mass = 3 kg
3. Thekinetic energy of an object of
mass, m moving with a velocity
of 5 ms-t is 25 J. What will be its
Toinetic energy when its velocity
is doubled? What will be its
Icinetic energy ruhen its velocity
is increased three times?
Answer:
1. 100 J
2. 225 J
Explanation:
We'll begin by calculating the mass of the object. This can be obtained as follow:
Velocity (v) = 5 ms¯¹
Kinetic energy (KE) = 25 J
Mass (m) =?
KE = ½mv²
25 = ½ × m × 5²
25 = ½ × m × 25
25 = 25m / 2
Cross multiply
25m = 25 × 2
25m = 50
Divide both side by 25
m = 50 / 25
m = 2 Kg
1. Determination of the kinetic energy when the velocity is doubled.
Mass (m) = 2 Kg
Velocity (v) = double the initial velocity
= 2 × 5 ms¯¹
= 10 ms¯¹
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 2 × 10²
KE = ½ × 2 × 100
KE = 100 J
2. Determination of the kinetic energy when the velocity increased three times.
Mass (m) = 2 Kg
Velocity (v) = three times the initial velocity
= 3 × 5 ms¯¹
= 15 ms¯¹
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 2 × 15²
KE = ½ × 2 × 225
KE = 225 J
. A car accelerates uniformly from 0 to 72 km/h in 11.5 seconds.
a) What is the acceleration of the car in m/s2? b) What is the position of the car by the time it reaches the velocity of 72 km/h?
Answer:
Explanation:
First job is to convert 72 km/hr to m/s:
[tex]72\frac{km}{hr}[/tex] × [tex]\frac{1000m}{1km}[/tex] × [tex]\frac{1hr}{3600s}[/tex] = 2.0 × 10¹ m/s
Now to find the acceleration which is
[tex]a=\frac{v_f-v_0}{t}[/tex] and filling in:
[tex]a=\frac{2.0*10^1-0}{11.5}=1.7\frac{m}{s^2}[/tex] That's part a. Part b want to know how far the car can get in 11.5 seconds (because that's the time it takes for the car to get to 72 km/hr). Since we know that the car can get 2.0 × 10¹ meters in 1 second, that means that in 11.5 seconds, the car can get 11.5(2.0 × 10¹) which is 230 meters.
A 5.0 kg block of ice is at rest at the top of a smooth inclined plane. The block is released and slides 2.0 m down the plane. Assuming there is no friction between the block and the surface, calculate
a) the gravitational potential energy at the top of the plane
b) the component of the weight parallel to the plane
c) the acceleration of the block
d) the velocity of the block at the bottom of the plane
e) the kinetic energy at the bottom of the plane.
Answer:
a) 98.1 Joules
b) 49.05 N × sin(θ)
c) 9.81 × sin(θ)
d) The velocity of the block at the bottom of the plane, v is approximately 6.264 m/s
e) 98.1 Joules
Explanation:
The given parameters of the block are;
The mass of the block, m = 5.0 kg
The distance down the plane the block slides, h = 2.0 m
The friction between the block and the surface = 0
Let θ represent the angle of inclination oof the plane
a) The gravitational potential energy, P.E. = m·g·h
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
∴ P.E. ≈ 5.0 kg × 9.81 m/s² × 2.0 m = 98.1 Joules
The gravitational potential energy, P.E. ≈ 98.1 Joules
b) The component of the weight of the block parallel to the plane, [tex]w_{\parallel}[/tex], is given as follows;
[tex]w_{\parallel}[/tex] = w × sin(θ) = m·g·sin(θ)
∴ [tex]w_{\parallel}[/tex] ≈ 5.0 kg × 9.81 m/s² × sin(θ) = 49.05 × sin(θ) N
The component of the weight of the block parallel to the plane, [tex]w_{\parallel}[/tex] ≈ 49.05 N × sin(θ)
c) The component of the weight along the inclined plane = The force with which the block moves along the inclined plane, therefore;
[tex]w_{\parallel}[/tex] = m·g·sin(θ) = m·a
Where a represents the acceleration of the block along the plane
Therefore, by comparison, we have;
g·sin(θ) = a
∴ a ≈ 9.81 × sin(θ)
d) Given that the motion of the block is 2.0 m downwards, we have;
The velocity of the block at the bottom of the plane, v² = 2·g·h
Therefore, v² ≈ 2 × 9.81 m/s²× 2.0 m = 39.24 m²/s²
v = √(39.24 m²/s²) ≈ 6.264 m/s
e) The kinetic energy at the bottom of the plane, K.E. = (1/2)·m·v²
∴ K.E. = (1/2) × 5.0 kg × 39.24 m²/s² = 98.1 J
define regular body in science
Answer:
All the sides and inner angles of a regular form must be equal. The sides and angles of an irregular form aren't the same. An equilateral triangle, for example, is a regular form because all of its sides and angles are the same length.
OAmalOHopeO
As a main sequence star exhausts hydrogen in its core, its surface becomes ___ and its energy output per second (luminosity) becomes ____.
Answer:
b
Explanation:
bc
an object that has lost its electrons become?
Compare the vertical and horizontal components of velocity for a ball launched at an angle on a level surface. How do the vertical and horizontal components of the ball's velocities change as it travels through the air?
Answer:
horizontal velocity remaing constan thorough out the motion but the vertical motion's velocity changes due to the gravity acting on it.
for everl 1 second the velocity decreases by 9.8 that is the gravity
Two parallel conducting plates are separated by 12.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude of the electric field strength between them, if the potential 5.6 cm from the zero volt plate is 450 V
Answer:
-8.036 kV/m
Explanation:
The electric field E = -ΔV/Δx where ΔV = change in electric potential = V - V' where V = electric potential at x = 5.6 cm = 450 V and V' = electric potential at x = 0 cm, = 0 V . So, ΔV = V - V' = 450 V - 0 V = 450 V.
Δx = distance between the 0 V plate and the 450 V point = 5.6 cm = 0.056 m
So, E = -ΔV/Δx
Substituting the values of the variables into the equation, we have
E = -ΔV/Δx
E = -450 V/0.056 m
E = -8035.7 V/m
E = -8.0357 kV/m
E ≅ -8.036 kV/m
Since the electric field between two parallel conducting plates is constant, the electric field between the plates is E = -8.036 kV/m