Explanation:
work a force is measured in newton-meters (N-m)
[tex] \huge \hookrightarrow \mathfrak{ \green{Answer}}[/tex]
we know,
[tex] \boxed{ \small{workdone = force \times displacement}}[/tex]
[tex]\longmapsto \: work \: done = N \times m[/tex]
[tex]work \: \: done \: \: is \: \: measured \: \: in \: \: \boxed {\color{gold}{Nm}}[/tex]
Nm = Newton meter
state the property of plastic that allows it to become electrostatically charged
(please answers ty)
Answer:
when a plastic rod is rubbed with a duster, electrons are transferred from one material to the other. The material that gains electrons becomes negatively charged. The material that loses electrons becomes positively charged.
A cannonball is launched with an initial speed of 20 m/s. Calculate the maximum height the ball could reach.
Answer:
During its trajectory the cannon ball will have kinetic and potential energy
E = KE + PE
The maximum value of PE will occur when the cannon ball was fired straight up - for other trajectories PE will be less because of translational KE
So E = M g h = 1/2 M v^2 for maximum height
h = v^2 / (2 * g) = 400 / (2 * 9.8) = 20.4 m
define area and give its SI unit
Answer:
area is the space bounded by certain lines, the SI unit of area is m²
area of rectangle =l×w×h
square=l²
a mango fruit drop down from top of its tree which is 5m high. How long does it take to reach the ground
Answer:
What is the mass tho?
You need the mass to answer this question
Answer:
[tex]distance = ut + \frac{1}{2} g {t}^{2} \\since \: the \: fruit \: is \: from \: rest : \: initial \: velocity \: is \: 0 \: (u = 0) \\ 5 = (0 \times t) + ( \frac{1}{2} \times 9.81 \times {t}^{2} ) \\ 5 = 4.905 {t}^{2} \\ t = \sqrt{1.0194} \\ t = 1 \: second[/tex]
When a body is accelerated under water, some of the surrounding water is also accelerated. This makes the body appear to have a larger mass than it actually has. For a sphere at rest this added mass is equal to the mass of one half of the displaced water. Calculate the force necessary to accelerate a 10 kg, 300-mm-diameter sphere which is at rest under water at the acceleration rate of 10 m/s2 in the horizontal direction. Use H2O
Solution :
Mass of the sphere, m = 10 kg
Diameter, D = 300 mm = 0.3 m
Volume of the sphere is [tex]$V=\frac{4}{3} \pi \left(\frac{0.3}{2}\right)^3$[/tex]
[tex]$V=0.01414 \ m^3$[/tex]
So the volume of displaced water, [tex]$V_w=V = 0.01414 \ m^3$[/tex]
Additional mass, [tex]$m_w = \frac{1}{2} \ \rho_w\times v_w$[/tex]
[tex]$=\frac{1}{2} \times 1000 \times 0.01414$[/tex]
[tex]$=7.0686 \ kg$[/tex]
So the total mass, [tex]$M = m_w+m$[/tex]
= 7.0686 + 10
= 17.0686
Force required, F = Ma
[tex]$F=17.0686 \times 10$[/tex]
= 170.686 N
The force that is necessary to accelerate when the acceleration rate of 10 m/s2 in the horizontal direction should be 170.686 N.
How to calculate the force?Since Mass of the sphere, m = 10 kg
Diameter, D = 300 mm = 0.3 m
Now volume of the sphere is
V = 4/π (diameter/3)^3
V = 4/3π(0.3/2)^3
= 0.01414m^3
Now the additional mass is
= 1/2* 1000 * volume
= 1/2 *1000*0.01414
= 7.0686
So, the total mass is
= Additional mass + force
= 7.0686 + 10
= 17.0686
Now the force is
= 17.0686 * 10
= 170.686
Hence, The force that is necessary to accelerate when the acceleration rate of 10 m/s2 in the horizontal direction should be 170.686 N.
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The planet Mars has a mass that is 0.11 times Earth’s mass and a radius that is
0.54 times Earth’s radius. a) How much would a 60.0-kg astronaut weigh if
she were to stand on the surface of Mars?
How far from the earth must a body be along a line toward the sun so that the sun’s
gravitational pull balances the earth’s? The sun is 9.3 × 107 miles away and its mass is
3.24 × 105 Me.
Answer:
r = 1.63×10^5 mi
Explanation:
Let r = distance of object from earth
Rs = distance between earth and sun
Ms = mass of the sun
= 3.24×10^5 Me (Me = mass of earth)
At a distance R from earth, the force Fs exerted by the sun on the object is equal to the force Fe exerted by the earth on the object. Using Newton's universal law of gravitation,
Fs = Fe
GmMs/(Rs - r)^2 = GmMe/r^2
This simplifies to
Ms/(Rs - r)^2 = Me/r^2
(3.24×10^5 Me)/(Rs - r)^2 = Me/r^2
Taking the reciprocal and then its square root, this simplifies further to
Rs - r = (569.2)r ----> Rs = 570.2r
or
r = Rs/570.2 = (9.3×10^7 mi)/570.2
= 1.63×10^5 mi
True or False question! ⬇⬇ Will give BRAINLIEST!
No matter the medium, light will always bend at the same angle when it passes from one medium to a different medium.
-True
-False
Answer:
the answer can be probably true since light can be refracted
Which of the following correctly describes the order of skin layers from the inside to the outside of the body?
A. dermis, epidermis, hypodermis
B. hypodermis, dermis, epidermis
C. epidermis, hypodermis, dermis
D. hypodermis, epidermis, dermis
Answer:
it is a
Explanation:
dermis,epidermis,hypodermis
Answer:
B. hypodermis, dermis, epidermis
this is because it is inside to outside
hypodermis is deeper tissue, dermis is beneath epidermis, and epidermis is the outermost layer
A man drags a 10.0 kg bag of mulch at a constant speed applying a 22.5 N force at a 32-degree angle. What is the friction force acting on the bag?
The
friction
force acting on the bag is 4.36µ.
What is friction force?
The
force
preventing sliding against one another of solid surfaces, fluid layers, and material components is known as friction.
Given,
Force =F=22.5 n
Mass=m= 10 kg
mu=µ=?
(1N= 1kg m/s²)
Gravitational force =g= 9.81 m/s²
As we know,
F= µ*m*g
µ= m*g/F
= 10*9.81/ 22.5
µ= 4.36 .
Therefore, the friction force acting on the bag is 4.36.
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The bag is subject to a friction force of 4.36. A man applies a 22.5 N force at a 32° angle while dragging a 10.0 kilogram bag of mulch at a steady speed.
How much friction is there?Friction refers to the force that resists hard surface, fluid layering, and materials ingredients from gliding past one another. Friction lessens a machine's mechanical advantage, or, to put it another way, friction decreases the output to input ratio.
Force = 22.5 N
Mass = 10 kg
⇒ mu = µ (1N= 1kg m/s²)
⇒ g = 9.81 m/s²
As we know, F= µ × m × g
⇒ µ = (m × g)/F
⇒ (10 × 9.81)/ 22.5
⇒ µ = 4.36 .
Therefore, the friction force acting on the bag is 4.36.
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Describe the image with a object distance of 65cm and a focal length of 20cm
Answer:
Imag distance = 28.9 cm
Explanation:
Given that,
Object distance, u = -65 cm
The focal length of the convex les, f = +20 cm
We need to find the image distance. The lens formula gives the relation between focal length, object distance and image distance is given by :
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{-65}\\\\v=28.9\ cm[/tex]
So, the image distance is 28.9 cm
Three point charges q1 = +2 μC, q2 = −3 μC, and q3 = +4 μC are located at the corners of a right angle triangle as shown below. Find the magnitude and direction of the resultant force on q3.
Answer:
Explanation:
4
The magnitude and direction of the resultant force on q3 is;
F_net = 32.2 N in the direction of q3
We are given;
q1 = +2 μC = 2 × 10^(-6) C
q2 = -3 μC = -3 × 10^(-6) C
q3 = +4 μC = 4 × 10^(-6) C
Now,from the given image, charges q1 and q2 attract charge q3.
Formula for force on a charge is;
F = (kq1 * q2)/r²
Where;
k = 9 × 10^(9) N.m²/C²
Thus;
F_y = kq1*q3/(r1)²
We have;
q1 = 2 × 10^(-6) C
r1 = 5 cm = 0.05 m
q3 = 4 × 10^(-6) C
Thus;
F_y = (9 × 10^(9) × 2 × 10^(-6) × 4 × 10^(-6))/(0.05^(2))
F_y = 28.8 N
Similarly;
F_x = kq2*q3/(r2)²
r2 = 5/(tan 30)
r2 = 8.66 cm = 0.0866 m
F_x = (9 × 10^(9) × -3 × 10^(-6) × 4 × 10^(-6))/(0.0866^(2))
F_x = -14.4 N
Thus, magnitude of resultant force is;
F_net = √((28.8)² + (-14.4)²)
F_net = √1,036.8
F_net = 32.2 N
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In velocity time graph of a body, two points are given as A(40,5) and B(20,2). What is the acceleration of the body
Answer:
0.15 m/s²
Explanation:
Acceleration is rate of change of velocity-time function. We can find acceleration by finding the slope. The slope of graph can be found by:
[tex] \displaystyle{a = \dfrac{ \Delta v}{ \Delta t} \: \to \: \dfrac{v_f - v_i}{t_f- t_i}}[/tex]
Substitute in:
[tex] \displaystyle{a = \dfrac{5 - 2}{40 - 20}} \\ \\ \displaystyle{a = \dfrac{3}{20}} \\ \\ \displaystyle{a = 0.15 \: \text{m/s}^{2} }[/tex]
Therefore, the acceleration is 0.15 m/s²
4. How many excess electrons does a balloon that was rubbed on a child's hair and
now has a charge of 4.0 x 107 Coulombs?
A 30-g bullet traveling horizontally at 300 m/s strikes a 1.0-kg block which is attached to a horizontal spring with a force constant of 2000 N/m and rests on a frictionless horizontal surface. The spring is on the far side of the block and aligned with the direction of travel of the bullet. The bullet becomes embedded in the block and, as a result of the impact, and the block slides against the spring. How far is the spring compressed before it reverses its direction of travel
Answer:
Explanation:
Using law of conservation of momentum during collision , velocity after collision can be found .
common velocity V = m v / ( m + M )
= .030 x 300 / ( .030 + 1 )
V= 8.7378 m /s
The kinetic energy of the bullet and block will be stored as elastic energy
1/2 ( M + m ) V² = 1/2 k A² where k is force constant and A is maximum compression in the spring .
( M + m ) V² = k A²
1.030 x 8.7378² = 2000 x A²
A² = .0393
A= .1982 m .
= 19.82 cm
Please need help ASAP, ASAP
If force is doubled, acceleration will also be doubled.
Newton’s second law of motion, unlike the first law of motion, pertains to the behaviour of objects for which all existing forces are unbalanced. The second law of motion is more quantitative and is used extensively to calculate what happens in situations involving a force.It states that: Force is equal to the rate of change of momentum. For a constant mass, force equals mass times acceleration.The acceleration of the body is directly proportional to the net force acting on the body and inversely proportional to the mass of the body. This means that as the force acting upon an object is increased, the acceleration of the object is increased.In this given question, force is doubled. We know that force is directly proportional to acceleration. As we increase the force on an object the acceleration increases proportionally. Therefore, if you double the force you double the acceleration.
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What are the horizontal and vertical components of a cat's displacement when the cat has climbed 5 m
directly up a tree © = 90°? Sin90=1
cos90=0
The horizontal component of the cat's displacement, Dx = 0 m
The vertical component of the cat's displacement, Dy = 12 m
What are the horizontal and vertical components of a vector?The horizontal component of a vector is the value of that vector acting in the horizontal direction.
The horizontal component of a vector is given as Dx.
Dx = Dcosθ
where;
D is the resultant of the vector
θ is the angle of the vector from the horizontal
The horizontal component of the cat's displacement is:
Dx = 12 * 0
Dx = 0 m
The vertical component of a vector is the value of that vector acting in the vertical direction.
The vertical component of a vector is given as Dy.
Dy = Dsinθ
where;
D is the resultant of the vector
θ is the angle of the vector from the horizontal
The vertical component of the cat's displacement is:
Dy = 12 * 1
Dy = 12 m
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True or False? 13. All living things are made of cells 14. All cells have DNA within their nucleus 15. The cell is the basic unit of life in all organisms 16. All cells in a living thing are similar in structure or function 17. Cells must develop from other cells
Answer:
1: True
2: False
3: True
4: False
5: True
A satellite is a large distance from a planet, and the gravitational force from the planet is the only significant force exerted on the satellite. The satellite begins falling toward the planet, eventually colliding with the surface of the planet. As the satellite falls, what claims is correct about how the force that the planet exerts on the satellite Fps changes and how the force that the satellite exerts on the planet Fsp changes, if at all? What reasoning supports this claim?
Answer:
The answer is "Option A".
Explanation:
Please find the complete question in the attached file.
The strength is equal due to the third law of Newton, and [tex]Fsp = Fps = \frac{Gmsmp}{r^2}\[/tex], therefore the force increases with decreased separation r.
how to prepare a trading stock account
Answer:
Opening stock details of raw material, semi-finished goods and finished goods.
Closing stock details of raw material, semi-finished goods, and finished goods.
Total purchases of goods fewer Purchase Returns.
Total sales of goods fewer Sales Returns.
The momentum of an object is determined to be 7.2×10-3 kg⋅m/s. Express this quantity as provided or use any equivalent unit. (Note: 1 kg = 1000 g
The mass of the moving item is referred to as momentum. There is magnitude and direction to momentum. The object has a momentum of 7.2 g-m/s.
What is the momentum of an object?Momentum is a vector quantity that is created during motion by an object's mass and velocity. Thus, momentum will be expressed in units of kg-m/s.
Considering that item has motion. Likewise, 1 kilogram equals 1000 g. Therefore, the momentum will transform into units of grams when we enter the value of 1 kg.
Momentum = 7.2 x 10⁻³ 1000g-m/s
Momentum = 7.2 g-m/s.
Thus, we may deduce that the object's momentum, measured in grams per second, is 7.2 g-m/s.
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Sound is not so distinct at day due to............
Explanation:
This is because of a phenomenon called refraction which effects the direction of sound.
A gas is placed in a 5.0-mL syringe. It exerts 141 mm Hg of pressure on the inside walls of the syringe. The syringe's plunger is pressed, reducing the volume of the syringe to 2.9 mL. The cap was not removed from the syringe, so none of the gas escapes. Assuming the temperature of the gas does not change, use Boyle's law (below) to determine the pressure of the compressed gas. P_1V_1 = P_2V_2
Answer: [tex]243.10\ mm\ \text{of}\ Hg[/tex]
Explanation:
Given
The initial volume of gas [tex]V_1=5\ ml[/tex]
Initial pressure is [tex]P_1=141\ mm\ \text{of}\ Hg[/tex]
Final volume is [tex]V_2=2.9\ ml[/tex]
As it is given [tex]P_1V_1=P_2V_2[/tex]
insert the values
[tex]141\times 5=P_2\cdot 2.9\\P_2=243.10\ mm\ \text{of}\ Hg[/tex]
How important do you think temperature, radiation, and geologic activity are to the habitability in general?
Answer:
Explanation:
All 3 are critical for habitability
The monster wants to make himself known to the family in the cottage in chapters 9-12 of Frankenstein. He decides he can't do that until he masters something. What skill
must he master?
A. How to dress properly
B. Language
C. The rules of etiquette
D. Music
Answer:
c
Explanation:
Two equally charged pith ball are 3cm apart in air and repel each other with a force of 4X105N - Comput the charge on each bali-
TheThe charge applied on each ball is 2 X 10∧-9 C.
Define force ?
An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
The pitch balls repent each other with force
F= 4 X 10^-5N
Let the charge on each ball be Q.
4 X 10^-5= k(Q)(Q) / (3 X 10^-2)^2
Q^2= 4*10^-5X 9*10^-4 / 9*10^9
Q^2 = 4 X 10^-18
Q= √4 X 10^-18
Q = 2 X 10^-9 C.
Therefore, the charge on each ball is 2X10^-9C.
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Each ball has been charged with 2 X 10⁹ C. When a product is held in electrostatic energy, its properties referred to as electric discharge allows it to feel force.
What is force?A force can compel a heavily armored item to increase or modify overall velocity. A push or a pull is a simple and effective method to explain force. That's because a force has both magnitude and direction it is a dimensionless number.
The pitch balls repent each other with force
F = 4 X 10⁻⁵N
4 X 10⁻⁵ = k(Q)(Q) / (3 X 10⁻²)²
⇒ Q² = 4 × 10⁻⁵ X 9 × 10⁻⁴ / 9 × 10⁹
⇒ Q² = 4 X 10⁻¹⁸
⇒ Q = √4 X 10⁻¹⁸
⇒ Q = 2 X 10⁻⁹ C.
Therefore, the charge on each ball is 2 X 10⁻⁹C.
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What is the meaning of the density of states function?
Ans-The density of states function describes the number of states that are available in a system and is essential for determining the carrier concentrations and energy distributions of carriers within a semiconductor.
Answer:
density equal to mass upon volume
simple definition
A force F making an angle with the horizontal is acting on an object resting on the table. Which statement is true for the motion of sliding the object on the table? A. It is independent of all the forces acting on the object. B. It depends only on the forces acting along the x-axis. C. It depends only on the forces acting along the y-axis. D. It depends only on the normal force acting on the object. E. It depends only on the frictional force acting on the object.
Answer:
C
Explanation:
it depends only on the forces acting along the y- axis
Answer:
The statement that is true for the motion of sliding the object on the table is : its independent of all forces acting on the object
Explanation:
50 POINTS !!!
Three particles of mass 8 kg, 3 kg, and 7 kg are connected by rigid rods of
negligible mass lying along the y axis and are placed at 5 m, -2 m, and -6 m,
respectively as in the figure. The system rotates about the x axis with an angular
speed of 1.65 rad/s.
a) Find the moment of inertia about the x axis. Consider the
masses to be point particles.
b) Find the total rotational energy.
c) Find the linear speed of the top particle of mass 8 kg in the figure.
Answer:
a) I = 464 kg m², b) K = 631 .6 J, c) v = 8.25 m / s
Explanation:
a) the moment of inertia of point particles is
I = ∑ m_i r_i²
in this case
I = 8 5² + 3 (-2) ² + 7 (-6) ²
I = 464 kg m²
b) The kinetic energy is
K = ½ I w²
K = ½ 464 1.65²
K = 631 .6 J
c) linear and angular velocity are related
v = w r
v = 1.65 5
v = 8.25 m / s
A single movable pulley is used to lift 4000N load to the height of 50 cm. If the efficiency of the pulley is 75%, Calculate effort distance,effort applied,output work and mechanical advantage.
Since the efficiency of the pulley is 75% and the single movable pulley is used to lift 4000N load to the height of 50 cm, the effort distance, effort applied, output work and mechanical advantage are as follows:
Effort Distance = 50 cm
Effort Applied = 5333.33 N
Output work = 2000 J
Mechanical Advantage = 0.75.
When a single movable pulley is used to lift a load, the effort applied is equal to the weight of the load being lifted, and the effort distance is the distance the effort is applied.
Load (output force) = 4000 N
Height lifted = 50 cm = 0.5 m
Efficiency of the pulley = 75%
Given the above information, we can calculate the effort applied, output work, and mechanical advantage as follows:
Effort applied (input force) = Load ÷ Efficiency
= 4000 N ÷ 0.75
= 5333.33 N
Output work = Load x Height lifted
= 4000 N x 0.5 m
= 2000 J
Effort distance = Height lifted
= 50 cm
Mechanical advantage = Load ÷ Effort applied
= 4000 N ÷ 5333.33 N
= 0.75
It is worth noting that pulley efficiency is the ratio of output work to the input work. And since the efficiency of the pulley is 75% so the input work is greater than the output work, which means some of the energy is wasted in overcoming friction and other losses.
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