Answer:
Newton's First Law of Motion states that a body will stay at rest or continue its path with constant velocity unless an external force acts upon it. Newton's Second Law of Motion states that the net force that acts upon a body is equal to the mass of the body multiplied by the acceleration due to the net force.
Who won the noble prize for physics. He led a coalition of scientists who argued to control nucleus weapons
Answer:
Linus Pauling
Explanation:
Prize motivation: "for his fight against the nuclear arms race between East and West." Linus Pauling received his Nobel Prize one year later, in 1963.
An elephant has a mass of 55,000 kg. What is the elephant’s weight on Saturn (g= 10.4)?
Answer:
Explanation:
F = ma
F = 55000 kg(10.4)
F = 572000 N
That is an amazingly large elephant, more like 11 elephants
A more typical mass for a single elephant would be 5000 kg with a weight on Saturn of 52000 N
A container holds 192 g of oxygen gas at a pressure of 8.00 atm.
How much heat Q is required to increase the temperature by
145 °C at constant volume?
The quantity of heat (Q) that is required to increase the temperature at constant volume is 18,082.95 Joules.
Given the following data:
Mass of oxygen gas = 192 gPressure = 8.00 atmTemperature = 145°CScientific data:
Ideal gas constant, R = 8.314 J/molKMolar mass of oxygen gas = 32 g/mol.To determine the quantity of heat (Q) that is required to increase the temperature at constant volume:
First of all, we would find the number of moles of oxygen gas.
[tex]Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{192}{32}[/tex]
Number of moles = 6 moles.
At constant volume, the heat capacity for a diatomic gas is given by:
[tex]C_v = \frac{5}{2} R\\\\C_v =\frac{5}{2} \times 8.314[/tex]
Heat capacity = 20.785 J/molK.
At constant volume, the quantity of heat (Q) is given by this formula:
[tex]Q = nC_v \Delta T\\\\Q=6 \times 20.785 \times 145[/tex]
Quantity of heat (Q) = 18,082.95 Joules
Note: [tex]\Delta T[/tex] = 145°C = 145 K (since the difference is the same).
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59. (II) The crate shown in Fig. 4-60 lies on a plane tilted at an angle A = 25.0° to the horizontal, with Mk 0.19. (a) Determine the acceleration of the crate as it slides down the plane. (b) If the crate starts from rest 8.15 m up along the plane from its base, what will be the crate's speed when it reaches the bottom of the incline?
Explanation:
a) We need to write down first Newton's 2nd law as applied to the given system. The equations of motion for the x- and y-axes can be written as follows:
[tex]x:\;\;\;\;\;mg\sin 25° - \mu_kN = ma\;\;\;\;\;\;(1)[/tex]
[tex]y:\;\;\;\;\;N - mg\cos 25° = 0\;\;\;\;\;\;\;\;\;(2)[/tex]
From Eqn(2), we see that
[tex]N = mg\cos 25°\;\;\;\;\;\;\;(3)[/tex]
so using Eqn(3) on Eqn(1), we get
[tex]mg\sin 25° - \mu_kmg\cos 25° = ma[/tex]
Solving for the acceleration, we see that
[tex]a = g(\sin 25° - \mu_k\cos 25°)[/tex]
[tex]\;\;\;\;= 2.45\:\text{m/s}^2[/tex]
b) Now that we have the acceleration, we can now solve for the velocity of the crate at the bottom of the plane. Using the equation
[tex]v^2 = v_0^2 + 2ax[/tex]
Since the crate started from rest, [tex]v_0 = 0.[/tex] Thus our equation reduces to
[tex]v^2 = 2ax \Rightarrow v = \sqrt{2ax}[/tex]
[tex]v = \sqrt{2(2.45\:\text{m/s}^2)(8.15\:\text{m})}[/tex]
[tex]\;\;\;\;= 6.32\:\text{m/s}[/tex]
(a / xy=2x+a)does anyone can solve this??
Answer:??
Answer:
Explanation:
a / xy = 2x + a
a = 2x²y + axy
a(1 - xy) = 2x²y
a = 2x²y / (1 - xy)
a / xy = 2x + a
a = 2x²y + axy
a = y(2x² + ax)
y = a / (2x² + ax)
a / xy = 2x + a
a = 2x²y + axy
a / xy = 2x + a
0 = 2yx² + ayx - a
x = (-ay ±√((ay)² - 4(2y)(-a))) / (2(2y))
x = (-ay ±√(a²y² + 8ay)) / 4y
You are driving your car around a roundabout when you get a flat tire and you decelerate at a constant rate to a stop. The diameter of the roundabout is 100m. It takes you 20 sec to come to a complete stop. While slowing down, you continue to drive in a circle and you stop halfway around the loop. What must your speed have been before the pop?
Answer:
2.5 meters per second
Explanation:
stops half way which is 50m and if its at a constant speed of 2.5 meters multiply that by the seconds and you get 50m
The speed depends on the distance and time. The speed before the pop is 2.5 m/s.
What is the speed?The speed of an object is defined as the total distance traveled by the object within a given time interval.
Given that the diameter of the roundabout is 100m. It takes you 20 sec to come to a complete stop.
While slowing down, you continue to drive in a circle and you stop halfway around the loop. It means that the half distance is 50 m. The speed is calculated as given below.
[tex]s = \dfrac {D}{t}[/tex]
[tex]s = \dfrac {50}{20}[/tex]
[tex]s = 2.5 \;\rm m/s[/tex]
Hence we can conclude that the speed before the pop is 2.5 m/s.
To know more about the speed, follow the link given below.
https://brainly.com/question/12759408.
Which car is experiencing negative acceleration?
A. a car sitting at a red light
B. a car slowing down as it approaches a red light
C. a car driving in a circle at a constant speed
D. a car changing its speed from 0 km/hr to 10 km/hr
Answer:B
Explanation:
Answer: b
Explanation: b as it slows down and decelerates
have a difinite shape and do not easily take the shape of their containers
Qué velocidad –en m/s– tiene un móvil, que recorre 15 km en 20 minutos
(es para hoy por faaaa)
Answer:
Explanation:
15 km(1000 m / km) / (20 min(60 s/min)) = 12.5 m/s
SOMEONE PLEASE HELP MEEEEEEE
Answer:
Weathering and erosion
Explanation:
Just like our bodies, Earth's cycles tend to maintain a balance or equilibrium .
A solid, uniform sphere with a mass of 2.5 kg rolls without slipping down an incline plane starting from rest at a vertical height of 19 m. If the sphere has a radius of 0.60 m, what is the angular speed of the sphere at the bottom of the incline plane
Answer:
1/2 m v^2 + 1/2 I ω^2 = m g h conservation of energy
I = 2/5 m R^2 inertia of solid sphere
1/2 m v^2 + 1/5 m ω^2 R^2 = m g h
1/2 v^2 + 1/5 v^2 = g h
v^2 = 10 g h / 7 = 1.43 * 9.80 * 19 m^2/s^2 = 266 m^2/s^2
v = 16.3 m/s
v = R ω
ω = 16.3 / .6 = 27.2 / sec
Observe and compare the forces acting on the turtle and the cat.
a turtle and a cat on the ground looking at each other and not moving
Public Domain
The forces are balanced on both animals because they are not moving.
The forces are unbalanced on both animals because they are not moving.
The forces are balanced on the turtle but unbalanced on the cat because it is heavier.
The forces are balanced on the cat but unbalanced on the turtle because the cat is larger.
Answer:
The forces are balanced on both animals because they are not moving
More importantly than not moving is not accelerating.
Explanation:
Answer: A! the animals are doing the same thing, the force is applied to both also there not moving
Explanation:
What are the two fundamental ways energy and momentum are moved from one place to another?
The two fundamental ways energy and momentum are moved from one
place to another are through the mass and velocity.
The formula for momentum is mass multiplied with velocity.
Momentum= mass × velocity
The two factors in which momentum and energy is moved from one place to
another is however through how much and how fast the object is moving
which validates the answer.
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two factors affecting the magnitude of force of gravity betwwn 2 objects are A. mass and matter B. mass and distance C. weight and mass D. distance and weight
Answer:
B. MASS & MATTER
Explanation:
Newton's law also states that the strength of gravity between any two objects depends on two factors: the masses of the objects and the distance between them. Objects with greater mass have a stronger force of gravity between them.
Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 1.60 cm from the axis to equal 300000 g (where g is the acceleration due to gravity).
Express your answer numerically in revolutions per minute.
Answer:
Explanation:
300000(9.8) = ω²(0.0160)
ω = 13555 rad/s
13555 rad/s (60 s/min/ 2π rad/rev) = 129,445 rev/min
What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s?
Answer:
Explanation:
T = 2π[tex]\sqrt{L/g}[/tex]
(T / 2π)² = L/g
g = 4π²L/T²
g = 4π²(0.75000)/(1.7357)²
g = 9.82814766...
g = 9.8281 m/s²
What initial speed v is required if the blocks m1 =2.5 kg and m2=1.5 kg are to travel a distance d =7.0cm before coming to rest? Assume the coefficient of kinetic friction between m1 and the tabletop is ųk=0.21
Answer:
OPTRIMUM PRIDE URGH URGH URGH
Explanation:
AHHAAHAHAHAHA
How much work does an elephant do while moving a circus wagon 60 meters with a pulling force of 240N
Answer:
14,400
Explanation:
You need to do 240x60 and you would get the answer as 14,000
You are on the roof of a building 30 m above the ground. Using hand crank 3 points
you are to lift a 300 kg dresser on to the roof. How much work must to
generated to lift the dresser? *
Answer:
The work will increase the potential energy.
W = PE = mgh = 300(9.8)(30) = 88200 J
You are investigating how objects move when they are dropped from different heights. To collect your data, you drop a 1 kg weight from different heights and record the time it takes for the object to hit the ground when dropped from different heights. The data you collect is shown below:
Height dropped (m)
Time to fall (s)
1.0
0.45
2.0
0.63
3.0
0.78
4.0
0.89
Next, you plan to drop a 5 kg weight from the same heights. How will your time values in your new data table for the 5 kg weight compare to the time values in the old data table for the 1 kg weight?
The time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.
The given parameters;
Mass of the first object, m1 = 1 kg
Mass of the second object, m2 = 5 kg
The final velocity of the objects during the downward motion is calculated as follows;
[tex]v_f = v_0 + gt\\\\v_f = 0 + gt\\\\\v_f = gt[/tex]
The time of motion of the object from the given height is calculated as;
[tex]h = v_0 t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} }[/tex]
The time of motion of each object is independent of mass of the object.
Thus, the time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.
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A block of mass m = 3.0 kg is pushed a distance d = 2.0 m along a frictionless horizontal table by
a constant applied force of magnitude F= 20.0 N directed at an angle 0= 30.0° below the horizontal
as shown in Figure. Determine the work done by (a) the applied force, (b) the normal force exerted
by the table, and (d) the net force on the block.
Explanation:
We apply the definition of work by a constant force in the first three parts, but then in the fourth part we add up the answers. The total (net) work is the sum of the amounts of work done by the individual forces, and is the work done by the total (net) force. This identification is not represented by an equation in the chapter text, but is something you know by thinking about it, without relying on an equation in a list.
The definition of work by a constant force is W=FΔrcosθ.
(a) The applied force does work given by
W=FΔrcosθ=(16.0N)(2.20m)cos25.00=31.9J
(b), (c) The normal force and the weight are both at 900 to the displacement in any time interval. Both do 0 work.
(d) ∑W=31.9J+0+0=31.9J
Which statement best describes wavelength?
wavelength is the speed at which a wave travels
wavelength is the height vertically of a wave
wavelength is the length of wave per sec
wavelength is the horizontal length of one complete wave cycle
Answer:
B
Explanation:
Wave length is the height perpendicular verically of a wave
Answer:
b
Explanation:
makes the most sense to me
Which statement about diffraction is correct?
A. Sound waves bend around the corners of various obstacles.
B. Sound waves can only travel in straight lines.
C. The amplitudes of two waves combine to appear as a wave smaller than the individual waves.
D. The amplitudes of two waves combine to appear as one big wave.
Answer:
I think A
Explanation:
Which statement about diffraction is correct?
A. Sound waves bend around the corners of various obstacles.
B. Sound waves can only travel in straight lines.
C. The amplitudes of two waves combine to appear as a wave smaller than the individual waves.
D. The amplitudes of two waves combine to appear as one big wave.
Is this right for the second one
Answer:
Yes.
Explanation:
There is no movement in magnetic, chemical, electrostatic, or nuclear (potential) energy. The other options for that question can't be right. Mechanical energy is a form of kinetic, so B cannot be true. Thermal energy is also kinetic, which makes C and D incorrect as well.
An apple is falling from its tree. While falling, its acceleration changes
but not direction. Explain.
it changes ever 9.8 m/sec2.
Telescopes work by using optics with true or false
Answer: true
its making me add 20 word so here they are g g g g g g g gg g t
Answer: its FALSE
Explanation:
Telescopes work by using optics with light is false, just took the quiz.
As the amount of charge on either of two charged objects increases, the electric force between the objects decreases.
O True O False
Answer:
False.
Explanation:
provided the distance between the forces remains the same, the force will increase with increased charge, whether attractive or repulsive.
A 2 kg object being pulled across the floor with a speed of 10 m/sec is suddenly
released and slides to rest in 5 sec. What is the magnitude of the frictional force
producing this deceleration?
Answer:
The frictional force producing this deceleration would have a magnitude of [tex]4\; \rm N[/tex].
Explanation:
The velocity of this object changed by [tex]\Delta v = (-10\; \rm m\cdot s^{-1})[/tex] in [tex]\Delta t = 5\; \rm s[/tex]. The acceleration of this object would be:
[tex]\begin{aligned}a &= \frac{\Delta v}{\Delta t} \\ &= \frac{-10\; \rm m\cdot s^{-1}}{5\; \rm s} = -2\; \rm m\cdot s^{-2}\end{aligned}[/tex].
Let [tex]m[/tex] denote the mass of this object. By Newton's Second Law of Motion, the net force on this object would be:
[tex]\begin{aligned}F &= m \, a \\ &= 2\; \rm kg \times (-2\; \rm m\cdot s^{-2}) \\ &= -4\; \rm N\end{aligned}[/tex].
([tex]1\; {\rm kg \cdot m \cdot s^{-2} = 1\; {\rm N}[/tex].)
If the floor is level, friction would be the only unbalanced force on this object. Thus, the magnitude of the frictional force on this object would also be [tex]4\; {\rm N}[/tex], same as the magnitude of the net force on this object.
This is an image of a satellite traveling around Earth. Explain what are the two forces that are keeping the satelite around Earth without flying off or hitting the ground.
Answer:
One force will be gravity & inertia.
Explanation:
Bioth are combine to keep Earth in orbit around the sun, and the moon in orbit around Earth
