If the coefficient of static friction is 0.3, then the minimum force required to get it moving is equal in magnitude to the maximum static friction that can hold the body in place.
By Newton's second law,
• the net vertical force is 0, since the body doesn't move up or down, and in particular
∑ F = n - mg = n - 50 N = 0 ==> n = 50 N
where n is the magnitude of the normal force; and
• the net horizontal force is also 0, since static friction keeps the body from moving, with
∑ F = F' - f = F' - µn = F' - 0.3 (50 N) = 0 ==> F' = 15 N
where F' is the magnitude of the applied force, f is the magnitude of static friction, and µ is the friction coefficient.
what is the closest to the order of magnitude of the energy rquired to split a nucleus in to its individual protons and neutrons
Answer:
More than two million electron volts.
Explanation:
More than two million electron volts energy are needed to break or split a deuteron into a proton and a neutron. Nuclear binding energy is the type of energy that is required to split an atom's nucleus into protons and neutrons. The deuteron is an isotope of hydrogen that is composed of a proton and a neutron and it is a stable particle. Very huge amount of energy is needed for the splitting of nucleus due to the presence of heavy particles i.e. proton and neutron.
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
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.
Capacitance (C) is measured in farads, where 1 farad is equivalent to 1 coulomb per volt. Resistance (R) is measured in ohms, and 1 ohm is equal to 1 volt per ampère. The coulomb is defined as the amount of charge passing when a current of 1 ampère flows for 1 second. What is the unit of RC expressed in its simplest form?
When fundamental units are combined, they result in derived units. RC (which means Resistance Capacitance) is a derived unit and its unit in the simplest form is Coulomb per Ampere (C/A)
Given that:
Capacitance (C) [tex]\to[/tex] Farads (f)
and
[tex]1f = 1\frac CV[/tex] ----- 1 farad = 1 capacitance per volt
Resistance (R) [tex]\to[/tex] Ohms [tex]\Omega[/tex]
[tex]1 \Omega = 1\frac{V}{A}[/tex]
The unit of RC is the product of the unit of R by the unit of C.
i.e.
[tex]RC = 1f \times 1\Omega[/tex]
Substitute [tex]1f = 1\frac CV[/tex]
[tex]RC = 1\frac CV \times 1\Omega[/tex]
Substitute [tex]1 \Omega = 1\frac{V}{A}[/tex]
[tex]RC = 1\frac CV \times 1\frac VA[/tex]
Cancel out volts (V)
[tex]RC = 1\frac CA[/tex]
[tex]\frac CA[/tex] means Coulomb per Ampere
Hence, the unit of RC is Coulomb per Ampere.
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Jake launches a water balloon at an angle of 35° above the horizontal. If he sends it flying with an initial velocity of 3 m/s, how far away does Fred (who is the same height as Jake) need to be for it to hit him (assuming Jake has a good aim)?
Answer:
R = 0.86 m
Explanation:
The formula for the range of the projectile motion can be used here:
[tex]R = \frac{v^2 Sin2\theta}{g}[/tex]
where,
R = Range of projectile = distance between Jake and Fred = ?
v = launch speed = 3 m/s
θ = Launch Angle = 35°
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]R = \frac{(3\ m/s)^2Sin[(2)(35^o)]}{9.81\ m/s^2}\\\\[/tex]
R = 0.86 m
Explain how you can use the factors of production to produce a fruit juice in a production company
Answer:
The factors of production include Land, Labour, Capital and Enterpreneurship
Explanation:
The fruit could be apple, orange , pineapple etc which are usually grown on land . They are tended to by people to ensure there is maximum yield. These people provide the required labour needed.
The cost of planting and payment of workers usually comes from the capital which is often used in running the business by the owner which makes certain decisions to ensure the fruit company is in place. All these factors work hand in hand to ensure production of fruit in a production company is possible.
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 gas at a pressure p is compressed to half it original volume and twice its original temperature. The new pressure is
Answer:
4p
Explanation:
If you halve the volume the pressure will double as they are inversely proportional. If you double the temperature the particles have double the kinetic energy so the pressure will double again.
So:
p×2×2 = 4p
Answer:
P V = n R T
P2 V2 / (P1 V1) = T2 / T1
P2 = (T2 / T1) (V1 / V2) P1 = 2 * 2 = 4
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.....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].
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.
A vehicle starts from rest with a uniform acceleration of 2m/s2.Find the final velocity of the vehicle after covering a distance of 400m
Answer:
40
Explanation:
vi=0
a=2
vf=?
d=400
vf^2=vi^2+2ad
vf^2=0+2×2×400=1600
vf=√1600=40m/s
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)
suppose the pilot starting again from rest opens the throttle part.way at constant acceleration the airboat then covers a distance of 60.0m in 10.0s find the net force action on the boat
Answer:
Acceleration is 1.2 m/s^2.
Explanation:
initial velocity, u = 0
distance, d = 60 m
time, t = 10 s
Let the acceleration is a.
use second equation of motion
[tex]s= u t +0.5 at^2\\\\60 = 0 + 0.5 \times a \times 10\times 10\\\\a = 1.2 m/s^2[/tex]
Now according to the Newton's second law
Force = mass x acceleration
Let the mass is m.
F = m x 1.2 = 1.2 m Newton
If displacement has a magnitude AND direction is it a scalar or vector quantity?
scalar
vector
Explanation:
Obviously vector quantity
A boy is moving a stone in a horizontal circle by mean of a thread attached to ut. The length if the threads us 1 m and the weight of the stone is 0.02 kg . The maximum tension of the thread can withstand us 1 ×10 ^-4 N. The boy is gradually increasing the speed if rotation. At what speed if the stone does the thread break ?
Answer:
Explanation:
This is a centripetal force problem where tension supplies the centripetal force needed to keep the stone moving in a circular manner. That formula for that is
[tex]F_c=T=\frac{mv^2}{r}[/tex] and filling in what we know:
[tex]1*10^{-4}=\frac{.02v^2}{1}[/tex] and solve that for v:
[tex]v=\sqrt{\frac{1*10^{-4}}{.02} }[/tex] which gives us a velocity of .07 m/s. Anything greater than this will break the string.
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
Having the correct posture and the degree or severity of braking, acceleration and steering inputs have a direct result on the ability of a motorist to effectively address changes in __________?
Answer: Vehicle Balance
Explanation:
Vehicle balance simply means how the weight of a vehicle is distributed across its tires which connects to the road
It should be noted that there'll be a shift in the balance of a vehicle when the braking, acceleration, or turning bring about a scenario in which the weight of th vehicle moves from one area to another.
6. traction a. friction between a tire and the road.b. pushes a moving object out of a curve and into a straight linec. the ability of a car to hold a straight lined. a road higher in the center than on either edge 7. kinetic energy a.a force that keeps objects moving in a straightb. reduces the effects of inertiac. energy of motiond. a road higher in the center than on either edge 8. force of impact a. a force that keeps objects moving in a straight line.b. a road higher in the center than on either edgec. the force with which a moving vehicle hits another objectd. perception distance, reaction distance, and braking distance
Answer:
6. a. friction between a tire and the road
7. c. energy of motion
8. c. the force with which a moving vehicle hits another object
Explanation:
6. As a car moves along the road, the tires push back against the ground. As tires push back against the ground, the road exerts and opposing force to the motion of the tires. This opposing force is the friction between the tires and the road. This opposing force between the tires and the rad is called traction.
So, the answer is a
7. As an object moves, it has energy. This energy due to its motion is called kinetic energy.
So, the answer is c
8. When a moving vehicle hits another object, it exerts a force on the object. The process of the vehicle hitting the other object is called impact and the force exerted on the object is called the force of impact.
So, the answer is c.
A. A
B. C
C. D
D. B
HELLLP GUYS!!!
Answer:
B
Explanation:
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]
A neutral object must have _______________ A. An equal amount of positive and negative charges B. no charges present at all C. more positive than negative charges D. more negative than positive
Answer:
A
Explanation:
Consider a question you may not have considered before. Suppose you have a mole of sodium in a container that contains nothing that it will react with. A mole of anything is 6.02 * 10^23 (in this case atoms).
Suppose that every one of those atoms has contributed 1 electron to something.
Do you think it would be safe to touch the container knowing that there are 6.02 * 10^23 positive charges all eager to get another electron, because they don't like repelling each other.
Safe or not? I'll give you a hint. A lightning bolt does not contain anywhere near 6.02*10^23 charges. No where near.
So -- since you never get a shock from just touching anything, There must not be electrons or ions present.
The answer is A
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)
. 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.
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 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.
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what is acceleration
[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]
[tex]\sf What \: is \: acceleration? \\ \\ \sf The \: rate \: of \: change \: of \: velocity \: of \: an \\ \sf object \: with \: respect \: to \: time \\ \sf is \: known \: as \: acceleration. [/tex]
What does the sun use for nuclear fusion?
Answer:
The sun uses hydrogen for nuclear fusion.
Answer:
Hydrogen
Explanation:
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
Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m each. If Margy is walking at 1.4 m/s and accelerates at 0.20 m/s2 during
Complete question is;
Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m each. If Margy is walking at 1.4 m/s and accelerates at 0.20 m/s² during one of the running portions, what is her final velocity at the end of the 100.0 m? Round your answer to the nearest tenth.
Answer:
6.5 m/s
Explanation:
We are told that she is walking at 1.4 m/s and accelerates at 0.20 m/s².
Thus;
Initial velocity; u = 1.4 m/s
Acceleration; a = 0.2 m/s²
Distance; s = 100 m
From Newton's equation of motion, we know that;
v² = u² + 2as
Where v is final velocity.
Thus;
v² = 1.4² + 2(0.2 × 100)
v² = 41.96
v = √41.96
v ≈ 6.5 m/s
