Answer:
30m³
Explanation:
Calculation for the volume the balloon would have
First step is to calculate the Constant
Constant=30pka*5m³
Constant=150m³
Now let calculate the Volume using this formula
Balloon Volume=Constant/ Pressure
Let plug in the formula
Balloon Volume=150m³/5kpa
Balloon Volume=30m³
Therefore the volume that the balloon would have when the pressure inside the ballon has fallen to 5kpa is 30m³
it serves as the fuctional unit of the nervous system
Answer:
i would say the neuron
HELLLPPPPPPPPPPPPPPPPPP
Answer:
If you going to ask a question ask it accurately.
Explanation:
A jet aircraft is traveling at 262 m/s in horizontal flight. The engine takes in air at a rate of 85.9 kg/s and burns fuel at a rate of 3.92 kg/s. The exhaust gases are ejected at 921 m/s relative to the aircraft. Find the thrust of the jet engine. Answer in units of N Find the delivered power. Answer in units of W.
Answer:
[tex]F_T=60132.52N[/tex]
[tex]P=15814852.76W[/tex]
Explanation:
From the question we are told that
Velocity of aircraft [tex]V=263m/s[/tex]
Engine air intake rate [tex]\triangle M_a=85.9kg/s[/tex]
Fuel burn rate [tex]\triangle M_f =3.92kg/s[/tex]
Velocity of exhaust gas [tex]V_e =921m/s[/tex]
Generally the Mass change rate of Rocket is mathematically given by
[tex]\triangle M = \triangle M_a+\triangle M_f[/tex]
[tex]\triangle M= 85.9+3.92[/tex]
[tex]\triangle M=89.82kg/s[/tex]
Generally the Trust of the rocket is given mathematically by
[tex]F_T=(\triangle M *V_e)-(dM_a/dt)*(V)[/tex]
[tex]F_T=(89.82 *921)-(85.9)*(263)[/tex]
[tex]F_T=60132.52N[/tex]
Generally the Rocket's delivered power is mathematically given by
Delivered power P
[tex]P=V*F_T[/tex]
[tex]P=263*60132.52N[/tex]
[tex]P=15814852.76W.[/tex]
Starting from rest, a sprinter reaches his top velocity in 3 seconds. He runs a distance of 24m in 3 seconds. What is his acceleration? (Assume his acceleration is uniform)
Answer:
[tex]a=5.34\ m/s^2[/tex]
Explanation:
Given that,
Initial velocity of a sprinter, u = 0
He runs a distance of 24m in 3 seconds.
We need to find his acceleration. Let a be his acceleration. Using second equation of kinematics to find it.
[tex]s=ut+\dfrac{1}{2}at^2\\\\24=0+\dfrac{1}{2}a(3)^2\\\\a=\dfrac{24\times 2}{9}\\\\a=5.34\ m/s^2[/tex]
So, the acceleration of the sprinter is [tex]5.34\ m/s^2[/tex].
The acceleration of the sprinter from rest to his top speed is 5.33m/s²,
Given the data in the question
Since the sprinter starts from rest,
Initial velocity; [tex]u = 0[/tex]Time taken; [tex]t = 3s[/tex]Distance covered; [tex]s = 24m[/tex]Acceleration; [tex]a = \ ?[/tex]
To determine the acceleration of the sprinter, we use the second equation of motion:
[tex]s = ut + \frac{1}{2}at^2[/tex]
Where s is distance covered, u is initial velocity, a is acceleration and t is time.
We substitute our given values into the equation
[tex]24m = [0 * 3s] + [\frac{1}{2} * a * (3s)^2]\\\\24m = \frac{1}{2} * a * 9s^2\\\\24m = a * 4.5s^2\\\\a = \frac{24m}{4.5s^2} \\\\a = 5.33m/s^2[/tex]
Therefore, the acceleration of the sprinter from rest to his top speed is 5.33m/s²
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The eagles suck, so do the giants and the jets and jaguars are irrelevant
anyone wanna argue
Naw ur pretty accurate, heck collage is the only football worth watching most the time. Hook'um horns!
The Periodic Table Question 4 of 10 If an element forms a 2+ ion, in which group of the periodic table would you expect to find it? D A. 2 B. 17 O C. 1 D. 18 7
Answer:
B is the difference قصدي لا انا اعرف بشير بس ما اعرف اخوه يعني *قصدي لا انا اعرف بشير بس ما اعرف اخوه يعني *قصدي لا انا اعرف بشير بس ما اعرف اخوه يعني *قصدي لا انا اعرف بشير بس ما اعرف اخوه يعني *قصدي لا انا اعرف بشير بس ما اعرف اخوه يعني *
HELP PLEASE!
A 700 kg race car makes one lap around a track. It has a velocity of 20 m/s with a centripetal force of 5,600 N. What is the radius of the track?
A speed skater goes around a turn with a 25 m radius. The skater has a velocity of 15 m/s and experiences a centripetal force of 720 N. What is the mass of the skater?
A 900-kg car moving at 5 m/s takes a turn around a circle with a radius of 30 m. Determine the net force acting upon the car.
An 800 kg race car makes one lap around a track. It has a velocity of 40 m/s with a centripetal force of 16,000 N. What is the radius of the track?
PLEASE EXPLAIN AND SHOW WORK!
The centripetal force is the force that keeps a body moving in a circular path.
The centripetal force is given by; F = mv^2/r
1) We have;
F = 5,600 N
v = 20 m/s
r =?
m = 700 kg
Making r the subject of the formula;
r =mv^2/F
r = 700 × (20)^2/5,600
r = 50 m
2) F = mv^2/r
F = 900 × (5)^2/30
F = 750 N
3) Making r the subject of the formula;
r =mv^2/F
r = 800 × (40)^2/ 16,000
r = 80 m
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Please what is the work done by a man who is pulling a box of 45kg of mass by means of rope which makes angle of 45 degrees ?
Answer:
No work is done since no distance is given
Explanation:
Since no distance is given, the force is not doing any work
No work is done by the man since we do not know the distance or displacement.
Work is only said to be done when the force applied on an object moves it through a particular distance.
Work done = Force x distance.
Since no distance is given in this problem, we can as well assume that the force applied is doing no work on the object.
How would I solve this? It's Newton's 2nd law
Answer:
5.65 m/s²
Explanation:
We'll begin by calculating the mass of PJ when in San Diego (i.e Earth). This can be obtained as follow:
Weight of PJ on Earth (Wₑ) = 545 N
Acceleration due to gravity (g) on Earth (gₑ) = 10 m/s²
Mass of PJ on Earth (mₑ) =.?
Wₑ = mₑ × gₑ
545 = mₑ × 10
Divide both side by 10
mₑ = 545 / 10
mₑ = 54.5 Kg
Thus, the mass of PJ on San Diego (i.e Earth) is 54.5 Kg
Finally, we shall determine the acceleration due to gravity of planet Koja. This can be obtained as follow:
Weight of PJ on Koja (Wₖ) = 308 N
Mass of PJ on Koja (mₖ) = mass of PJ on Earth (mₑ) because mass is constant irrespective of location.
Mass of PJ on Earth (mₑ) = 54.5 Kg
Mass of PJ on Koja (mₖ) = 54.5 Kg
Acceleration due to gravity of on Koja (gₖ) =?
Wₖ = mₖ × gₖ
308 = 54.5 × gₖ
Divide both side by 54.5
gₖ = 308 / 54.5
gₖ = 5.65 m/s²
Thus, the acceleration due to gravity on planet Koja is 5.65 m/s²
A dog starts at the origin and runs forward at 6m/s for 1.5s and then turns around to fetch the ball by running backward at 7m/s for 3s. If the dog runs back to the origin at 4m/s the how much time has elapsed between the start and when he returns with the ball
Answer:
Total time elapsed between the start and when he returns with the ball is 7.5s
Explanation:
From the question,
- The dog starts at the origin and runs forward at 6m/s for 1.5s. First, we will determine the distance covered while running forward.
From
Speed = Distance / Time
Distance = Speed × Time
Speed = 6m/s
Time = 1.5s
∴ Distance = 6m/s × 1.5s
Distance = 9m
That is, the dog covered a distance of 9m while running forward.
- The dog turns around and runs backward at 7m/s for 3s. Now, we will also determine the distance the dog covered backwards.
Distance = Speed × Time
Speed = 7m/s
Time = 3s
Distance = 7m/s × 3s
Distance = 21m
The dog's displacement from the origin is 21m - 9m = 12m
Now, to calculate how much time has elapsed between the start if the dog runs back to the origin at 4m/s, we will first determine the time the dog spent back to the origin and then add to the time spent for the first two distances.
To get back to the origin, the dog needs to cover 12m
From
Speed = Distance / Time
Time = Distance / Speed
Distance = 12m
Speed = 4m/s
∴ Time = (12m) / (4m/s)
Time = 3s
Therefore, the dog spent 3s to run back to the origin.
Hence, total time elapsed = 1.5s + 3s + 3s
Total time elapsed = 7.5s
A 50Kg girl jumps off a 5-meter-high diving board. What is her kinetic energy right before she
hits the water?
A. 0 J
B. 25 J
C. 1225 J
D. 2450 J
Answer:
D is the correct answer
Explanation:
Ek=m*g*h=50*9.8*5=2450
The kinetic energy right before she hits the water is 2450J. So, the correct option is D.
What is Kinetic energy?Kinetic energy is defined as the energy that is due to the motion of an object. If we want to accelerate an object, we must apply a force, by applying a force we need to do work. After the work is done, energy has been transferred to the object, and the object will continue to move with a new constant speed.
A 50Kg girl jumps off a 5-meter-high diving board.
We need to find the kinetic energy of the girl before she enters the water which means that the kinetic energy becomes equal to the potential energy such that,
P.E.=K.E. = mgh
where, m=mass of the object
g= acceleration due to gravity [tex](9.8m/s^2)[/tex]
h= height
So, K.E= 50* 9.8*5 = 2450 J
Thus, the kinetic energy right before she hits the water is 2450J. So, the correct option is D.
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A solar panel gives our 250 Watts
of power in 2 seconds. How much
work did the solar panel do?
An 69-kg jogger is heading due east at a speed of 1.6 m/s. A 63-kg jogger is heading 14 ° north of east at a speed of 1.5 m/s. Find (a) the magnitude and (b) the direction of the sum of the momenta of the two joggers. Describe the direction as an angle with respect to due east.
Answer:
[tex]P=203.3819375kg.m/s\\P\approx203.4kg.m/s[/tex]
[tex]\theta=6.4541\textdegree North East\\\theta\approx6.5\textdegree North East[/tex]
Explanation:
From the question we are told that
Mass of first jogger [tex]M=69kg[/tex]
Speed[tex]v_1=1.6m/s[/tex]
Direction [tex]d= East[/tex]
Mass of 2nd jogger [tex]M=63kg[/tex]
Speed [tex]v_2=1.5m/s[/tex]
Direction [tex]d= 14\textdegree north[/tex]
Generally equation for momentum along the the horizontal is mathematically given as
[tex]P_x=m_1v_1 +m_2v_2cos\theta[/tex]
[tex]P_x=(69)*(1.6) +(63)*(1.5)cos14[/tex]
[tex]P_x=202.0929461kg.m/s[/tex]
Generally equation for momentum along the the vertical is mathematically given as
[tex]P_y=m_2v_2cos\theta[/tex]
[tex]P_x=22.86161913kgm/s[/tex]
a)Generally the magnitude of momentum is mathematically given by
[tex]P=\sqrt{(P_x)^2+(P_y)^2}[/tex]
[tex]P=\sqrt{(202.0929461)^2+(22.86161913)^2}[/tex]
[tex]P=\sqrt{41364.21249}[/tex]
[tex]P=203.3819375kg.m/s[/tex]
[tex]P\approx203.4kg.m/s[/tex]
b) Generally the angle [tex]\theta[/tex] is mathematically given by
[tex]\theta=tan^-^1\frac{py}{px} \\[/tex]
[tex]\theta=tan^-^1\frac{22.86161913}{202.0929461}[/tex]
[tex]\theta=tan^-^1(0.1131242805)[/tex]
[tex]\theta=6.4541\textdegree North East[/tex]
What state of matter can easily be compressed (squeezed into a much smaller space)? *
gases
solids
solids, liquids, and gases
liquids
Answer:
gases.
Explanation:
gases have the highest volume. if you compress (increase pressure) a gas, the volume decreases and it will take up less space as a liquid, and if you compress more, the liquid will become a solid.
additionally info: this is generally true, but as you get into more difficult subjects, this rule does not hold. for example, some substances cannot obtain a certain phase of matter when they are a certain temperature and or pressure. but that could be way more out of this questions league.
Jeff throws a ball straight up. For which situation is the vertical velocity zero?
a. on the way up
b. at the top
c. on the way back down
d. none of the above
Answer:
b. at the top
Explanation:
At the top of the climb, the vertical velocity of Jeff's throw has a value of zero.
Vertical velocity in this instance is acting against the acceleration due to gravity.
At the top of the thrown, the force of the throw becomes zero and the velocity also becomes zero. As the body finds it way downward, it begins to fall freely. The fall is due to the acceleration due to gravity which is experienced by every object on earth.Which change of state takes place when a gas loses energy?
A. condensation
B. evaporation
C. molting
D. solidification
Answer:
Condensation (((((((((((((
A piece of glass weights 25 g in air, 16.77g in water at 4C and 16.89g in water at 60C. Find the mean coefficient of cubical expansion of water between 4C anf 60C, taking the coefficient of linear expansion of glass as 8*10^6
sorry i think so i dont exactly know i am sorry
Explanation:
As a certain amount of water is cooled from room temperature until it reaches 4 °C, its volume drops. The density reduces as the volume increases below 4 °C. Therefore, water's greatest density occurs at 4 degrees Celsius.
What coefficient of cubical expansion of water?Water has an unusual trait called anomalous expansion, which causes it to enlarge rather than compress when the temperature drops from 4 °C to 0 °C, and it becomes less dense.
At temperatures not far below zero degrees Celsius, water ice is unique for having low coefficients of static and dynamic friction that range from 0.04-0.02, but as the temperature drops, these numbers rise.
Therefore, When a metal sheet's temperature rises by one degree, its coefficient of area expansion is calculated as the increase in surface area per unit of original surface area.
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my friend wore a smart watch to bed,when he woke up in the morning it showed 300 steps taken how did this happen?
Answer:
Your tracker will detect your arm movement and will interpret it as steps so this is what it likely happening. Something similar happens while driving, if you go through a bumpy road or your car has a stiff suspension then this can cause the tracker to detect movement and count it as steps. Usually while driving and if you notice some vibration on your car or steering wheel you can try having your tracker a little bit more tight around your wrist and this will help decrease the amount of steps detected.
Explanation:
(Via AlvaroFitbit)
what happens if I increase the height of a skater
Answer:
The higher the skater is the more potential energy he has. As his height decreases, his potential energy decreases and his kinetic energy increases.
Explanation:
Make me BRAINLIST please
Answer:
The higher the skater is the more potential energy he has. As his height decreases, his potential energy decreases, and his kinetic energy increases. ... Explore how the skater's change in speed relates to the potential and kinetic energy of the skater.
Explanation:
(Via Uteach)
A force of 2 kN is applied to an object to make it move 3.6 m in the direction of the force. Select the correct value of work done on the object.
Answers to choose from:
1.8J
7200J
7.2J
555.6J
Answer:
W= F × d
W= 2kn × 3.6
W= 7.2 J
Work is measured in Joules!
Hello, what does the first condition mean?
Answer:
An object at rest is described by Newton's First Law of Motion. An object in static equilibrium has zero net force acting upon it. The First Condition of Equilibrium is that the vector sum of all the forces acting on a body vanishes.
Explanation:
Answer:
see below
Explanation:
the first condition means that the light ray inside the optically denser medium hits the surface of the optically less dense medium.
this in simple words means, light must be travelling from am optically denser medium to an optically less dense medium
hope this helps :)
A 5.75 g bullet is fired with a velocity of 1.50 x 102 m/s toward a stationary solid block resting on a frictionless surface. The bullet embeds but the block does not move. 1. What is the change in momentum of the bullet if it embeds in the block? 2. What is the change in momentum of the bullet if it bounces off the block in the opposite direction with a speed of 100 m/s?
Answer:
1. -0.863 kgm/s 2. -1.438 kgm/s
Explanation:
1. What is the change in momentum of the bullet if it embeds in the block?
Since the block does not move, the velocity of the bullet after hitting the block , v is zero. That is v = 0 m/s
Now, the momentum change of the bullet ΔP = m(v - u) where m = mass of block = 5.75 g = 5.75 × 10⁻³ kg, u = initial velocity of bullet = 1.50 × 10² m/s and v = final velocity of bullet after hitting the block = 0 m/s (since it embeds in the block and the block does not move).
So, ΔP = m(v - u)
= 5.75 × 10⁻³ kg(0 m/s - 1.50 × 10² m/s)
= 5.75 × 10⁻³ kg(- 1.50 × 10² m/s)
= -8.625 × 10⁻¹ kgm/s
= -0.8625 kgm/s
≅ -0.863 kgm/s
2. What is the change in momentum of the bullet if it bounces off the block in the opposite direction with a speed of 100 m/s?
If it bounces off the block in the opposite direction with a speed of 100 m/s, then its final velocity is v = -100 m/s.
So, our momentum change ΔP' = m(v - u) where m = mass of block = 5.75 g = 5.75 × 10⁻³ kg, u = initial velocity of bullet = 1.50 × 10² m/s and v = final velocity of bullet after hitting the block = -100 m/s = -1 × 10² m/s
So, ΔP = m(v - u)
= 5.75 × 10⁻³ kg(-1 × 10² m/s - 1.50 × 10² m/s)
= 5.75 × 10⁻³ kg(-2.50 × 10² m/s)
= -14.375 × 10⁻¹ kgm/s
= -1.4375 kgm/s
≅ -1.438 kgm/s
Why does Farm Bureau and other advocacy organization oppose any mandated labeling of biotech crops?
Answer:
I’m. Nog sure
Explanation:
Compare between Convex lens and concave lens
Answer:
A convex lens or converging lens focuses the light rays to a specific point whereas a concave lens or diverging lens diverges the light rays. When these lenses are combined, they produce sharper images. Most of the eyeglass lenses use combinations of convex and concave lenses.
The phases of the moon are caused by which of the following
A. Earths rotation around its axis
B. Earths orbit around the sun
C. The moons orbit around the earth
D. Earths shadow falling on the moon
Answer: C: The moons orbit around the earth.
Explanation:
Calculate the temperature change when 1000J of heat is supplied to 100g of water.
Please explain
Explanation:
Given
heat supplied (Q) = 1000 J
mass(m) = 100 g = 0.1 kg
specific heat capacity of water(s) = 4200 J/kg°C
change in Temperature (dt) = ?
We know we have the relation
Q = m * s * dt
Or, 1000 = 0.1 * 4200 * dt
Or, 1000 = 420 dt
Or, dt = 1000/ 420
Therefore dt = 2.38°
Hope it will help :)
The temperature changes when 1000J of heat is supplied to 100g of water is dt = 2.38°
What is temperature?The average kinetic energy of a system is measured by its temperature. The kinetic energy of a system starts to rise as the particle's velocity increases, which raises the system's temperature.
The energy that is transferred when two bodies with different surface temperatures come in touch is referred to as heat.
The change is when 1000J of heat is supplied to 100g of water.
Given that:
heat supplied (Q) = 1000 J
mass(m) = 100 g = 0.1 kg
specific heat capacity of water(s) = 4200 J/kg°C
Change in Temperature (dt) = ?
We know we have the relation
Q = m x s x dt
Or, 1000 = 0.1 x 4200 x dt
Or, 1000 = 420 dt
Or, dt = 1000/ 420
Therefore, the temperature changes when 1000J of heat is supplied to 100g of water is dt = 2.38°.
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(97-R) Dos cargas puntuales iguales están separadas por una distancia d. A) ¿Es nulo el campo eléctrico total en algún punto? Si es así, ¿cuál es la posición de dicho punto? B) Repita el apartado anterior suponiendo que las cargas fueran de distinto signo.
Answer:
Explanation:
El campo eléctrico provocado por una carga puntal sobre determinado punto viene dado por la ecuación:
E = K * Q /r² en esa ecuación E es el campo eléctrico (un vector), K es una constante K = 1/4*π*ε₀ Q es la carga que origina el campo, y r la distancia entre la carga y el punto donde se quiere determinar el campo. La fuerza de origen eléctrico y el campo eléctrico van en la dirección de la recta de unión entre carga y punto ( es decir según r) y esta fuerza sera repulsiva o de atracción según que la carga que la origina sea positiva o negativa (respectivamente, ya que en la determinación de fuerza y campo, se asume que se coloca una carga de prueba positiva en el punto ).
Siendo así, si dos cargas elécticas son iguales y están separadas una distancia r ; en el punto medio (r/2) los campos producidos por cada una de las cargas serán de la misma magntud.
Carga Q₁ = Carga Q₂ = Q
E₁ = K * Q₁ / (r/2)² y E₂ = K * Q₂/(r/2)²
Y además tendran direcciones totalmente opuestas, por lo que se anularan
En el segundo caso las cargas son iguales en magntud pero de diferentes signos ( una positiva y la otra negativa ). En este caso los campos eléctricos continuaran siendo de la msma magnitud pero dado que la carga positiva repele la carga de muestra (teorica) y la negativa la atrae los dos campos coinciden en su dirección y resultará una campo de magnitud doble. La dirección del campo será en la tendencia de acercarse a la caga negativa, así si la carga negativa está a la derecha la fuerza de origen eléctrico y el campo van hacia la derecha y si la carga negativa está a la izquierda pues esa será la diercción del campo
A vector
of magnitude 5 units and another vector ū of magnitude 3 units point in directions
differing by 60°. Find out (a) the scalar product of the two vectors and (b) the vector product of the two vectors
Answer:
7.5 units
13 units
Explanation:
[tex]|v|=5\ \text{units}[/tex]
[tex]|u|=3\ \text{units}[/tex]
[tex]\theta[/tex] = Angle between the vectors = [tex]60^{\circ}[/tex]
Scalar product is given by
[tex]u\cdot v=|u||v|\cos\theta\\ =3\cdot 5\cdot \cos60^{\circ}\\ =7.5\ \text{units}[/tex]
The scalar product of the vectors is 7.5 units.
Vector product is given by
[tex]u\times v=|u||v|\sin\theta\\ =3\times 5\sin60^{\circ}\\ =13\ \text{units}[/tex]
The vector product of the vectors is 13 units.
A flight attendant pulls her 70 N flight bag a distance of 318 m along a level airport floor at a constant velocity. The force she exerts is 41 N at an angle of 57° above the horizontal.
Complete question :
A flight attendant pulls her 70 N flight bag a distance of 318 m along a level airport floor at a constant velocity. The force she exerts is 41 N at an angle of 57° above the horizontal.
(a) Find the work she does on the flight bag.
(b) Find the work done by the force of friction on the flight bag.
(c) Find the coefficient of kinetic friction between the flight bag and the floor.
Answer:
7107 J ; - 7107 J ; 0.55
Explanation:
Given that :
Distance, d = 318m
Applied force = 41 N
θ = 57°
A.) Workdone = Force exerted along direction of motion
Workdone = applied Force * distance * cosθ
Workdone = 41 * 318 * cos57 = 7101.0037
Workdone = 7,101 J
B.) Workdone by force of friction on flight bag:
- 7,101 J (since the body moves at constant velocity)
C.)
Coefficient of kinetic friction (μ) = Frictional force / normal reaction)
μ = F / N
Frictional force, F = Workdone by friction / distance
F = 6200 / 318
F = 19.47N ;
Ff = weight of Flight bag = 70 N
N = Ff - Fsinθ
N = 70 - applied Force sinθ
N = 70 - 41sin57
N = 70 - 34.385493
N = 35.614506
μ = 19.47 / 35.614506
μ = 0.5466873
1. How much heat must be absorbed by 375 grams of water to raise its
temperature by 25° C?(Cp of water is 4.184)
Answer:
39225J
Explanation:
Given parameters:
Mass of water = 375grams of water
Change in temperature = 25°C
Specific heat capacity of water = 4.184J/g°C
Unknown:
Amount of heat absorbed = ?
Solution:
To solve this problem, we use the expression below:
H = m c Ф
H is the heat absorbed
m is the mass
c is the specific heat capacity
Ф is the change in temperature
Insert the parameters and solve;
H = 375 x 4.184 x (25) = 39225J