Answers
Answer: This is not easy lol
Explanation:
Related Questions
Which change will always result in an increase in the gravitational force between two objects?
O increasing the masses of the objects and increasing the distance between the objects
O decreasing the masses of the objects and decreasing the distance between the objects
O increasing the masses of the objects and decreasing the distance between the objects
• decreasing the masses of the objects and increasing the distance between the objects
Answers
Answer:
increasing the masses of the objects and decreasing the distance between the objects
Explanation:
Pls quickly brainliest to the first to anwser
Answers
Answer:
8m/s^2
Explanation:
hope it helps........
Explanation:
you're supposed to know the formula of acceleration which is velocity of a time then you can solve the question
Scientists have concluded that the uppermost part of the mantle is partially-molten. Which observation helped them reach this conclusion?
Answers
Answer:
P and S waves slow down when they reach this layer. The asthenosphere, also known as the magma chamber, is the uppermost component of the mantle. This layer is partially molten and is a ductile zone in a tectonically poor state.
It's almost hard and seismic waves move through the asthenosphere at a slow rate. The fragile lithosphere and the uppermost portion of the asthenosphere are assumed to be rigid.
seismic waves travel more quickly through denser materials and therefore generally travel more quickly with the depth it moves more slowly through a liquid than a solid. Molten areas within the Earth slow down P waves and stop S waves because their shearing motion cannot be transmitted through a liquid. Partially molten areas may slow down the P waves and attenuate or weaken S waves.
hope this helps...
S and P wave slow down and stop in the uppermost part of the mantle. - For this, scientists have concluded that the uppermost part of the mantle is partially-molten.
What is mantle?A planetary body's mantle is a layer that is surrounded by the crust on top and the core underneath. The largest and most substantial layer of a planetary body, mantles are often comprised of rock or ice. Planetary bodies that have undergone density differentiation typically have mantles. Mantles are found on all terrestrial planets (including Earth), many asteroids, and a few planetary moons.
Between the crust and the outer core, there is a silicate rock layer known as the Earth's mantle. Despite being mostly solid, it behaves like a viscous fluid over geological time. Oceanic crust is created by the partial melting of the mantle at mid-ocean ridges, and continental crust is created by the partial melting of the mantle at subduction zones.
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Two identical loudspeakers are driven in phase by the same amplifier. The speakers are positioned a distance of 3.2 m apart. A person stands 5.0 m away from one speaker and 6.2 m away from the other. Calculate the second lowest frequency that results in destructive interference at the point where the person is standing. Assume the speed of sound to be 330 m/ s. A) 183 Hz B) 275 Hz C) 413 Hz D) 137 Hz E) 550 Hz
Answers
Answer:
C) 413 Hz
Explanation:
For destructive interference, the path difference ΔL = (n + 1/2)λ where ΔL = L₂ - L₁ where L₁ = person's distance from one speaker (the closer one) = 5.0m and L₂ = person's distance from other speaker (the farther one) = 6.2 m and λ = wavelength = v/f where v = speed of sound = 330 m/s and f = frequency
So, ΔL = (n + 1/2)λ
L₂ - L₁ = (n + 1/2)v/f
f = (n + 1/2)v/(L₂ - L₁)
At the second lowest frequency that results in destructive interference at the point where the person is standing, n = 1.
So,
f = (1 + 1/2)v/(L₂ - L₁)
f = 3v/2(L₂ - L₁)
Substituting the values of the variables into the equation, we have
f = 3v/2(L₂ - L₁)
f = 3(330 m/s)/2(6.2 m - 5.0 m)
f = 3(330 m/s)/2(1.2 m)
f = 990 m/s ÷ 2.4 m)
f = 412.5 Hz
f ≅ 413 Hz
A hand dryer blows heated air downwards out of the exit duct at a velocity of 4 m/s. The temperature and density of the ambient air at the inlet are 15 C and 1.23 kg/m3, while at the outlet it has temperature 35 C and density 1.15 kg/m3 The blower power is 10.0 W and the heater power is 715 W. Consider the inlet to be at the large mass of ambient air which has negligible velocity.
a) What is the pressure at the outlet? 4 m/s, 35 C
b) You will be applying the energy equation. Why can you ignore any height differences in this situation?
c) If the specific heat of air C-1000 J/(kg K), where Δυ-C Δ T, find the change in internal energy per unit mass from the inlet to outlet.
d) Find the mass flow rate through the dryer.
e) What is the power loss in the system?
f) What is the loss in the system?
g) What is the head loss in the system?
h) What is the total loss coefficient of the system, referred to the outlet velocity?
i) If there were no heater, would the temperature of gas at the outlet be higher, the same, or lower than the inlet? Explain why.
Answers
Answer:
nzkdjdksishdjsdjjdjnzkskejeoueeuieeoyrie ryrhdhcksodopdncndnszdkhfoeosheiwuef wokxkzdkjdoeehxjbxbdkeiehdhdhddjjddjdhhdhdhhhjdhfjdjjfjddhdheudiwiehdbejwowud
Explanation:
isos
Please solve for 15 points. Please don’t input a link.
Answers
Answer:
a). Single replacement.
Explanation:
Because one element replaces another element in a compound
types of wave interactions include
Answers
Refraction
Diffraction
Hope this helps :))
(Need help pls)!!!!!
Answers
Answer:
I assume they want answer A, but it's a silly question.
Explanation:
Well, if we assume that the girls are applying equal force, as they are the same size and strength (apparently also the same resolve etc.).
Equal force in opposite directions means the composite force is null. An object that no force is applied to continues moving in the same direction, so if the doll was moving somewhere, it would keep doing so, except all of this makes no sense the doll is not in gravityless vacuum what are we even talking about.
The answer A is the least ridiculous.
At noon, ship A is 110 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 4:00 PM
Answers
Answer:
[tex]4.47\ \text{km/h}[/tex]
Explanation:
[tex]\dfrac{da}{dt}[/tex] = Rate at which the distance between A and starting point of B is changing = -20 km/h
[tex]\dfrac{db}{dt}[/tex] = Rate at which the distance of B is changing = 15 km/h
[tex]\dfrac{dc}{dt}[/tex] = Rate at which the distance between A and B is changing
Time after which the rate at which the distance between A and B is changing is 4 hours
Distance covered by A in 4 hours = [tex]20\times 4=80\ \text{km}[/tex]
a = Distance remaining to the start point of B = [tex]110-80=30\ \text{km}[/tex]
b = Distance covered by B in 4 hours = [tex]15\times 4=60\ \text{km}[/tex]
Distance between A and B after 4 hours
[tex]c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}[/tex]
[tex]c^2=a^2+b^2[/tex]
Differentiating with respect to time we get
[tex]c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}[/tex]
The rate at which the distance between the ships is changing at 4 PM is [tex]4.47\ \text{km/h}[/tex].
When you look at the backside of a shiny teaspoon held at arm's length, do you see yourself upright or upside down? (b) When you look at the other side of the spoon, do you see yourself upright or upside down? Assume in both cases that the distance between you and the spoon is greater than the focal length of the spoon.
Answers
Answer:
a) The back spoon gives a right image (upright)
b) the front gives an inverted image
Explanation:
The spoon is a curved metallic object, when we see ourselves from the back we have a convex mirror, in this type of mirror when the law of reflection is applied the rays diverge therefore the eye-brain system forms the image with the prolongation of the rays, therefore the image is straight and smaller than the object.
When we look through the deep side of the spoon, we have a concave mirror and as the object (we) is further away than the distance, the rays converge to a point, so the image is real, inverted smaller than the object.
In summary.
a) The back spoon gives a right image (upright)
b) the front gives an inverted image
A 0.70-kg disk with a rotational inertia given by MR 2/2 is free to rotate on a fixed horizontal axis suspended from the ceiling. A string is wrapped around the disk and a 2.0-kg mass hangs from the free end. If the string does not slip then as the mass falls and the cylinder rotates the suspension holding the cylinder pulls up on the mass with a force of______
Answers
Answer:
The force will be "9.8 N".
Explanation:
The given values are:
mass,
m = 0.7 kg
M = 2
g = 9.8
Now,
⇒ [tex]\tau = T \alpha[/tex]
then,
⇒ [tex]\frac{1}{2}mR^2(\frac{1}{R}\frac{dv}{dt}) =M(g-a_t)R[/tex]
⇒ [tex]\frac{1}{2}m \ a_t=m(g-a_t)[/tex]
⇒ [tex]a_t=\frac{2g}{(\frac{m}{M} +2)}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{2\times 9.8}{\frac{0.7}{2} +2}[/tex]
⇒ [tex]=8.34 \ m/s[/tex]
hence,
⇒ [tex]T=mg+M(g-a_t)[/tex]
On substituting the values, we get
⇒ [tex]=0.7\times 9.8+2(9.8-8.34)[/tex]
⇒ [tex]=6.86+2(1.46)[/tex]
⇒ [tex]=6.86+2.92[/tex]
⇒ [tex]=9.8 \ N[/tex]
What is known as the amount of mass in
a given volume?
A. density
B. pressure
C. surface tension
D. mass pressure
Answers
Light rays enter a transparent material. Which description best describes what happens to the light rays
Answers
When light encounters transparent materials, almost all of it passes directly through them. Glass, for example, is transparent to all visible light. ... Most of the light is either reflected by the object or absorbed and converted to thermal energy. Materials such as wood, stone, and metals are opaque to visible light.
What is a transfer of energy called?
A. Displacement
B. Acceleration
C. Work
D. Torque
Answers
a train has an initial velocity of 30 m/s. If the train accelerates uniformly at a rate of 6.3 m/s ^ for 2.8 seconds what is the trains final velocity?
Answers
T
Answer:
the velocity is a second final to initial velocity of 39
Which runner finished the 100 m race in the least amount of time?
Ming
Which runner stopped running for a few seconds during the race?
At what distance did Anastasia overtake Chloe in the race?
Answers
1: Ming
2: Chloe
3: 40m
A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If both objects have the same kinetic energy, what is the ratio of the speed of the cylinder to the speed of the spherical shell
Answers
Answer:
vcyl / vsph = 1.05
Explanation:
The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.The traslational part can be written as follows:[tex]K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)[/tex]
The rotational part can be expressed as follows:[tex]K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)[/tex]
where I = moment of Inertia regarding the axis of rotation.ω = angular speed of the rotating object.If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:[tex]v = \omega * R (3)[/tex]
For a solid cylinder, I = M*R²/2 (4)Replacing (3) and (4) in (2), we get:[tex]K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)[/tex]
Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:[tex]K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)[/tex]
Repeating the same steps for the spherical shell:[tex]I_{sph} = \frac{2}{3} * M* R^{2} (7)[/tex]
[tex]K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)[/tex]
[tex]K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)[/tex]
Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.Rearranging, and taking square roots on both sides, we get:[tex]\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)[/tex]
This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.The Brazilian rain forest is an area with significant biodiversity. As the rain forest is replaced with agricultural land, it is reasonable to predict a reduction in -
Answers-
A: consumption of solar energy.
B: sustainability over time.
C: precipitation levels.
D: average daily temperature.
Answers
Answer:
Bb
Explanation:
Daryl ties a rope to a brick and lifts the brick straight up. The free-body
diagram below shows the brick when it is suspended above the ground.
Force 1
Force 2
What is force 1 in this diagram?
O A. Friction
OB. Tension
O C. Normal force
O D. Weight
Answers
The force 1 is tension force.
To find the correct statement among all the options, we need to know more about friction, tension, normal force and weight.
What is friction?Friction force is found between two surfaces when one is kept or moved on another surface.It is directed opposite to the direction of motion.What is tension force?When any object is hanged by an thread or rope, that object exerts a force on that rope. This force is called as tensional force.It's directed from along the rope towards the point of hanging.What is normal force?When an object is kept on a surface, the surface exerts a force on the object to oppose the weight of the object which is the normal force.It's perpendicular to the surface that an object contacts.What is weight?Weight is the gravitational force exerted by earth on that object. It's always directed towards the center of the earth.Thus, we can conclude that the correct option is (B).
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If an athlete runs the triathlon of 10 km in 2 hours, what is her average speed in kilometers per hour?
Answers
Answer: 5 km per hour
Explanation:
if in 10 km there is 2 hours, then 10 divided by 2 is 5.
Find the wavelength of light which is capable of ionizing a hydrogen atom?
Answers
Answer:
The correct answer is - 91.4 nm
Explanation:
According to Bohr's model, the minimum wavelength to ionize Hydrogen atom from n= 1 state is expressed as:
(h×c)/λ=13.6eV
here,
h - Planck constant
c - the speed of light
λ - wavelength
Placing the value in the formula for the wavelength
(6.626×10^−34J.s × 3×10^8 m/s)/λ = 13.6 ×1.6 × 10^−19 J
λ≈91.4nm
Thus, the correct answer would be = 91.4 nm
explain the refraction of light on a glass slab
Answers
Answer:
refraction is the change in direction of a wave passing from one medium to another or from a gradual change in the medium.
If a 75 W lightbulb is 15% efficient, how many joules of light energy does the bulb produce every minute?
Answers
Answer:
1 W = 1 J / sec Definition of watt is 1 joule / sec
So if a bulb uses 75 J / sec it must use
75 J/s * 60 sec / min = 4500 J/min energy used by bulb
If bulb is 15% efficient then the light delivered is
P = 4500 J / min * .15 = 675 J / min
Two experiments are performed on an object to determine how much the object resists a change in its state of motion while at rest and while in motion. In the first experiment, the object is pushed with a constant known force along a horizontal surface. There is negligible friction between the surface and the object. A motion sensor is used to measure the speed of the object as it is pushed. In a second experiment, the object is tied to a string and pulled upward with a constant known force, and a motion sensor is used to measure the speed of the object as it is pulled upward. The student uses the data collected from the motion sensor to determine the mass of the object in both experiments.
Required:
What classifies the type of mass that was determined in each experiment?
Answers
Answer:
In the first experiment, the mass is inertial mass and in the second experiment, the mass is a gravitational mass.
Explanation:
It is given that a student performs two types of experiment to see how change in its resistance while in the state of motion and in rest.
In the first experiment, an object is pushed with a force against a horizontal surface and the speed is measured using a sensor. Here, work is done against the inertia of the object as it is pushed from rest. So the mass is inertial mass.
In the second experiment, an object is pushed or thrown upwards with a force and speed is measured. Here, the mass is gravitational mass as the work done in the second experiment is against the gravity or against the weight of the object.
In the first experiment, the mass is inertial mass and in the second experiment, the mass is a gravitational mass.
As per the given problem, the student performs two types of experiment to see how change in its resistance while in the state of motion and in rest.
In the first experiment, an object is pushed with a force against a horizontal surface and the speed is measured using a sensor. Here, work is done against the inertia of the object as it is pushed from rest. So the mass is inertial mass. In the second experiment, an object is pushed or thrown upwards with a force and speed is measured. Here, the mass is gravitational mass as the work done in the second experiment is against the gravity or against the weight of the object.Thus, we can conclude that the in the first experiment, the mass is inertial mass and in the second experiment, the mass is a gravitational mass.
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An object, with mass 64 kg and speed 14 m/s relative to an observer, explodes into two pieces, one 2 times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer's reference frame
Answers
Answer:
K_f = 1881.6 J
Explanation:
To solve this exercise, let's start by finding the velocities of the bodies.
We define a system formed by the initial object and its parts, with this the forces during the explosion are internal and the moment is conserved
initial instant. Before the explosion
p₀ = M v₀
final instant. After the explosion
p_f = m₁ v + m₂ 0
the moeoto is preserved
p₀ = p_f
M v₀ = m₁ v
v = [tex]\frac{m_1}{M}[/tex] v₀
in the exercise they indicate that the most massive part has twice the other part
M = m₁ + m₂
M = 2m₂ + m₂ = 3 m₂
m₂ = M / 3
so the most massive part is worth
m₁ = 2 M / 3
we substitute
v = ⅔ v₀
with the speed of each element we can look for the kinetic energy
initial
K₀ = ½ M v₀²
Final
K_f = ½ m₁ v² + 0
K_f = ½ (⅔ M) (⅔ v₀)²
K_f = [tex]\frac{8}{27}[/tex] (½ M v₀²)
K_f = [tex]\frac{8}{27}[/tex] K₀
the energy added to the system is
ΔK = Kf -K₀
ΔK = (8/27 - 1) K₀
ΔK = -0.7 K₀
K_f = K₀ + ΔK
K_f = K₀ (1 -0.7)
K_f = 0.3 K₀
let's calculate
K_f = 0.3 (½ 64 14²)
K_f = 1881.6 J
A 64.0 cm long cord is vibrating in such a manner that it forms a standing wave with two antinodes. (The cord is fixed at both ends.) Which harmonic does this wave represent
Answers
Answer:
the wave represents the second harmonic.
Explanation:
Given;
length of the cord, L = 64 cm
The first harmonic of a cord fixed at both ends is given as;
[tex]f_o = \frac{V}{2L}[/tex]
The wavelength of a standing wave with two antinodes is calculated as follows;
L = N---> A -----> N + N ----> A -----> N
Where;
N is node
A is antinode
L = N---> A -----> N + N ----> A -----> N = λ/2 + λ/2
L = λ
The harmonic is calculated as;
[tex]f = \frac{V}{\lambda} \\\\f = \frac{V}{L} = 2(\frac{V}{2L} ) = 2(f_o) = 2^{nd} \ harmonic[/tex]
Therefore, the wave represents the second harmonic.
L = λ
Light containing two different wavelengths passes through a diffraction grating with 1,250 slits/cm. On a screen 17.5 cm from the grating, the third-order maximum of the shorter wavelength falls midway between the central maximum and the first side maximum for the longer wavelength. If the neighboring maxima of the longer wavelength are 8.44 mm apart on the screen, what are the wavelengths in the light
Answers
Answer:
[tex]\lambda_s =6.43*10^-4m[/tex]
Explanation:
From the question we are told that:
Diffraction grating [tex]N=1250slits/cm[/tex]
Distance b/w Screen and grating length [tex]d_{sg}=17.5 cm[/tex]
Distance b/w neighboring maxima and Screen [tex]d_{ms}=8.44[/tex]
Generally the equation for grating space is mathematically given by
[tex]d(g)=\frac{1}{N}[/tex]
[tex]d(g)=\frac{100}{1250}[/tex]
[tex]d(g)=0.08[/tex]
Generally the equation for small angle approximation is mathematically given by
[tex]\triangle y=\frac{\lambda d}{L}[/tex]
Therefore for longest wavelength
[tex]\lambda _l=\frac{8.44*10^{-3}*(0.08)}{0.175m}[/tex]
[tex]\lambda _l=3.858*10^{-3}[/tex]
Therefore the third order maximum equation for the shorter wavelength as
[tex]\lambda_s =\frac{1}{6} \lambda_l[/tex]
[tex]\lambda_s =\frac{1}{6} (3.858*10^-^3)[/tex]
[tex]\lambda_s =6.43*10^-4m[/tex]
The wavelengths in the light is given as
[tex]\lambda_s =6.43*10^-4m[/tex]
Please answer this for 15 points please don’t put in a link.
Answers
Answer:
c. Double Replacement
Explanation:
As in Double Replacement reaction exchanges the cations (or the anions) of two ionic compounds.
Here, in BaCl2 , Ba has replaced with NO3 to form Ba(NO3)2
and in 2AgNo3 , Ag has replaced with Cl to form 2AgCl.
In a nuclear fusion reaction, atoms:
split apart.
combine.
explode.
cool down.
Answers
a Ferris wheel with a diameter of 35 m starts from rest and achieves its maximum operational tangential speed of 2.3 m/s in a time of 15 s. what is the magnitude of the wheels angular acceleration?
b. what is the magnitude of the tangential acceleration after the maximum operational speed is reached?
Answers
ω = v / r = 2.20m/s / ½*35.0m = 0.126 rad/s
and so prior to reaching that final speed
the angular acceleration
α = Δω / Δt = 0.126rad/s / 15.0s = 0.00838 rad/s² ◄
and the tangential acceleration
a = Δv / Δt = 2.20m/s / 15.0s = 0.147 m/s² ◄
OR
a = αr = 0.00838rad/s² * ½*35.0m = 0.147 m/s² ◄
(a) The magnitude of the wheels angular acceleration is 0.0088 rad/s².
(b) The magnitude of the tangential acceleration after the maximum operational speed is reached is 0.153 m/s².
Angular acceleration of the wheel
The angular acceleration of the wheel is calculated as follows;
α = ω/t
ω = v/r
α = v/(rt)
α = (2.3)/(17.5 x 15)
α = 0.0088 rad/s²
Tangential acceleration of the wheela = v/t
a = (2.3)/15
a = 0.153 m/s²
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