ANSWER
[tex]A\text{.}50N[/tex]EXPLANATION
Parameters given:
Mass of crate, m = 20 kg
Force of push, F = 200 N
Acceleration of crate, a = 7.5 m/s²
To find the force of friction, we have to apply the net force acting on a body. The sum of all forces acting on the crate is equal to the product of its mass and its acceleration.
Therefore, we have:
[tex]F_{\text{net}}=F-F_f[/tex]where Ff = force of friction
Therefore, we have that:
[tex]\begin{gathered} m\cdot a=200-F_f \\ 20\cdot7.5=200-F_f \\ 150=200-F_f \\ \Rightarrow F_f=200-150 \\ F_f=50N \end{gathered}[/tex]That is the force of friction. (Option A)
A large truck with mass 30 m crashes into a
small sedan with mass m. If the truck exerts a
force F on the sedan, what force will the
sedan exert on the truck?
A. F
B. 2F
C. F/30
D. 30F
Option A. F . The force will be the same as the large truck exerts on the small sedan but in opposite direction. according to Newton's third law of motion.
In physics, a force is an influence that can exchange the motion of an item. A pressure can cause an item with mass to trade its velocity, i.e., to boost up. pressure can also be defined intuitively as a push or a pull. A pressure has both significance and course, making it a vector quantity.
The word 'Force' has a specific which means. At this stage, it is completely suitable to describe a force as a push or a pull. Pressure isn't something that an object consists of or 'has in it. Pressure is exerted on one item with the aid of every other. The idea of a force isn't always restrained to living matters or non-living things.
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Find the answer to the problem
Box accelerates at a rate of 3.46 m/s.
How does acceleration work?Acceleration is the rate at which velocity changes over time in both the direction and the speed. Anything is said to have been accelerated when it goes faster or slower in a straight line. Even travel on a circle accelerates steadily since the direction is continuously changing.
What metrics are used to quantify acceleration?meters per second per second (m/s2).
Formula,
F NET = m, F NET = F1 + F2, etc.
F NET equals 33 + 3 = 36 .
Then,
a = F NET/m, where F NET = m. a
a = 36 /10.4
a = 3.46 m/s.
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A ball is thrown up in the air, it reaches its maximum height and falls back. It’s acceleration is:A - same up and down but zero at the topB- less on the way up than on the way downC- the same ally all points of motionD- less on the way down than on the way up
When the ball is thrown up in the air, it is called a Freefall motion.
In freefall motion, the acceleration remains constant at all times because the acceleration due to gravity is always constant.
The acceleration due to gravity is 9.81 m/s^2
The ball has zero velocity (not acceleration) at the maximum height.
Therefore, the correct answer is option C
"Its acceleration is the same at all points of motion".
Change 72km/h into m/s
Answer:
20 m/s
Explanation:
72 km/hr * 1 hr/ 3600sec * 1000m/km = 20 m/s
Solving For Temperature (T)
7. How many degrees would the temperature of a 450 g piece of iron increase if 7600 J of
energy are applied to it? (The specific heat of iron is0.4494 J/g x °C)
Temperature is a unit used to represent how hot or cold something is. It can be stated using the Celsius or Fahrenheit scales, among others. Temperature shows which way heat energy will naturally flow, i.e., from a hotter (body with a higher temperature) to a colder (body with a lower temperature) (one at a lower temperature)
Additionally, we are provided the iron's particular heat as well as the iron skillet's mask.
We must determine the required quantity of heat based on this information. It will be Q = c x mass x delta P using this method.
That is the temperature change. Therefore, we can replace the value here. The c represents the iron's particular temperature. That comes to 0.4499.
The supplied mass is 5.07, and the resulting temperature change will be 250 - 25 °C. If we can solve this, we can determine what Q is worth.
Alternately, we might state that the required amount of heat is 512.19 Joules.
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Two identical metallic spheres A and
B, each carrying a charge of q C, are fixed.
They repel each other with a force of
2x10-5 N. Another identical uncharged
sphere C is made to touch A, moved to
touch B, and placed halfway between A
and B. What is then the electric charge
and the electric force (in N) on each
sphere in terms of q?
The electric charge and the electric force on each sphere in terms of q is 2
The Electric Charge and Electric ForceThe ability of particles or objects to attract or repel one another without coming into contact is known as electric charge. Incompatiblely charged particles are drawn to one another. Like-charged particles repel one another. Electric force is the name given to the force of attraction or repulsion.
Two metallic spheres, a and b, with a charge of half and a force of two tens of five
F1=k(1/2)(1/2)/r2
Equation F1=k/4r
A and C are two metallic spheres, respectively, so
Equation F2=k/2rsquare
equ1/equ2
2×10^-5/F2=k/4r^2×2r^2/k
F2=4×10^-5
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A vector A has components Ax = −5.50 m and Ay = 7.50 m. Find the magnitude (in m) and the direction (in degrees counterclockwise from the +x-axis) of the vector.magnitude mdirection ° counterclockwise from the +x-axis
The x-component of vector A is Ax = -5.50 m
The y-component of vector A is Ay = 7.50 m
We are asked to find the magnitude and the direction of the vector (counterclockwise from the +x-axis)
The magnitude of vector A is given by
[tex]\begin{gathered} |A|=\sqrt[]{A_x+A_y} \\ |A|=\sqrt[]{(-5.50)^2+(7.50)^2_{}} \\ |A|=\sqrt[]{30.25+56.25} \\ |A|=\sqrt[]{86.50} \\ |A|=9.3\; m \end{gathered}[/tex]So, the magnitude of vector A is 9.3 m
First, let us draw a diagram to better understand the problem.
The direction of vector A is given by
[tex]\begin{gathered} \tan \theta=\frac{A_y}{A_x} \\ \tan \theta=\frac{7.50}{-5.50} \\ \theta=\tan ^{-1}(\frac{7.50}{-5.50}) \\ \theta=-53.75\degree \end{gathered}[/tex]From the diagram, notice that θ lies in the second quadrant.
So, the counterclockwise direction from the positive x-axis is given by
[tex]180\degree-53.75\degree=126.25\degree[/tex]So, the direction is 126.25° counterclockwise from the +x-axis.
A storage tank 20 m high is filled with pure water. (Assume the tank is open and exposed to the atmosphere at the top.)
(a) Find the gauge pressure at the bottom of the tank.
(b) Calculate the magnitude of the net force that acts on a square access hatch at the bottom of the tank that measures 0.6 m by 0.6 m.
PLEASE HELP ASAP!!!
A. The overpressure at the bottom of the tank is 297500 N/m².
B. The magnitude of the net force acting on the square access hatch at the bottom of the tank is 107100 N.
Multiplying the flow rate by the minimum operating time of the pump gives the descending capacity. As a rule of thumb for a minimum run time, a pump operating at 10 gallons per minute GPM or less should produce a run time of 1 gallon per minute. For example, 10 GPM flow rate x 1 = 10 gallon drawdown capacity.
This means that the gauge pressure is equal to the absolute pressure minus the atmospheric ambient pressure. If the absolute pressure is greater than the atmospheric pressure it is called positive overpressure. If the absolute pressure is lower than atmospheric pressure it is called negative overpressure. The hydrostatic pressure at any height below water is calculated by P = hdg. where h is the height below open water.
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A wave is represented by the equation y=0.5sin 0.4π(x-60t) where the distance x is measured in cm and time(t) in seconds. what is the wave length of the wave,amplitude and frequency
The amplitude = 0.5 m
The wavelength = 5 m
The frequency = 12 Hz
Explanation:The given wave function is:
y=0.5sin 0.4π(x-60t)
This can be rewritten as:
y=0.5sin (0.4πx-24πt)................(1)
Comparing the equation above with the general wave function:
y = A sin(kx - wt).................(2)
where A represents the amplitude
k represents the wave number
w represents the angular frequency
By comparing equations (1) and (2):
the amplitude, A = 0.5 m
k = 0.4π
Also k = 2π/λ
0.4π = 2π/λ
λ = 2π/0.4π
λ = 5 m
The wavelength = 5 m
w = 24π
Also, w = 2πf
Therefore, 2πf = 24π
f = 24π/2π
f = 12 Hz
The frequency = 12 Hz
L../IN10. An object of mass 60.986 kg is sliding on a horizontalsurface with a uniform speed. The coefficientof kinetic fiction of the surfaces is 0.15. Calculatethe force of friction exerted by the surface on the object. (1 point)A. O 101.798 NB. O114.077 NC. 050.2 ND. 145.451 NE. O 89.649 N
Given:
The mass of the object is,
[tex]m=60.986\text{ kg}[/tex]The coefficient of friction is,
[tex]\mu=0.15[/tex]The object is sliding on the horizontal surface at a uniform speed.
To find:
the force of friction exerted by the surface on the object
Explanation:
If we draw the free body diagram of the object we see,
The normal reaction is R, the applied force is F, and the frictional force isf.
As the object is continuing its constant speed, we can say the object is in equilibrium.
So,
[tex]R=mg[/tex]and
[tex]\begin{gathered} f=F=\mu R \\ f=\mu mg \\ f=0.15\times60.986\times9.8 \\ f=89.649\text{ N} \end{gathered}[/tex]Hence, the friction force exerted by the surface is 89.649 N.
A puck of mass m = 0.065 kg is moving in a circle on a horizontal frictionless surface. It is held in its path by a massless string of length L = 0.65 m. The puck makes one revolution every t = 0.55 s. what is the magnitude of the tension in the string while the pack is revolving?if the string breaks how fast will the pack move away?
The centripetal acceleration of an object in circular motion with tangential velocity v if the radius of the trajectory is r, is:
[tex]a_c=\frac{v^2}{r}[/tex]On the other hand, if the period of one revolution is T, then the distance traveled is 2πr (which is the circumference of the trajectory) and the tangential velocity is:
[tex]v=\frac{2\pi r}{T}[/tex]In this case, the radius of the trajectory is the length of the string, 0.65m, and the period is T=0.55s. Substitute these data into the formula to find v:
[tex]v=\frac{2\pi(0.65m)}{0.55s}=7.4256\frac{m}{s}[/tex]Plug in v=7.4256m/s and r=0.65m to find the centripetal acceleration of the puck:
[tex]a_c=\frac{(7.4256\frac{m}{s})^2}{0.65m}=84.83\frac{m}{s^2}[/tex]According to Newton's Second Law of Motion, there must be a net force acting on the object responsible for the centripetal acceleration. That force F is the tension of the string. Then:
[tex]F=m\cdot a_c[/tex]Substitute the value for a and m=0.065kg to find the tension of the string:
[tex]\begin{gathered} F=0.065\operatorname{kg}\times84.83\frac{m}{s^2} \\ =5.51N \end{gathered}[/tex]If the string breaks, the puck will move as fast as its tangential velocity, which is 7.4 m/s.
Therefore, the answers are:
Tension in the string: 5.51N
Speed of the puck if the string breaks: 7.43 m/s
Light normally travels in straight lines. What term is used to describe the bending of light when it crosses from one medium to another?
The term refraction is used to describe the bending of light when it crosses from one medium to another. The bending of light happens due to the different densities of the two substances.
A 150-pound person uses 6.2 calories per minute when walking at a speed of 4 mph. How long must a person walk at this speed to use at least 230 calories?
The time taken by person to use 6.2 calories is 1 min.
Therefore, the time taken to use 230 calories can be calculated as,
[tex]\begin{gathered} t=\frac{(230\text{ calories)(1 min)}}{(6.2\text{ calories)}} \\ \approx37.1\text{ min} \end{gathered}[/tex]Thus, the time taken to use 230 calories is 37.1 min.
2. Name the force or forces that cause these objects to
experience changes in motion. Neglect air resistance. Rw
(a) A ball falls toward the ground.
(b) A person accelerates up in an elevator.
(c) A car gradually slows down while approaching a red
light.
a) A ball falls toward the ground:
The force of gravity is acting on the falling ball. On the surface of the earth, the direction of this force is always downward, towards the ground. It pulls on all objects with mass.
b)A person accelerates up in an elevator:
On the person in the elevator, there are two forces:
1) Normal Force upward, exerted by the floor on the person perpendicular to the surface (“normal” means perpendicular).
2) Gravitational Force downward, exerted by the Earth on the person. This is also called the person’s “weight”.
c) A car gradually slows down while approaching a red light.:
Friction may slow the car down as it moves along the road, but it also is the force that enables the car to move forward at all. It is the friction force that keeps the tires from sliding on the road. By the same token, it is friction that makes the car come to a stop when the brakes are applied.
What is Force ?
An object's push or pull is seen as exerting a force. The interaction of the objects produces push and pull. You can also use words like stretch and squeeze to describe force. The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude. The application of force is the location at which force is applied, and the direction in which the force is applied is known as the direction of the force. A spring balance can be used to calculate the Force. Newton is the SI unit of force (N).
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An automobile with an initial speed of 4.30 m/s accelerates uniformly at the rate of 3.0 m/s/s. Find the final speed and the displacement after 5.0s
The final speed of the automobile with an initial speed of 4.30 m/s that rates uniformly at the rate of 3.0 m/s/s for 5 s is 19.3 m / s and its displacement will be 59 m
v = u + at
v = Final velocity
u = Initial velocity
a = Acceleration
t = Time
u = 4.3 m / s
a = 3 m / s / s
t = 5 s
v = 4.3 + ( 3 * 5 )
v = 19.3 m / s
s = ut + 1 / 2 at²
s = Displacement
s = ( 4.3 * 5 ) + ( 1 / 2 *3 * 5 * 5 )
s = 21.5 + 37.5
s = 59 m
Therefore,
Final speed = 19.3 m / sDisplacement = 59 mTo know more about Displacement
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Place the red, 100 g ball at the 60 cm mark on the ramp. How far did the box move? a90 cm b0cm cno more that 100 cm
Answer:
place it on the 50 cm mark
A 1,116 kg space vehicle is traveling along a straight-line at a constant speed of 800 m/s. What is the magnitude of the net force on the space vehicle?
The magnitude of the net force on this space vehicle moving at constant speed is zero (0) Newton.
What is a net force?A net force can be defined as the vector sum of all the forces that are acting on a physical object or body. This ultimately implies that, a net force is a single force that substitutes the effect of all the forces acting on a physical object or body.
Generally speaking, the amount of net force that keeps a physical object or body moving at constant speed and in the same direction is always equal to zero (0) Newton in accordance with Newton's First Law of Motion.
In this context, we can reasonably infer and logically deduce that the net force on a physical object or body would be equal to zero (0) Newton when it moves at constant speed.
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Consider two homogeneous bodies of the same volume is it necessary for them to have same mass why
No, two homogeneous bodies of the same volume do have not the same mass.
How do two bodies of the same volume have the different same mass?
No, two objects don't need to have same volume to have the same mass because the density of the objects they are made up of can be different from one another. Mass and volume are independent, two objects with the same volume can have different amounts of masses. So that's why the objects can have different densities. Density is defined as the mass of the substance divided by its volume.
Density = mass /volume
Mass is defined as the amount of matter that is present in an object, while on the other hand, volume is the space occupied by an object. For example, a bowling ball and a basketball have the same volume but the bowling ball has more mass.
So we can conclude that two homogeneous bodies of the same volume do have not the same mass because of the difference in their volume.
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If Mercury's average orbital distance from the sun is 0.312 AU and k for our sun is 1.00 AU3/yr2, then what is Mercury's orbital period?
Take into account that by the Kepler's Third Law, you can use the following formula:
[tex]k^{}=\frac{a^3}{T^2}[/tex]where,
T: period of Mercury
a: average distance from Mercuty to Sun = 0.312AU
k: Kepler's constant = 1.00 AU^3/yr^2
Then, by replacing the previous values of the parameters, you obtain for T:
[tex]T=\sqrt[]{\frac{a^3}{k}}=\sqrt[]{\frac{(0.312AU)^3}{1.00\frac{AU^3}{yr^2}}}\approx0.56yr[/tex]Hence, the Mercury's orbital period is approximately 0.56 yr
The position of a particle is given by the vector < 3.3t2 , 3.2 , - 2.4t >. What is the magnitude of the acceleration at t=0?
The position of a particle is given by the vector < 3.3t² , 3.2 , - 2.4t >. The magnitude of the acceleration at t=0 is 3.3 m/sec².
What is acceleration?The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration.
Given in the question The position of a particle is given by the vector
< 3.3t² , 3.2 , - 2.4t >. The magnitude of the acceleration at t=0.
Differentiating position vector for two times we get acceleration vector,
so acceleration is 3.3 m/sec²
.The position of a particle is given by the vector < 3.3t² , 3.2 , - 2.4t >. The magnitude of the acceleration at t=0 is 3.3 m/sec²
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Find the answer to the problem
The box is moving at a 2.06 m/s rate.
What causes acceleration?Acceleration is the rate at which velocity changes over time in both the direction and the speed. Anything is said to have been accelerated when it goes faster or slower in a straight line. Even travel on a circle accelerates steadily since the direction is continuously changing.
What metrics are used to quantify acceleration?Acceleration is measured in meters per second per second (m/s2).
Formulas for
F NET are F NET = m and F NET = F1 + F2.
3 + 3 + 7 Equals 13 when F NET is used.
F NET = m * a then,
where a = F NET/m and
a = 13N / 6.3 kg and 2.06m/s respectively.
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A particle changed q= 3,10x10^-6 C stays still on point P, a second particle with the same value q and a mass of 1,80x10^-2 kg is initially in repose at a distance of r1 = 9,00 cm from point P. This is second particle is released and repelled by the first. Determine the velocity in which this second particle finds itself at a distance r2 = 25,0 cm from point P.
note 1: from what i know i have to use the potential energy formula (U= mV^2/2), and possible calculate the electric force ? but from there i get lost
note 2: sorry if this isn't grammatically correct, i translated it from spanish
The velocity in which this second particle finds itself at a distance r2 = 25,0 cm from point P is 17.22 m/s.
What is the electric force between the two particles?The electric force between the two particles is calculated from Coulomb's law of electrostatic force.
F = kq₁q₂/r²
where;
k is Coulomb's constantq₁ is magnitude of first chargeq₂ is magnitude of second charger is the distance between the two chargesF = (9 x 10⁹ x 3.10 x 10⁻⁶ x 3.10 x 10⁻⁶) / (0.09 m)²
F = 10.68 N
The velocity in which this second particle finds itself at a distance r2 = 25,0 cm from point P is calculated by applying the principle of conservation of energy.
W = ΔK.E
where;
W is work doneΔK.E is kinetic energy of the particleFr = ¹/₂mv²
2Fr = mv²
v² = 2Fr/m
v = √(2Fr/m)
v = √(2 x 10.68 x 0.25 / 1.8 x 10⁻²)
v = 17.22 m/s
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9. How much change in temperature would the addition of 35 000 Joules of heat have ona
538.0 gram sample of copper? (Look up specific heat of copper)
The change in temperature would be 168.97 °C
Q = cm∆t
Where
Q = Amount of heat energy
c = specific heat capacity of copper metal = 0.385 J/g°C
m = mass of copper
∆t = change in temperature
From the question,
Given:
m = 538 g
c = 0.385 J/g°C
Q = 35 000 Joules
Q = cm∆t
35000 = 0.385 x 538 x ∆t
∆t = 168.97 °C
Hence, change in temperature would be 168.97 °C
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With what angular speed, in rad/s, does the Earth rotate about its axis?
We know that it takes the earth approcimately 24 hours to complete one rotations about its axis.
This means that we can calcuate the angular speed using the following formula:
[tex]\omega=\frac{\theta}{t}[/tex]Where:
[tex]\begin{gathered} \theta=\text{ angle in radians} \\ t=\text{ time} \end{gathered}[/tex]We will convert the time from hours to second using the following conversion factor:
[tex]1h=3600s[/tex]Now, we mutiply noth sides by 24 and we get:
[tex][/tex]An ocean fishing boat is drifting just above a school oftuna on a foggy day. Without warning, an engine backfireoccurs on another boat at a distance of d = 1.55 km(Figure 1). The speed of sound in air is 343 m/s, and insea water is 1560 m/s.How much time elapsed before the backfire is heard by the fishermen? Assume that the fishermen hears the sound that travels only through air.
For the fisher man use speed of sound in air as v = 343
d= vt
t= d/v
where:
t= time
d= distance = 1.55 km = 1550 m
v= speed= 343 m/s
t= 1550m / 343 m/s = 4.52 s
Biggs and Smalls play tug-of-war with a 12-meter rope in P.E. class. Both are in their socks on a waxed gym floor, which is nearly frictionless. After a brief time, they meet. If Biggs has twice the mass of Smalls, how far has Biggs moved?
After a brief time, Biggs and smalls would meet at 4m from Biggs' side if the rope is 12 meter long and the floor is frictionless.
[tex]X_{cm}[/tex] = ( m1 x1 + m2 x2 ) / ( m1 + m2 )
[tex]X_{cm}[/tex] = Center of mass
After a brief time, Biggs and smalls would meet at the point of centre of mass. Let Biggs be 1 and smalls be 2 and the distances be measured from Biggs' side assuming that Biggs is standing on the left side.
m1 = 2 * m2
x1 = 0 m
x2 = 12 m
[tex]X_{cm}[/tex] = [ ( 2 * m2 * 0 ) + ( m2 * 12 ) ] / [ ( 2 * m2 ) + m2 ]
[tex]X_{cm}[/tex] = 12 m2 / 3 m2
[tex]X_{cm}[/tex] = 4 m
Center of mass is the average position of all the objects on a system, weighed according to the mass of individual objects. It is a position relative to an object.
Therefore, after a brief time, Biggs and smalls would meet at 4 m from Biggs' side.
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Draw a supply graph for pizzas. Show what would happen to the supply of pizzas if two more pizzerias open up in town. Identify the determinant of supply that caused the shift.
Pizza delivery when two more pizzerias in the city start changing their product prices.
All other things being equal an increase in demand will increase the equilibrium price. You will have more to deliver. A decrease in demand leads to a decrease in the equilibrium price. Less quantity to deliver. An increasing change in supply shifts the supply curve to the right, and a decreasing change in supply shifts the supply curve to the left.
Essentially it is an increase or decrease in quantity offered with a higher or lower offer price. As the price of pizza slices rises consumers are more likely to purchase one of these alternatives as a replacement for pizza. This reduces the amount of pizza requested. Higher prices tend to increase supply as more profits are made available.
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A car accelerates from rest at -1.60 m/s2. (Include the sign in your answers.)(a) What is the velocity at the end of 5.6 s? m/s(b) What is the displacement after 5.6 s?
Given data
*The given initial velocity of the car is u = 0 m/s
*The given acceleration of the car is
[tex]undefined[/tex]Science AssignmentQuestion: Give 20 example each on scalar & vector quantities...
Scalar: Temperature, mass, distance, time, speed, volume, area, voltage, current, resistance, work, density, energy, amount of substance, luminous intensity, area density, chemical potential, electric current density, capacitance, inductance.
Vector: Displacement, velocity, acceleration, force, position, eletric field, magnetic field, linear momentum, angular momentum, torque, angular velocity, angular acceleration, electric field density, magnetic field density, roll, pitch, yaw, surge, sway, heave.
please help me with this homework question I really need help I have till tomorrow to do it
Relate scientific theories to the law of parsimony
In science, the law of parsimony is used to select between related scientific theories to those that contain a minor number of assumptions.
What is the law of parsimony?The law of parsimony is a well-sustained body of information that shows how the theories that contain a minor number of assumptions before experimentation are generally the correct ones.
Therefore, with this information, we can conclude that the law of parsimony can be used to select scientific theories and this approach is generally accurate to obtain the correct theory to be used due to assumptions that may hamper the scientific results.
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