Answer:
true
Explanation:
Which is heavier: 30 kilogram or 300 milligrams?
Answer:
30 killograms is 85.9803
and 300 milligrams is 0.000661387 pounds
Explanation:
I really dont know physics
Answer:
3
Explanation:
I had this question before
A washing machine is being dragged with a starting force of 380.8 N to the West, but this is reduced by friction. If the mass of the washing machine is 37 kg
and the coefficient of friction is 0.54, what was the Net acceleration experienced by the washing machine?
A 10.27 m/s^2 South
B 5 m/s^2 South
C 19.01 m/s^2 South
D 7.6 m/s^2 South
Answer:
B. [tex]5\,\frac{m}{s^{2}}[/tex] South
Explanation:
The correct statement is:
A washing machine is being dragged with a starting force of 380.8 N to the South, but this is reduced by friction. If the mass of the washing machine is 37 kg and the coefficient of friction is 0.54, what was the Net acceleration experienced by the washing machine?
The net force of the washing machine moving on the horizontal ground is equal to the external force minus the friction force, where the normal force from the ground on the washing machine is equal to its weight. The free body diagram of the washing machine is presented below. The equation of equilibrium associated with the acceleration of the washing machine is:
[tex]\Sigma F = P -\mu_{k}\cdot m \cdot g = m\cdot a[/tex] (1)
Where:
[tex]P[/tex] - External force exerted on the washing machine, measured in newtons.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]a[/tex] - Net acceleration of the washing machine.
If we know that [tex]P = 380.8\,N[/tex], [tex]m = 37\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\mu_{k} = 0.54[/tex], then the net acceleration experienced by the washing machine is:
[tex]a = \frac{P}{m}-\mu_{k}\cdot g[/tex]
[tex]a = \frac{380.8\,N}{37\,kg}-(0.54)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]a = 4.996\,\frac{m}{s^{2}}[/tex]
The correct answer is B.
A hot air balloon holds 7.40 x 107 liters of gas. In order for the hot air balloon to rise into the sky, it must be heated to 120°C. Once the balloon is up in the air, the air needs to be cooled in order for the balloon to land. If the temperature of the balloon is decreased to 40.0°C, what will the new volume of the gas be?
Remember to pay close attention to the units of temperature before beginning your calculations.
Answer:
2.467×10^7 litres
Explanation:
Using the formula V1/T1 = V2/T2
V1 and V2 are the initial and final volume respectively
T1 and T2 are their respective temperatures
Given
V1 = 7.4×10^7litres
T1 = 120°C
T2 = 40°C
Required
V2
From the formula
V2 = V1T2/T1
V2 = 7.4×10^7(40)/120
V2 = (7.4×10^7)/3
V2 = 2.467×10^7 litres
Hence the new volume of the gas will be 2.467×10^7 litres
how to create or build a homemade golf club. is this possible? If so, what suggestion did you find? Do you have an original idea on how to build a homemade golf club? what types of materials and tools would you need? what skills are necessary to construct the club?
Answer:
All you need is golfing equipment make a small hole in the ground and have like something to cover the sun off of you and boom you got a golfing spot and a golf club no skills needed. Oh and you need a small flag for your gold hole but then your good and you can have fun
Explanation:
A pinball bangs against a bumper of a pinball machine with a speed of 0.46 m/s. If the ball has a mass of 0.058 kg, what is the ball's kinetic energy?
This needs to be done in 3 mins
Answer:
Okay I might be wrong and sorry if i am either its 12 or 0.402
Explanation:
* I recommend the second one cause its more logical
Be aware I might be wrong
How high can a 40 N force move a load, when 395 J of work is done? (Show work)
A 40 N force would move a load 9.875 m high when 395 J of work is done on the load.
What is work done?The total amount of energy transferred when a force is applied to move an object through some distance
The work done is the multiplication of applied force with displacement.
The SI unit of work done by an object is Joules.
IJ is the work done when a force of 1 N displaces an object by 1 m .
Work Done = Force * Displacement
For the given problem Force =40N and work done = 395 J
By using the above formula
Work Done = Force * Displacement
395 j = 40 N × height
height =9.875 m
Thus a 40 N force supported by a work done of 395 J moves the load up to 9.875 m height
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A man speeding at 40m/s decides to outrun the cops and starts to
accelerate at a rate of 2.5m/s2 for 12 seconds. What is the criminal's new
speed?
Answer:
70 m/s.
Explanation:
Given that,
Initial speed, u = 40 m/s
Acceleration = 2.5 m/s²
Time, t = 12 s
We need to find criminal's new speed. Let it is v. Using equation of motion to find it as follows :
v = u +at
Substitute all the values
v = 40 + 2.5(12)
v = 70 m/s
So, the new speed is 70 m/s.
As you enter the lab, you find two bottles labeled "concentrated ammonium phosphate." In one paragraph, using your own words, describe the steps you could take to change one of the bottles into a diluted solution, and one of the bottles into a saturated solution.
Answer:
Butane is a four member carbon chain.
The properties of butane are as follows.
Butane is less denser than water.
In the presence of air, it readily burns to form water vapour and carbon dioxide. The reaction is as follows.
It has a boiling point of .
Butane has faint petroleum like odor.
Butane does not react with water.
Butane has weak dispersion force and water has strong hydrogen bonding. The weak dispersion force is unable to break the hydrogen bonding hence, butane does not dissolve in water.
Thus, we can conclude that the chemical properties butane burns readily in air and does not react with water are true out of the given list of options.
Explanation:
The law of __________________________ says that people think of similar objects as belonging together.
Answer:
Explanation:is the law means that the police are the law
A child kicks a ball horizontally with a speed of 4.8 m/s off a deck 3.5 m off the
ground. How long will it take, in seconds, for the ball to hit the ground?
Answer:
0.73 sExplanation:
The time taken can be found by using the formula
[tex] t = \frac{d}{v} \\ [/tex]
d is the distance
v is the velocity
From the question we have
[tex]t = \frac{3.5}{4.8} \\ = 0.72916...[/tex]
We have the final answer as
0.73 sHope this helps you
A boy throws a stone vertically in the air with an initial speed of 40m/sec.At the instant the stone is thrown,a monkey at the top of a tree h m high falls freely.How long will it take for the stone to hit the monkey if they meet at a point mid way between the ground and the top of tree?
∞infinaty stones∞
because why not
A wagon having a mass of 42 kg is accelerated across a level road at 1.50 m/s2. What net force acts on the wagon horizontally?
Answer:
The net force acting on the wagon is 63 N
Explanation:
Net Force
According to the second Newton's law, the net force exerted by an external agent on an object of mass (m) is:
F = m.a
Where (a) is the acceleration of the object.
The wagon has a mass of m=42 Kg is accelerated at [tex]a=1.5\ m/s^2[/tex]. The net force is:
F = 42*1.5
F = 63 N
The net force acting on the wagon is 63 N
Which of the following is an example of light energy being changed to chemical energy?
Group of answer choices
A flashlight with rechargeable batteries
A solar panel that charges a battery
Food that is eaten to provide mechanical energy
Answer:
A solar panel
Explanation:
A machine that uses 200 w of power moves an object a distance of 15 m in 25 sec. Find
the force needed and the work done by this machine,
Answer:dummy
Explanation:
Dum dum
Using the statistical definition of entropy, what is the entropy of a system where W = 4?
A. 1.87 × 10–23 joules/kelvin
B. 1.56 × 10–23 joules/kelvin
C. 1.91 × 10–23 joules/kelvin
D. 2.07 × 10–23 joules/kelvin
Answer: S = 1.91x10^(-23) J/K
Explanation:
The statistical definition of entropy says that:
S = k*Ln(W)
where:
W is the number of macrostates, in this case W = 4.
k is a constant:
k = 1.38x10^(-23) J/K
then the entropy of this system will be:
S = (1.38x10^(-23) J/K)*ln(4) = 1.91x10^(-23) J/K
Then the correct option is C.
Answer:
C. 1.91 × 10–23 joules/kelvin
can i be marked brainiest, please? Explanation:
A force of 30 N stretches a very light ideal spring 0.73 m from equilibrium. What is the force constant (spring constant) of the spring
The forces constant (spring constant) of the spring will be 41.09 N/m.
What is spring force?The force required to extend or compress a spring by some distance scales linearly concerning that distance is known as the spring force. Its formula is;
F = kx
The given data in the problem is;
F is the spring force = 30 N
K is the spring constant= ?
x is the displacement of spring = 0.73 m
The spring constant is;
K =F/x
K=30/0.73
K=41.09 N/m
Hence the force constant (spring constant) of the spring will be 41.09 N/m.
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Which of the following examples of energy transfer CANNOT happen?
A toy car rolling down a hill increases speed and then rolls to a stop.
A bunger jumper jumps from a bridge and goes up higher with each recoil.
A child playing on a swing moves fastest at the bottom of the swing's path.
A rock dropped from the top of a bridge increases in speed until it hits the water.
The energy transfer which cannot be happened is option c , at the bottom of the swing, the kinetic energy gets converted completely into its potential energy thus, it have no speed when it reached the bottom.
What is kinetic energy ?Kinetic energy of a body is generated by virtue of its motion. Kinetic energy increases as its mass and velocity increases. Potential energy of a body is the energy stored on it when it comes to rest.
When an object is falling freely to the ground its kinetic energy is converting to the potential energy up to the point it reaches the ground. When it touches the ground, the kinetic energy converts back to potential energy.
Therefore, when the child in the swing comes down from a height its speed is increasing and have greater kinetic energy but this is converted to potential energy when it reach the bottom. Hence no speed it have at the bottom.
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A bus accelerates forward. If an apple were on the floor of the bus it would move forward.
A: True
B: False
Answer:
False
Explanation:
Since it is on the bus, it would not move forward because the outside acceleration cannot be considered.
What kind of waves are light waves and radio waves
Answer:
Electromagnetic waves
Explanation:
A train going 14m/s moves 250 m while accelerating to a stop. What is the train’s deceleration?
Answer:
-0.056 is the deceleration
In raising an object vertically at a constant speed of 2 m / s 10 watts of power is developed the weight of the object is
Answer:
5NExplanation:
Power = Workdone/Time
Since workdone = Force * distance = mgh
Power = mgh/t
Given
Power = 10watts
speed = 2m/s
Power = Weight * velocity
Substitute
10 = Weight * 2
Weight = 10/2
Weight = 5N
Hence the weight of the object is 5N
Asha walks 15 m west, then 20 m north, then 15 m east. Calculate the distance
covered by Asha.(a) (Numbering problem)_
Distance walked to the west= 15m
" " " " north= 20m
" " " " east = 15m
Total Distance= (15+ 20+ 15) m
= 50m
Asha covered 50m distance in total.
i) The energy stored in wound watch spring is
a) K.E
b) P.E
c) Heat Energy
d) Chemical Energy
Answer:
Springs when non compressed has potential energy.
A weightlifter takes 0.5s to raise a barbell with 2 free-weights from the ground to a height of 0.9m. If each free-weight has a mass of 30kg and the barbell has a mass of 20kg, calculate how much power the weightlifter produces when lifting the barbell?
Answer:
P = 1412.82 [Watt]
Explanation:
First we must calculate the total mass that makes up the barbell and the two free weights at each end of the barbell.
[tex]m=30+30+20\\m= 80 [kg][/tex]
Weight is now defined as the product of mass by gravitational acceleration.
[tex]w=m*g[/tex]
where:
m = mass = 80 [kg]
g = gravity acceleration = 9.81 [m/s²]
[tex]w=80*9.81\\w=784.8[N][/tex]
Now using the product of the weight by the distance traveled we can calculate the work.
[tex]W=w*d[/tex]
where:
W = work [J]
w = weight = 784.8 [N]
d = distance = 0.9 [m]
[tex]W=784.8*0.9\\W=706.41[J][/tex]
And power is defined as the relationship of work at a certain time. The potency is expressed by means of the following formula.
[tex]P=W/t\\P=706.41/0.5\\P=1412.82[W][/tex]
URGENTE : Un cuerpo de 8m³ flota en el agua. Si el volumen sumergido es de 5m³. Determina el modulo de la fuerza de empuje (en kn) que presenta
Answer:
La magnitud del empuje hacia arriba o fuerza hacia arriba que actúa sobre el cuerpo es 48,853 kN
Explanation:
Los parámetros dados son;
El volumen del cuerpo = 8 m.³
El volumen del cuerpo sumergido = 5 m³
Por el principio de Arquímedes, que establece que el empuje hacia arriba, [tex]F_b[/tex], actuar sobre un cuerpo en un fluido es igual al peso del fluido desplazado por el cuerpo
Matemáticamente, tenemos;
[tex]F_b[/tex] = ρ × g × V
Dónde;
ρ = La densidad del fluido = La densidad del agua = 997 kg / m³
g = La aceleración debida a la gravedad = 9,8 m / s²
V = El volumen del fluido desplazado = El volumen del cuerpo sumergido = 5 m.³
∴ [tex]F_b[/tex] = 997 kg / m³ × 9,8 m / s² × 5 m³ = 48,853 N = 48,853 kN (hacia arriba)
La magnitud del empuje hacia arriba o la fuerza hacia arriba que actúa sobre el cuerpo = 48,853 kN.
At a sports event, the car starts from iosi, in 5.0 siis acceleration is 5.
Calculate the distance travelled by ban
100 m
25 m
63 m
O bag 6.
The Score Tower hingga laglomery 4
werissa 13
Question:
At a sports event, the car starts from rest. in 5.0 s its acceleration is 5.0 m/s2. Calculate the distance travelled by car.
Answer:
62.5 metres
Explanation:
Given
[tex]u = 0[/tex] -- Initial Velocity
[tex]a = 5.0m/s^2[/tex] --- acceleration
[tex]t =5.0s[/tex] -- time
Required
Calculate the distance
This question will be solved using the following equation of motion
[tex]S = ut + \frac{1}{2}at^2[/tex]
Where S represents the distance
Substitute values for u, t and a
[tex]S = 0 * 5.0 + \frac{1}{2} * 5.0 * 5.0^2[/tex]
[tex]S = 0 + \frac{1}{2} * 5.0 * 25.0[/tex]
[tex]S = \frac{1}{2} * 5.0 * 25.0[/tex]
[tex]S = \frac{1}{2} * 125.0[/tex]
[tex]S = 62.5m[/tex]
Hence, the distance travelled is 62.5m
A 75 kg ball carrier is running to the right at 6.5 m/s. An 80 kg defender is chasing the ball carrier running at 7.0 m/s. The defender catches the ball carrier in a completely inelastic collision. What was the kinetic energy of the ball carrier/defender system BEFORE the tackle?
Answer:
3,544.375Joules
Explanation:
Kinetic energy is the energy possessed by a body by virtue of its motion, It is expressed as;
Kinetic energy = 1/2mv²
m is the mass of the body
v is the velocity
For the ball carrier;
KE = 1/2(75)(6.5)²
KE = 3168.75/2
KE = 1584.375Joules
For the defender;
KE = 1/2(80)(7)²
KE = 3920/2
KE = 1960Joules
The kinetic energy of the ball carrier/defender system BEFORE the tackle = KE for the carrier + KE for the defender
kinetic energy of the ball carrier/defender system BEFORE the tackle= 1584.375+1960 = 3,544.375Joules
pls help me with this
Answer:
A screw
Explanation:
Cuatro niños jalan de manera separada un mismo carrito de baleros con un mismo desplazamiento, pero uno con un ángulo de 45°, otro con un ángulo de 35°, otro de 20° y el ultimo con un ángulo de 0°, medidos con respecto al eje horizontal. ¿Cuál de los cuatro realiza un mayor trabajo, si la fuerza que se aplica cada uno de ellos es la misma? Selecciona el ángulo:
Answer:
El niño que aplica una fuerza a un ángulo de 0º realiza el mayor trabajo.
Explanation:
Si consideramos que las fuerzas aplicadas en el carrito de baleros es constante tanto en magnitud como en dirección, el trabajo realizado por cada niño ([tex]W[/tex]), medido en joules, puede definirse mediante la siguiente expresión:
[tex]W =F\cdot x\cdot \cos \alpha[/tex] (1)
Donde:
[tex]F[/tex] - Magnitud de la fuerza aplicada por el niño, medida en newtons.
[tex]x[/tex] - Distancia desplazada por el carrito de baleros, medida en metros.
[tex]\alpha[/tex] - Dirección de la fuerza aplicada, medida en grados sexagesimales.
De (1) tenemos que la fuerza efectiva que empuja el carro es:
[tex]F_{e} = F\cdot \cos \alpha[/tex]
La fuerza aplicada por cada niño es despejada:
[tex]F = \frac{F_{e}}{\cos \alpha}[/tex]
A mayor ángulo, tenemos un menor valor del coseno y, por ende, mayor valor de la fuerza aplicada. Ahora, el mayor trabajo realizado se realiza con el menor ángulo posible y bajo esta consideración, el niño que aplica una fuerza a un ángulo de 0º realiza el mayor trabajo.