Answer:
Fnet = 5 Newton
Explanation:
Fnet = 10 N - 5 N
Fnet = 5 N
Carrie is driving a car. Which factors determine the kinetic energy of her car
at this point
Answer:
The two main factors that affect kinetic energy are mass and speed.
Explanation:
Kinetic energy is the energy that is caused by the motion. The kinetic energy of an object is the energy or force that the object has due to its motion. Your moving vehicle has kinetic energy; as you increase your vehicle's speed, your vehicle's kinetic energy increases.
Have a great day! :D
convert -40⁰ c to Fahrenheit scale
Answer:
104f
Explanation:
(40°C × 9/5) + 32 = 104°F
8. An airplane of mass 8500 kg dives has an altitude of 15,000 m. It then dives steeply to an
altitude of 11,000 m. What was the change in potential energy?
Answer:
333.5 MJ
Explanation:
ΔV = m·g·Δh
= 8500 · 9.81 · (15000-11000)
= 8500 · 9.81 · 4000
= 333 540 000
≈ 333.5 ·10⁶J = 333.5 MJ
The change in potential energy is 333.5 MJ
What is Potential Energy?
Potential energy is defined as the energy possessed by an object because of its position relative to other objects, tension within itself, its electric charge, or other factors. Any object that is raised above its rest position has energy stored in it, so it is called potential energy because it has the potential to do work when released.
It can be expressed as:
P.E.= mgh
where, m is the mass of an object measured in grams 'g'
g is the acceleration due to gravity which is [tex]9.8 m/s^2[/tex]
h is the height measured in meter 'm'
Here,
the given information is
m= 8500kg
Δh= [tex]h_2 - h_1[/tex] = 15000-11000 = 4,000m
So, P.E. = 850 x 9.8 x 4000 = 333 540 000= 333.5 ·10⁶J = 333.5 MJ
Thus, the change in potential energy is 333.5 MJ
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How much force is required to raise a 0.2 kg mass?
Answer:
1.96 N
Explanation:
⇒This is your full answer
How do mechanical waves compare with electromagnetic waves?
Answer:
Main Difference Between Mechanical and Electromagnetic waves
A wave is composed of some kind of disturbance that propagates. We can classify waves into many different types based on their properties. One of the properties of the waves depends on whether they need a medium to propagate or not. The primary difference between electromagnetic and mechanical waves is also based on this property. Mechanical waves need a medium, while electromagnetic waves do not need a medium to propagate. Electromagnetic waves can travel through a vacuum. The other differences between mechanical and electromagnetic waves are given below:
Electromagnetic waves can travel through a vacuum, that is an empty space, whereas mechanical waves cannot. They need a medium to travel such as water or air. Ripples in a pond are an example of mechanical waves whereas electromagnetic waves include light and radio signals, which can travel through the vacuum of space.
Mechanical waves can be classed as elastic waves because their transmission depends on the medium's (water, air etc.) elastic properties.
Electromagnetic waves are caused because of the varying magnetic and electric fields. They are produced by the vibration of the charged particles.
Because of these differences, the speed of each type of wave varies significantly. Electromagnetic waves travel at the speed of light but mechanical waves are far slower.
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HELP ASAP!!!!! Choose all the answers that apply. Technology A)influences science
B)helps scientists observe fast phenomena
C)is the same as science
D) influences history
E)helps scientists observe slow phenomena
20 cm long 10 cm wide and 5 cm thick as a mass of 500 g determine the greatest pressure that can be exerted by block on the flat surface?
100000 Pascal
Explanation:
pressure= force/area
Max pressure= force/min area
so f=5
min area= 5×10^-5
5÷5*10^-5 = 100000pascal
How long do elephants live?
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).
I need the answer to this question
Answer:
I don't see anything so you should repost
An airplane accelerates from a speed of 33 m/s at the constant rate of 3.0 m/s2 over a distance of 500 m. What the final velocity?
Answer:
Explanation:
v² = u² + 2as
v² = 33² + 2(3.0)(500)
v² = 4089
v = 63.9452...
v = 64 m/s
Comparing energy resources
A cylindrical rod formed from silicon is 16.8 cm long and has a mass of 2.17 kg. The density of silicon is 2.33 g/cm3 . What is the diameter of the cylinder
Answer:
We are given the length, the mass and the density of the cylinder. First let us calculate for the volume by dividing the mass by the density.
volume = mass /density
where mass = 2.17 kg = 2170 g, therefore:
volume = 2170 g / (2.33 g/cm^3)
volume = 931.33 cm^3
We know that the volume of a cylinder has the formula:
volume = π r^2 h
since h = 16.8 cm, therefore calculating for radius:
931.33 cm^3 = π r^2 (16.8 cm)
r^2 = 17.646 cm^2
r = 4.2 cm
Hence the diameter (d) is:
d = 2 r
d = 8.4 cm
Explanation:
The diameter of the cylindrical rod is approximately 0.382 cm.
To find the diameter of the cylindrical rod, we can use the formula for the volume of a cylinder and then solve for the diameter.
The formula for the volume of a cylinder is:
V = π[tex]r^{2}[/tex]h,
where V is the volume, r is the radius, and h is the height (or length) of the cylinder.
In this case, we know the length of the cylinder (h) is 16.8 cm. We need to find the radius (r) in order to calculate the diameter.
The mass of the cylinder can be related to its volume and density using the formula:
m = ρV,
where m is the mass, ρ is the density, and V is the volume.
Rearranging this formula, we can solve for V:
V = m / ρ.
Now we have two equations:
V = π[tex]r^{2}[/tex]h,
V = m / ρ.
Setting these two equations equal to each other, we can solve for r:
π[tex]r^{2}[/tex]h = m / ρ.
Substituting the given values:
π[tex]r^{2}[/tex] * 16.8 cm = 2.17 kg / (2.33 g/[tex]cm^3[/tex]).
Let's solve this equation for r:
[tex]r^{2}[/tex] = (2.17 kg / (2.33 g/[tex]cm^3[/tex])) / (π * 16.8 cm).
[tex]r^{2}[/tex] ≈ 0.036775 [tex]cm^2[/tex].
Taking the square root of both sides:
r ≈ 0.191 cm.
Finally, we can find the diameter (d) by multiplying the radius by 2:
d ≈ 2 * 0.191 cm.
d ≈ 0.382 cm.
Therefore, the diameter of the cylindrical rod is approximately 0.382 cm.
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when an object is immersed in liquid at rest,why is the net force on the object is the horizontal direction equal to zero?
Based on Newton's third law, the pressure is the same on points that are at the same level but on opposite sides.
Why is the net force in the horizontal direction on an object immersed in a liquid at rest equal to zero?When an object is immersed in a fluid at rest,
the fluid exerts a force on its surface which always normal to the object's surfacethe pressure is the same on points that are at the same level but on opposite sides.Therefore, the net force along horizontal firection is equal to zero.
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A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it. (a) What is the force constant of the spring
Answer:
Explanation:
The change in force = 9.81(1.95 - 0.300) = 16.2 N
The change in length is 0.750 - 0.200 = 0.550 m
K = ΔF/Δx = 16.2/0.550 = 29.4 N/m
Heat is transferred from molecules with more kinetic energy to molecules with _________kinetic energy.
Answer:
low
Explanation:
the higher the kinetic energy, the More the vibration of molecules, thus heat is more on the side with highly vibrating molecules
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]
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Answer:
I think that the answer is 4.318 as a=f/p
A projectile is launched with a horizontal velocity of 20 m/s and an initial vertical velocity of 20 m/s. What is the projectile's acceleration in the Horizontal direction? Verticle direction?
Answer:
Vertical acceleration 9.8 m/s² downward
Horizontal acceleration 0.0 m/s²
assuming no air resistance.
What is Rest in physics ?
Rest in physics generally refers to the state in which an object is stationary.
Rest in physics refers to a situation in which an object does not move from one point to another. Usually, an object is at rest when it is acted upon by equal and opposite forces. For example, a book lying on a table.In a nutshell, the state of rest in physics generally refers to the state in which an object is stationary and does not translate.Hence, in Physics, an object is in a state of rest when it does not move from one point to another.
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A 220 g mass is on a frictionless horizontal surface at the end of a spring that has force constant of 7.0
Nm-1. The mass is displaced 5.2 cm from its equilibrium position and then released to undergo simple
harmonic motion.
At what displacement from the equilibrium position is the potential energy equal to the kinetic energy
The concept of conservation of energy and harmonic motion allows to find the result for the power where the kinetic and potential energy are equal is:
x = 0.135 cm
Given parameters
The mass m = 220 g = 0.220 kg The spring cosntnate3 k = 7.0 N / m Initial displacement A = 5.2 cm = 5.2 10-2 mTo find
The position where the kinetic and potential energy are equal
A simple harmonic movement is a movement where the restoring force is proportional to the displacement, the result of this movement is described by the expression.
x = A cos wt + fi
w² = [tex]\frac{k}{m}[/tex]
Where x is the displacement from the equilibrium position, A the initial amplitude of the system, w the angular velocity t the time, fi a phase constant determined by the initial conditions, k the spring constant and m the mass.
The speed is defined by the variation of the position with respect to time.
v = [tex]\frac{dx}{dt}[/tex]
let's evaluate
v = - A w sin (wt + Ф)
Since the body releases for a time t = 0 the velocity is zero, therefore the expression remains.
0 = - A w sin Ф
For the equality to be correct, the sine function must be zero, this implies that the phase constant is zero
x = A cos wt
Let's find the point where the kinetic and potential energy are equal.
K = U
½ m v² = m g x
we substitute
½ A² w² sin² wt = g A cos wt
sin² wt = [tex]\frac{2g}{A}[/tex] cos wt
let's calculate
w = [tex]\sqrt{\frac{7}{0.220} }[/tex]
w = 5.64 rad / s
sin² 5.64t = 2 9.8 / 0.052 cos 5.64t
sin² 5.64t = 376.92 cos 5.64 t
1 - cos² 5.64t = 376.92 cos 5.64t
cos² 5.64t -376.92 cos564t -1 = 0
we make the change of variable
x = cos 5.64t
x²- 376.92 x - 1 = 0
x = 0.026
cos 5.64t = 0.026
Let's find the displacement for this time
x = 5.2 10-2 0.026
x = 1.35 10-3 m
In conclusion Using the concepts of conservation of energy and harmonic motion we can find the result for the could where the kientic and potential enegies are equal is:
x = 0.135 cm
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does gravity cause a change in an objects movement
Answer:
yes
Explanation:
because gravity moves things
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
what were your preparetion before going the different physical fitness test?
Answer:
Avoid heavy strenuous exercise for the 24 hours prior to testing. Do not exercise at all on the day of testing to ensure you are well rested. Wear appropriate clothing for the conditions (e.g. shorts/track pants and t-shirt/singlet/sports top) and non-slip athletic footwear with laces securely fastened
2. Conner flips a coin up in the air (to determine if he or his sister needs to do the dishes) at an upward velocity of 4.00 m/s. He fails to catch it on the way down and it falls down an additional 1.6 m to the floor. Calculate the maximum height and maximum magnitude (could be positive or negative) velocity of the coin.
Answer:
5.6
Explanation:
Not so sure
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 are some physical factors we deal with for sleeping?
Answer:
we deal our body movement during sleeping
1- A soccer ball was kicked from the ground with an initial speed of 12m/s at an
angle 32° above the horizontal. What are the x and y positions of the ball
0.50s after it is kicked?
Answer:
Y = 0.5 × 6.36 m/s = 3.18 meters
X = 0.5 × 10.18 m/s = 5.09 meters
Explanation:
X and Y components:
Y = 6.36 m/s
X = 10.18 m/s
Position after 0.5 seconds:
Y = 0.5 × 6.36 m/s = 3.18 meters
X = 0.5 × 10.18 m/s = 5.09 meters
The x and y positions of the ball 0.50s after it is kicked are 5.09 meter and 1.95 meter respectively.
What is velocity?
The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction. SI unit of velocity is meter/second.
Given parameters:
Initial speed of the soccer ball at an angle 32° above the horizontal is: u = 12m/s.
Time interval: t = 0.50 second.
Now, x component of initial velocity = u cosθ = 12 × cos 32° m/s = 10.17 m/s.
y component of initial velocity = u sinθ = 12 × sin32° =6.36 m/s.
Hence,
after 0.50 s, x-position of the ball = u cosθ t
= 10.17 × 0.50 meter
= 5.09 meter.
after 0.50 s, y-position of the ball = u sinθ t -gt²/2
= 6.36 × 0.50 - (9.8×0.50²/2) meter
= 1.95 meter.
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Question below...............
Answer:
friction force
Explanation:
force of friction is opposite to the force applied it resist the motion
A 12.0 kg box resting on a horizontal, frictionless surface is attached to a 5.00 kg weight by a thin, light wire that passes over a frictionless pulley (Figure 1). The pulley has the shape of a uniform solid disk of mass 2.60 kg and diameter 0.500 m. A) After the system is released, find the tension in the horizontal segment of the wire. B) After the system is released, find the tension in the vertical segment of the wire. C) After the system is released, find the x-component of the acceleration of the box. D) After the system is released, find the x- and y-components of the force that the axle exerts on the pulley.
Answer:
Explanation:
Assuming x direction is horizontal any y direction is vertical
As the moment of inertia of the pulley is I = (½)mR²
The pulley mass will appear as ½ its actual mass in the system.
F = ma
5.00(9.81) = (5.00 + 2.60/2 + 12.0)a
a = 49.05 / 18.3
a = 2.6803278...
a = 2.68 m/s²
A) T = 12.0(2.68) = 32.16... = 32.2 N
B) T = 5.00(9.81 - 2.68) = 35.65 = 35.7 N
C) aₓ = 2.68 m/s²
D) Fx = 32.2 N
Fy = 35.7 + 2.60(9.81) = 61.2 N
A)The tension in the horizontal segment of the wire will be 32.2 N
B)The tension in the vertical segment of the wire will be 35.7 N
What is acceleration?Acceleration is defined as the rate of change of the velocity of the body. Its unit is m/sec².It is a vector quantity. It requires both magnitudes as well as direction to define.
The given data in the problem is;
m is the mass box resting on a horizontal =12.0 kg
the uniform solid disk of mass of 2.60 kg
d is the diameter = 0.500 m.
A)The tension in the horizontal segment of the wire will be 32.2 N
The acceleration is found as;
[tex]\rm F = ma \\\\ a= \frac{49.05}{18.3} \\\\ \rm a= 2.68 \ m/sec^2[/tex]
The tension in the horizontal segment of the wire is found as;
T=ma
T= 12.0 ×2.68
T=32.2 N
Hence the tension in the horizontal segment of the wire will be 32.2 N
B)The tension in the vertical segment of the wire will be 35.7 N
T=m(g-a )
T= 5.00 × (9.8-2.68)
T=35.7 N
Hence the tension in the vertical segment of the wire will be 35.7 N
C) the x-component of the acceleration of the box. will be 2.68 m/s².
aₓ = 2.68 m/s²
D)The x- and y-components of the force that the axle exerts on the pulley will be 32.2 and 61.2 N respectively.
Fx = 32.2 N
Fy = 35.7 + 2.60(9.81) = 61.2 N
Hence the x- and y-components of the force that the axle exerts on the pulley will be 32.2 and 61.2 N respectively.
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