The increase in the temperature of the soil when bag fall from the given height is 0.065 ⁰C.
The given parameters:
mass of the bag, m = 15.53 kgheight of fall, h = 5.5 mspecific heat capacity of soil, C = 0.200 kcal/kg ⁰CApply the principle of conservation of energy as follows;
[tex]mgh = mC \Delta T\\\\gh = C \Delta T\\\\\Delta T= \frac{gh}{C}[/tex]
Convert the value of C in kcal/kg ⁰C to J/kg ⁰C
1 kcal = 4180 J
[tex]0.200 \ kcal/kg ^0 C= 0.2 \times 4180 (J/kg ^0C) = 836 \ J/kg^0 C[/tex]
The increase in the temperature of the soil is calculated as follows;
[tex]\Delta T= \frac{gh}{C}\\\\\Delta T= \frac{9.8 \times 5.5}{836} \\\\\Delta T= 0.065 \ ^0C[/tex]
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At what speed must the electron revolve round the nucleus of
the hydrogen in its ground state in order that it may not be pulled into the
nucleus by electrostatic attraction
Explanation:
I think this is it, give it a try
Which of the following statements are true?
(a) An object can move even when no force acts on it.
(b) If an object isn't moving, no external forces act on it.
(c) If a single force acts on an object, the object accelerates.
(d) If an object accelerates, a force is acting on it.
(e) If an object isn't accelerating, no external force is acting on it.
(f) If the net force acting on an object is in the positive x-direction, the object moves only in the positive x-direction.
I can't understand how A is true.
Answer: a, c, d
Explanation: a is true because the object will continue to move even without any force because of inertia (so yh thats why a is true). c is true because an object can accelerate if a single force acts on it. to accelerate (not move), it needs a force to act on it
Would appreciate brainly <3
Explanation:
inertia ...
you push or throw something, and you apply some force to it at that moment, but then it moves and keeps moving even long after you have no more connection to it, and no more force is applied to it.
please consider : we are only talking about moving. not about acceleration.
so, yes, (a) is true.
(c) is true.
(d) is true.
What is the volume of 150g of a substance that has a density of 150g of a substance that has a density of 11.3g/cm3
Answer:
25.0 cm3
Explanation:
The volume is 25.0 cm3 .
If it takes 50.0 seconds to lift 10.0 Newtons of books to a height of 7.0
meters, calculate the power required to lift it?
Figure 1 shows that the two vectors in different direction respectively. Determine the resultant for the two vectors
Answer:
15-45/2 = -7.5 downwards
Explanation:
cos(60) x 45 = z
z = 45/2
sin (30) x 70 = y
y = 15
15-45/ 2 = -7.5 downwards
what is a capacitor and capacitance.
Answer:
Capacitor is a device which used to store energy .
Capacitance is the ratio of the change in an electric charge in a system to the corresponding change in its electric potential.
Explanation:
Jimmy walks to school by traveling 2.0 miles east and 3.0 miles north from his starting point. What are the magnitude and direction of Jimmy's displacement with respect to his original position?
Answer:
3.6 miles at 56.3 degrees north of east.
Explanation:
If you draw a diagram to calculate Jimmy's displacement, you end up with a triangle.
To solve for the magnitude of Jimmy's displacement with respect to his original position, we need to use the Pythagorean Theorem formula.
Pythagorean Theorem Formula = a^2 + b^2 = c^2
Step by Step to solve for the magnitude of Jimmy's displacement:
1) a^2 + b^2 = c^2
2) 2.0^2 + 3^2 = c^2
3) 4 + 9 = c^2
4) 13 = c^2
5) To find what "c" equals to we need to find the square root of 13.
6) √13 = 3.6 | c = 3.6 miles
To solve for the direction of Jimmy's displacement with respect to his original position, we need to use the following formula:
tan^-1 ∅ (3/2)
tan^-1 ∅ (1.5)
∅ = 56.309° north of east
Therefore, your answer is 3.6 miles at 56.3 degrees north of east.
Question below...............
Answer:
friction force
Explanation:
force of friction is opposite to the force applied it resist the motion
Croquet ball A moving at 8.3 m/s makes a head on collision with ball B of equal mass and initially at rest. Immediately after the collision ball B moves forward at 6.4 m/s .
What fraction of the initial kinetic energy is lost in the collision?
Answer:
0.25
Explanation:
That is the right answer.
The fraction of the initial kinetic energy is lost in the collision is 35.3%.
The given parameters:
Initial velocity of ball A = 8.3 m/sInitial velocity of ball B = 0Final velocity of ball B = 6.4 m/sThe initial kinetic energy of the system collision is calculated as follows;
[tex]K.E_i = \frac{1}{2} mv_1_i^2 + \frac{1}{2} mv_2_i^2\\\\K.E_i = \frac{1}{2} (m)(8.3)^2 + \frac{1}{2} (m) (0)^2\\\\K.E_i = 34.445 m[/tex]
The final velocity of ball A after collision is calculated as follows;
[tex]u_1 + v_1 = u_2 + v_2\\\\8.3 + v_1 = 0 + 6.4\\\\v_1 = 6.4 - 8.3\\\\v_1 = -1.9 \ m/s[/tex]
The final kinetic energy of the system after collision is calculated as follows;
[tex]K.E_f = \frac{1}{2} m(-1.9)^2 + \frac{1}{2} m(6.4)^2\\\\K.E_f = 22.285 \ m \[/tex]
The fraction of the initial kinetic energy is lost in the collision is calculated as follows;
[tex]= \frac{K_i - K_f}{K_i} \\\\= \frac{34.445 - 22.285}{34.445} \\\\= 0.353\\\\= 35.3\%[/tex]
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Starting from rest, a racecar moves 112 m in the first 7 s of uniform acceleration. What is the car's acceleration?
Answer:
215mph
Explanation:
multiply 112milws by 2 and multiple that by seven, then divide it by two
builder places a 3kg hammer on the top of a ladder, which is 4m above the ground. Calculate the gravitational potential energy of the hammer while on the ladder.
Answer:
[tex]E=mgh[/tex]
[tex]m=3kg[/tex]
[tex]h=4m[/tex]
[tex]g=9.8m/s^{2}[/tex]
[tex]E= 3*4*9.8=117.6J[/tex]
Explanation:
Only substitute amounts to formula.
Hope this helps ;)
Cheers :D
7. A 1.0 kg metal head of a geology hammer strikes a solid rock with a velocity of 5.0 m/s. Assuming all the energy is retained by the hammer head, how much will its temperature increase
The increase in temperature of the metal hammer is 0.028 ⁰C.
The given parameters:
mass of the metal hammer, m = 1.0 kgspeed of the hammer, v = 5.0 m/sspecific heat capacity of iron, 450 J/kg⁰CThe increase in temperature of the metal hammer is calculated as follows;
[tex]Q = K.E\\\\mc \Delta T = \frac{1}{2} mv^2\\\\\Delta T = \frac{v^2}{2 c}[/tex]
where;
c is the specific heat capacity of the metal hammer
Assuming the metal hammer is iron, c = 450 J/kg⁰C
[tex]\Delta T = \frac{5^2}{2 \times 450} \\\\\Delta T = 0.028 \ ^0C[/tex]
Thus, the increase in temperature of the metal hammer is 0.028 ⁰C.
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Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a horizontal surface of negligible friction, releasing the block, and measuring the velocity v of the block as it leaves the spring, as shown in Figure 1. The experiments indicate that as x increases, so does v in a linear relationship. The surface is now lifted so that the surface is at an angle θ above the horizontal. Which of the following indicates how the relationship between v and x changes?
Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.
The statement that indicates how the relationship between v and x changes is; As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x.Reasons:
The energy given to the block by the spring = [tex]\mathbf{0.5 \cdot k \cdot x^2}[/tex]
According to the principle of conservation of energy, we have;
On a flat plane, energy given to the block = [tex]0.5 \cdot k \cdot x^2[/tex] = kinetic energy of
block = [tex]0.5 \cdot m \cdot v^2[/tex]
Therefore;
0.5·k·x² = 0.5·m·v²
Which gives;
x² ∝ v²
x ∝ v
On a plane inclined at an angle θ, we have;
The energy of the spring = [tex]\mathbf{0.5 \cdot k \cdot x^2}[/tex]
The force of the weight of the block on the string, [tex]F = m \cdot g \cdot sin(\theta)[/tex]
The energy given to the block = [tex]0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta)[/tex] = The kinetic energy of block as it leaves the spring = [tex]\mathbf{0.5 \cdot m \cdot v^2}[/tex]
Which gives;
[tex]0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta) = 0.5 \cdot m \cdot v^2[/tex]
Which is of the form;
a·x² - b = c·v²
a·x² + c·v² = b
Where;
a, b, and c are constants
The graph of the equation a·x² + c·v² = b is an ellipse
Therefore;
As x increases, v increases, however, the value of v obtained will be lesser than the same value of x as when the block is on a flat plane.Please find attached a drawing related to the question obtained from a similar question online
The possible question options are;
As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of xThe relationship is no longer linear and v will be more for the same value of xThe relationship is still linear, with lesser value of vThe relationship is still linear, with higher value of vThe relationship is still linear, but vary inversely, such that as x increases, v decreasesLearn more here:
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What’s Newton’s second law? Explain and mention some examples in daily life
Answer:
Newton's second law states that .
The rate of change of linear momentum is directly proportional to the force applied.Formulically
F=maF=Force
m=mass
a=acceleration
The best example is hitting a tennis ball.
What is the potential energy when a 100 kg object is raised 4.00 m straight up?
Answer:
100kg x 4 x 10 = 4000J
Explanation:
2. An auditorium has 58 seats in the first row, 62 seats in the second row, 66 seats in the third row, and so
on.
a)Find the explicit formula of this arithmetic sequence.
B) find the number of seats in the twentieth row.
A solenoid 91.0 cm long has a radius of 1.50 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magnitude of the magnetic field inside the solenoid.
The magnitude of the magnetic field inside the solenoid is [tex]6.46 \times 10^{-3} \ T[/tex].
The given parameters;
length of the solenoid, L = 91 cm = 0.91 mradius of the solenoid, r = 1.5 cm = 0.015 mnumber of turns of the solenoid, N = 1300 current in the solenoid, I = 3.6 AThe magnitude of the magnetic field inside the solenoid is calculated as;
[tex]B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\[/tex]
where;
[tex]\mu_o[/tex] is the permeability of frees space = 4π x 10⁻⁷ T.m/A
[tex]B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T[/tex]
Thus, the magnitude of the magnetic field inside the solenoid is [tex]6.46 \times 10^{-3} \ T[/tex].
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what are people words in english
Answer: Plenty
Explanation:
some words are Hi, Banana, Dude and many more
Which quantity or quantities is/are increasing for the object represented by line B?
Answer:
C. Velocity and Position
Explanation:
The quantities that are increasing for the object represented by line B are velocity and position. The correct option is b.
What is velocity?The direction of a body or object's movement is defined by its velocity. In its basic form, speed is a scalar quantity. In essence, velocity is a vector quantity. It is the speed at which distance changes. It is the displacement change rate.
Velocity is the pace and direction of an object's movement, whereas speed is the time rate at which an object is traveling along a path. In other words, velocity is a vector, whereas speed is a scalar value.
The graph is given which represents the velocity and time with terms A, B, and C. As opposed to the position-time graph, which describes an object's motion over time, the velocity-time graph reveals an object's speed.
Therefore, the correct option is b. velocity and position.
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The question is incomplete. Your most probably complete question is given below:
Velocity and acceleration
velocity and position
velocity only
velocity, position, and acceleration
A fat wire, radius a, carries a constant current I, uniformly distributed over its cross section. A narrow gap in the wire, of width w << a, forms a parallel-plate capacitor. Find the magnetic field in the gap, at a distance s < a from the axis.
The magnetic field in the gap is [tex]\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}[/tex]
In a circuit, a magnetic flux is circulated or followed via a confined environment or passage called a magnetic field gap. The narrow air gap is a non-magnetic component of a magnetic circuit that is normally connected to the remainder of the circuit magnetically in series. This enables a significant amount of magnetic flux to pass via the gap.
The magnetic field in the gap at a distance s < a can be computed by using the formula:
[tex]\mathbf{ \oint Bdl = \mu_oI_{enclosed}}[/tex]
where;
Magnetic flux density = Bdistance = d[tex]\mathbf{B( 2 \pi d) = \mu _o \oint _s J_d da }[/tex]
where;
[tex]\mathbf{J_d}[/tex] = drift current density[tex]\mathbf{B( 2 \pi d) = \mu _o J_d \oint _sda }[/tex]
[tex]\mathbf{B( 2 \pi d) = \mu _o J_d (\pi d^2) }[/tex]
Making the magnetic flux density the subject, we have:
[tex]\mathbf{B =\dfrac{ \mu _o J_d (\pi d^2) }{( 2 \pi d)}}[/tex]
[tex]\mathbf{B =\dfrac{ \mu _o J_dd}{ 2 }}[/tex]
Recall that, the drift current density [tex]\mathbf{J_d = \dfrac{I}{\pi a^2}}[/tex]
[tex]\mathbf{B = \dfrac{\mu_o d}{2}(\dfrac{I}{\pi a^2})}[/tex]
Recall that distance in question is said to be (s);
∴
[tex]\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}[/tex]
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when a torque is acting on a fly wheel the angular velocity of the fly wheel changes from 10rad/sec to 25rad/sec in 5sec.what will be the magnitudes of the angular acceleration of the fly wheel?
Hello!
We can use the following angular kinematic equation to solve:
α = Δω/Δt, or (ωf-ωi)/t
Plug in the given values:
α = (25 - 10)/5 = 15/5 = 3 rad/sec²
A stone is allowed to fall from the top of the tower 100m high at the same moment another stone is projected vertically upward with a velocity of 25m/s. Where and when will the two cross each other
Answer:
both the stone will meet at a distance of 80 m from the top of tower.
Explanation:
let "t" = time after which both stones meet
"S" = distance travelled by the stone dropped from the top of tower
(100-S) = distance travelled by the projected stone.
◆ i) For stone dropped from the top of tower
-S = 0 + 1/2 (-10) t²
or, S = 5t²
◆ ii) For stone projected upward
(100 - S) = 25t + 1/2 (-10) t²
= 25t - 5t²
Adding i) and ii) , We get
100 = 25t
or t = 4 s
Therefore, Two stones will meet after 4 s.
◆ ¡¡¡) Put value of t = 4 s in Equation i) , we get
S = 5 × 16
= 80 m.
Thus , both the stone will meet at a distance of 80 m from the top of tower.
(Hope this helps can I pls have brainlist (crown)☺️)
Which of the following methods of pest control used in IPM systems is effective because of the increase in biological diversity it provides? timed crop planting O weed suppression composted soil amendments deep tilling soil
Answer:
Decline in soil health is a serious worldwide problem that decreases complexity and stability of agricultural ecosystems, commonly making them more prone to outbreaks of herbivorous insect pests. Potato (Solanum tuberosum L., Solanaceae) and onion (Allium cepa L., Amaryllidaceae) production is currently characterized by high soil disturbance and heavy reliance on synthetic inputs, including insecticides. Evidence suggests that adopting soil conservation techniques often (but not always) increases mortality and decreases reproductive output for the major insect pests of these important vegetable crops. Known mechanisms responsible for such an effect include increases in density and activity of natural enemy populations, enhanced plant defenses, and modified physical characteristics of respective agricultural habitats. However, most research efforts focused on mulches and organic soil amendments, with additional research needed on elucidating effects and their mechanisms for conservation tillage, cover crops, and arbuscular mycorrhizae.
Introduction
Soil erosion is a serious problem worldwide (Amundson et al., 2015). Although it is often overshadowed in public discourse by other concerns, such as climate change and invasive species, soil deterioration receives considerable attention from the scientific community (Montgomery, 2007; Borrelli et al., 2017; Berhe et al., 2018). In addition to the loss of agricultural productivity, soil erosion has been linked to increased emissions of greenhouse gases and reduced water quality (Amundson et al., 2015; Berhe et al., 2018). Global soil erosion is forecasted to increase in the near future because of cropland expansion, especially in the least economically developed areas (Borrelli et al., 2017).
The coefficient of kinetic friction for a 22 kg bobsled on a track is 0.10. What force is required to push it down a 5.0 degree incline and achieve a speed of 62 km/h at the end of 75 m
Hi there!
We must begin by converting km/h to m/s using dimensional analysis:
[tex]\frac{62km}{1hr} * \frac{1hr}{3600sec}*\frac{1000m}{1km} = 17.22 m/s[/tex]
Now, we can use the kinematic equation below to find the required acceleration:
vf² = vi² + 2ad
We can assume the object starts from rest, so:
vf² = 2ad
(17.22)²/(2 · 75) = a
a = 1.978 m/s²
Now, we can begin looking at forces.
For an object moving down a ramp experiencing friction and an applied force, we have the forces:
Fκ = μMgcosθ = Force due to kinetic friction
Mgsinθ = Force due to gravity
A = Applied Force
We can write out the summation. Let down the incline be positive.
ΣF = A + Mgsinθ - μMgcosθ
Or:
ma = A + Mgsinθ - μMgcosθ
We can plug in the given values:
22(1.978) = A + 22(9.8sin(5)) - 0.10(22 · 9.8cos(5))
A = 46.203 N
Danny is competing in the high jump. When he is in the air, his body has _______ energy due to its height, and it has _______ energy due to its motion.
Answer:
Gravitational potential energy
kinetic energy
Which would most likely form a homogenous mixture?
Answer:
B) a pinch of sugar mixed with cup of water
Answer:
A pinch of sugar mixed with a cup of water
Explanation:
A homogeneous mixtures have a uniform appearance where the parts are pretty evenly spaced out throughout the mixture. Picking out the individual pieces should be hard, ,which is why sugar in water is the best choice.
calculate 18% of 2758 correct to 4 significant figure
Answer:
......the answer is 496.4
What is First Aid.
I mark u brainliest answer
Answer:
First aid refers to the emergency or immediate care you should provide when a person is injured or ill until full medical treatment is available.
Explanation:
In the popular game "Angry Birds" (refer Figure), for the red bird to hit the green bird with their range 55.0 m, what should be the launch angle (with respect to the horizontal) of the slingshot if the time of flight is 2.50 s?
Answer:
Explanation:
What are we to use as gravity in this imaginary world? I will ASSUME 9.81 m/s² and assume no air resistance.
horizontal velocity component
vx = 55.0/2.50 = 22 m/s
vertical initial velocity component
(12.0 - 10.0) = 0 + vy₀(2.50) + ½(-9.81)2.50²
vy₀ = 13.0625 m/s
θ = arctan(vy₀/vx) = arctan(13.0625/22) = 30.6997225... = 30.7°
Entra vapor a una tobera adiabática con un flujo másico de 250 kg/h. Al entrar el vapor
tiene una energía interna específica de 2510 kJ/kg, una presión de 1378 kPa, un
volumen específico de 0.147 m3
/kg y una velocidad de 5 m/s. Las condiciones de salida
son P= 138.7 kPa, volumen específico de 1.099 m3
/kg y energía interna específica de
2263 kJ/kg. Determine la velocidad de salida.
Este problema describe el funcionamiento de una tobera adiabática (sin flujo de calor), la cual tiene una corriente de entrada que difiere de la de salida espacial y energéticamente, pero que conservan el mismo flujo másico de 250 kg/h. De este modo, usamos un balance de energía con el fin the determinar la velocidad a la que sale el fluido, según es requerido en el problema:
[tex]mu_1+mP_1v_1+mgh_1+\frac{1}{2} mv_1^2=mu_2+mP_2v_2+mgh_2+\frac{1}{2} mv_2^2[/tex]
En la que se tiene que la incógnita es la velocidad de salida, [tex]v_2[/tex], y es posible simplificar el flujo másico, [tex]m[/tex], al estar como factor común en ambos lados y despreciar la energía potencial (mgh), ya que no hay diferencia de altura significativa entre la entrada y salida de la tobera.
De este modo, es posible reemplazar los valores dados para obtener la siguiente expresión:
[tex]2510\frac{kJ}{kg} +1378kPa*0.147\frac{m^3}{kg} +\frac{1}{2} (5\frac{m}{s} )^2=2263\frac{kJ}{kg}+138.7kPa*1.099\frac{m^3}{kg} +\frac{1}{2} v_2^2[/tex]
Y así, hallar la velocidad de salida como sigue:
[tex]2725.066\frac{kJ}{kg}=2415.431\frac{kJ}{kg} +\frac{1}{2} v_2^2\\\\v_2=\sqrt{2(2725.066\frac{kJ}{kg}-2415.431\frac{kJ}{kg} )} \\\\v_2=24.9\frac{m}{s}[/tex]
Para revisar:
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