The answer is compression because when you are squishing a marshmello you are compressing it.
Imagine you are in a small boat on a small pond that has no inflow or outflow. If you take an anchor that was sitting on the floor of the boat and lower it over the side until it sits on the ground at the bottom of the pond, will the water level rise slightly, stay the same, or lower slightly
Answer:
The pond's water level will fall.
Explanation:
Archimedes principle explains that a floating body will displace the amount of water that weighs the same as it, whereas a body resting on the bottom of the water displaces the amount of water that is equal to the body's volume.
When the anchor is in the boat it is in the category of floating body and when it is on the bottom of the pond it is in the second category.
Since anchors are naturally heavy and denser than water, the amount of water displaced when the anchor is in the boat is greater than the amount of water displaced when the anchor is on the bottom of the pond since the way anchors are doesn't make for them to have considerable volume.
When the anchor is dropped to the bottom of the pond, the water level will therefore fall. If the anchor doesn't reach the bottom it is still in the floating object category and there will be no difference to the water level, but once it touches the bottom of the pond, the water level of the pond drops.
Select the correct answer.
Two charged objects, A and B, are exerting an electric force on each other. What will happen if the charge on A is increased?
Which of the following absorbs the energy required by photosynthesis?
Answer:
There are no options, so....
Explanation:
Chlorophyll a absorbs its energy from the Violet-Blue and Reddish orange-Red wavelengths, and little from the intermediate (Green-Yellow-Orange) wavelengths.
Hannah tests her new sports car by racing with Sam, an experienced racer. Both start from rest, but Hannah leaves the starting line 1.00 s after Sam does. Sam moves with a constant acceleration of 3.50 m/s2, while Hannah maintains an acceleration of 4.90 m/s2. Find (a) the time at which Hannah overtakes Sam, (b) the distance she travels before she catches him, and (c) the speeds of both cars at the instant Hannah overtakes Sam.
Answer:
a) t = 6.46 s, b) x = 72.98 m, c) v₁ = 26.75 m / s, v₂ = 22.61 m / s
Explanation:
This is an exercise in kinematics, let's write the expressions for each person
Hanna
leaves a time t₀o = 1s after Sam's output, both with zero initial velocity and acceleration of a₁ = 4.90 m / s²
x₁ = 0 + ½ a₁ (t-t₀) ²
v₁ = 0 + a₁ (t-t₀)
Sam
with an acceleration of a₂ = 3.50 m / s² and with an initial velocity of zero
x₂ = 0+ ½ a₂ t²
v₂ = 0 + a₂ t
a) at the point where the position of the two is found is the same
x₁ = x₂
½ a₁ (t-t₀) ² = ½ a₂ t²
let's solve
t-t₀ = [tex]\sqrt{\frac{a_2}{a_1} }[/tex] t
t (1 - [tex]\sqrt{ \frac{a_2}{a_1} }[/tex]) = t₀
t = [tex]\frac{t_o}{ 1-\sqrt{ \frac{a_2}{a_1} } }[/tex]
let's calculate
t = [tex]\frac{ 1}{1- \sqrt{\frac{3.50}{4.90} } }[/tex]
t = [tex]\frac{1}{1- 0.845}[/tex] 1 / 1- 0.845
t = 6.46 s
b) the distance traveled is
x = ½ a₂ t²
x = ½ 3.5 6.46²
x = 72.98 m
c) Hanna's speed
v₁ = 4.9 (6.46 -1)
v₁ = 26.75 m / s
sam's speed
v₂ = a2 t
v₂ = 3.50 6.46²
v₂ = 22.61 m / s
Si un electrón recorre el acelerador lineal de Stanford de 2 millas de longitud a 99% de la velocidad de la luz, ¿Cuál es la longitud del acelerador según el electrón?
Calculate the energy in electron volts of X-rays that have a frequency of 6.00 x 1016 Hz.
207 eV
228 eV
249 eV
Answer:
228
Explanation:
What is the voltage drop across an alarm clock that is connected to a circuit with a current of 1.10A and a resistance of 90Ω?
V = 99 volts
Explanation:
The voltage drop can be calculated using Ohm's law:
V = IR
= (1.10 A)(90 Ω)
= 99 volts
Explain how ultrasound devices use the Doppler effect to create images of body parts.
HELP ME ASAP!! Will give brainliest answer!!
Answer:
Transmits high-frequency (1 to 5 megahertz) sound pulses into your body using a probe. The sound waves travel into your body and hit a boundary between tissues (e.g. between fluid and soft tissue, soft tissue and bone).
Explanation:
Doppler ultrasound works by measuring sound waves that are reflected from moving objects, such as red blood cells.
Answer:
The guy above is pretty good
Explanation:
I'd go with that answer, give them brainliest
A 500-nm wavelength light in vacuum illuminates a soap film with an index of refraction of 1.33. Air (n=1.00) is on both sides of the film. If the light strikes the film nearly perpindicularly, what is the smallest film thickness such that the film appears bright?
ANS --> 94.0
Please show your work as to how to end up with this answer.
Answer:
Wavelength of light in film (let y = lambda)
y = 500 nm / (4/3) = 375 nm
There will be a phase change at the air/film interface (not the other side)
S = 4 t thickness of film = S/4 where S equals 1 wavelength
This is because of the phase change at one surface
375 nm = 4 * t
t = 93.8 nm
a charge particle moves along a circle under the action of possible electric and magnetic field
Answer:
The correct answer is - B ≠ 0, E = 0.
Explanation:
A force q E is exerted by the electric field on the charged particle that accelerates always, that is, it increases the speed of the particle. The particle can never be rotated in a rotation in the circle in an electric field.
The magnitude of the velocity cannot be changed by the magnetic field but changes only the velocity of the direction.
If the particle moves in a circle it means that the speed should begin for a year. The only direct velocity is constant and only the change be b can be achieved. So, B ≠ 0 and E = 0.
Imagine an alternate universe where all of the quantum number rules were identical to ours except m_{s} had three allowed values (rather than two as it does in our universe). If this were the case, and the Pauli Exclusion Principle still applies, how many electrons would be allowed in each orbital
Answer:
so in a given orbital there can be 3 electrons.
Explanation:
The Pauli exclusion principle states that all the quantum numbers of an electron cannot be equal, if the spatial part of the wave function is the same, the spin part of the wave function determines how many electrons fit in each orbital.
In the case of having two values, two electrons change. In the case of three allowed values, one electron fits for each value, so in a given orbital there can be 3 electrons.
A thermodynamic system consists of an ideal gas at a volume of 3.50 L and initial pressure of 6.2 × 104 Pa. As the volume is held constant, the pressure is increased to 8.2 × 104 Pa. What work is involved in this process?
Answer:
0 J
Explanation:
Since work done W = PΔV where P = pressure and ΔV = change in volume.
Since the volume is constant, ΔV = 0
So, Work done, W = PΔV = P × 0 = 0 J
So, the work done is 0 J.
If a car travelled for 20 m/s and then 10 seconds later it was moving 10 m/s, its acceleration would be?
Answer:
This is the answer. hope this help u
What are moana's hobbies
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dipole moment are used to calculate the
Answer:
Explanation:
ádasdasdasd
If two charged balloons are 24cm apart and they feel a force of electrical repulsion of 20N, what would the force of electrical repulsion become if the balloons were moved closer together
to a distance of 4cm?
Answer:
Soory
Explanation:
I really dont know but i will send you wait
The plum pudding model of the atom states that
Answer:
According to this model, the atom is a sphere of positive charge, and negatively charged electrons are embedded in it to balance the total positive charge.
Explanation:
Hope this helps you
Answer:
The plum pudding model of the atom states that had negatively-charged electrons embedded within a positively-charged "soup."
Explanation:
Thomson's plum pudding model of the atom had negatively-charged electrons embedded within a positively-charged "soup." Rutherford's gold foil experiment showed that the atom is mostly empty space with a tiny, dense, positively-charged nucleus. Based on these results, Rutherford proposed the nuclear model of the atom.
The 2nd maximum of a double slit diffraction pattern makes an angle 20 degrees when a wavelength of 600 nm is used. What is the angle for the 5th maximum when the slit separation d is increased to 1.5d keeping the wavelength the same. g
Answer:
θ = 34.77°
Explanation:
From diffraction equation:
[tex]m\lambda = dSin\theta[/tex]
where,
m = order of diffraction
λ = wavelength of light used
d = slit separation
θ = angle
Therefore, for initial case:
m = 2
λ = 600 nm = 6 x 10⁻⁷ m
d = slit seperation = ?
θ = angle 20°
Therefore,
[tex](2)(6\ x\ 10^{-7}\ m)=d(Sin\ 20^o)\\\\d = \frac{12\ x 10^{-7}\ m}{0.342}\\\\d = 3.5\ x\ 10^{-6}\ m[/tex]
Now, for the second case:
m = 5
λ = 600 nm = 6 x 10⁻⁷ m
d = slit seperation = (1.5)(3.5 x 10⁻⁶ m) = 5.26 x 10⁻⁶ m
θ = angle = ?
Therefore,
[tex](5)(6\ x\ 10^{-6}\ m) = (5.26\ x\ 10^{-6}\ m)Sin\theta\\\\Sin\theta = \frac{(5)(6\ x\ 10^{-7}\ m)}{(5.26\ x\ 10^{-6}\ m)}\\\\\theta = Sin^{-1}(0.5703)[/tex]
θ = 34.77°
help me out guys
I'm stuck in this
Answer: The answer will be R1 = B
Explanation:
Cause Here x component of A = A
X component of 2B = 2BcosФ
X component of Resultant = 0
I.e. A + 2BcosФ = 0
cosФ = -A/2B
also R1 = √(A² + B² + 2ABcosФ)
=√ [A² + B² + 2AB(-A/2B)]
=√A² + B² -A²
R1 = √B²
R1 = B so R1 = B is correct.
A gauge is attached to a pressurized nitrogen tank reads a gauge pressure of 28 in of mercury. If atmospheric pressure is 14.4 psia, what is the absolute pressure in the tank
Answer:
The absolute pressure is 28.15 psi.
Explanation:
Gauge pressure = 28 inch of Mercury
Absolute pressure, Po = 14.4 psi
The absolute pressure is the sum of the gauge pressure and the absolute pressure.
gauge pressure = 28 inch = 0.7112 m of Mercury
= 0.7112 x 13.6 x 1000 x 9.8 = 94788.736 Pa
= 13.75 psi
The absolute pressure is
P = 14.4 + 13.75 = 28.15 psi
3. A 5 gm/100 ml solution of drug X is stored in a closed test tube
at 25°C. If the rate of degradation of the drug is 0.05 day-1,
calculate the time required for the initial concentration to
drop to (a) 50% (half-life) and (b) 90% (shelf-life) of its initial
value.
Answer:
See explanation
Explanation:
The degradation of the drug is a first order process;
Hence;
ln[A] = ln[A]o - kt
Where;
ln[A] = final concentration of the drug
ln[A]o= initial concentration of the drug = 5 gm/100
k= degradation constant = 0.05 day-1
t= time taken
When [A] =[ A]o - 0.5[A]o = 0.5[A]o
ln2.5 = ln5 - 0.05t
ln2.5- ln5 = - 0.05t
t= ln2.5- ln5/-0.05
t= 0.9162 - 1.6094/-0.05
t= 14 days
b) when [A] = [A]o - 0.9[A]o = 0.1[A]o
ln0.5 = ln5 -0.05t
t= ln0.5 - ln5/0.05
t= -0.693 - 1.6094/-0.05
t= 46 days
A student is comparing the speed of sound in air and water. She measures the frequency, f, and wavelength, λ, of sound waves in both air and water. The results are shown in the following table. Based on her data, how does the speed of sound in water compare to the speed of sound in air?
A.The speed of sound through water is 4.3 times faster than sound through air.
B.The speed of sound through water is 2.6 times slower than sound through air.
C.The speed of sound through water is 8.4 times faster than sound through air.
D.The speed of sound through air is approximately equal to the speed of sound through water.
Answer:
Option A. The speed of sound through water is 4.3 times faster than sound through air.
Explanation:
To answer the question correctly, we shall determine the speed of the wave in both cases. This is illustrated below:
For Air:
Frequency (fₐ) = 195 Hz
Wavelength (λₐ) = 1.76 m
Velocity (vₐ) =?
vₐ = λₐfₐ
vₐ = 1.76 × 195
vₐ = 343.2 m/s
For Water:
Frequency (fᵥᵥ) = 195 Hz
Wavelength (λᵥᵥ) = 7.6 m
Velocity (vᵥᵥ) =?
vᵥᵥ = λᵥᵥfᵥᵥ
vᵥᵥ = 7.6 × 195
vᵥᵥ = 1482 m/s
Finally, we shall compare the speed in water to that of air. This can be obtained as follow:
Velocity in air (vₐ) = 343.2 m/s
Velocity in water (vᵥᵥ) = 1482 m/s
Water : Air
vᵥᵥ : vₐ =>
vᵥᵥ / vₐ = 1482 / 343.2
vᵥᵥ / vₐ = 4.3
Cross multiply
vᵥᵥ = 4.3 × vₐ
Thus, the speed in water is 4.3 times the speed in air.
Option A gives the correct answer to the question.
A mass with a charge of 4.60 x 10-7 C rests on a frictionless surface. A compressed spring exerts a force on the mass on the left side. Three centimeters to the right of the mass is a 7.50 x 10-7 C fixed charge. How much is the spring compressed if its spring constant is 14 N/m
Answer:
The compression of the spring is 24.6 cm
Explanation:
magnitude of the charge on the left, q₁ = 4.6 x 10⁻⁷ C
magnitude of the charge on the right, q₂ = 7.5 x 10⁻⁷ C
distance between the two charges, r = 3 cm = 0.03 m
spring constant, k = 14 N/m
The attractive force between the two charges is calculated using Coulomb's law;
[tex]F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(4.6\times 10^{-7})(7.5\times 10^{-7})}{(0.03)^2} \\\\F= 3.45 \ N[/tex]
The extension of the spring is calculated as follows;
F = kx
x = F/k
x = 3.45 / 14
x = 0.246 m
x = 24.6 cm
The compression of the spring is 24.6 cm
A meterstick of negligible mass is placed on a fulcrum at the 0.4 m mark, with a 1 kg mass hung at the zero mark and a 0.5 kg mass hung at the 1.0 m mark. The meterstick is held horizontal and released. Immediately after release, the magnitude of the net torque on the meterstick about the fulcrum is most nearly:________
a. 1 Nm
b. 2 Nm.
c. 2.5 Nm.
d. 7 Nm
e. 7.5 Nm
Answer:
The net torque is 0.98 Nm.
Explanation:
The torque is given by
Torque = force x perpendicular distance
The clock wise torque is taken as negative while the counter clock wise torque is taken as positive.
Take the torques about the fulcrum.
Torque = 1 x 9.8 x 0.4 - 0.5 x 9.8 x 0.6
Torque = 3.92 - 2.94 = 0.98 Nm
9) An electrical appliance has a resistance of 25 N. When this electrical ap-
pliance is connected to a 230 V supply line, the current passing through it
will be:
d) 92 A
a) 0.92 A
c) 9.2 A
b) 2.9 A
(C)
Explanation:
From Ohm's law,
V = IR
Solving for I,
I = V/R
= (230 V)/(25 ohms)
= 9.2 A
g Select True or False: Boyle's Law states that the volume of a fixed amount of gas maintained at constant temperature is directly proportional to the gas pressure
Answer: The given statement is False.
Explanation:
Boyle's law states that pressure is inversely proportional to the volume of the gas at constant temperature and the number of moles.
Mathematically,
[tex]P\propto \frac{1}{V}[/tex] (At constant temperature and number of moles)
OR
[tex]P_1V_1=P_2V_2[/tex]
where,
[tex]P_1\text{ and }V_1[/tex] are the initial pressure and volume of the gas
[tex]P_2\text{ and }V_2[/tex] are the final pressure and volume of the gas
Hence, the given statement is False.
Select the correct answer.
A force of 100 newtons is applied to a box at an angle
acceleration of the box?
36° with the horizontal. If the mass of the box is 25 kilograms, what is the horizontal
OA
1.52 meters/second2
3.24 meters/second
C.
5.48 meters/second
Ο Ο Ο Ο
6.87 meters/seconda
7.15 meters/second
Answer:
3
Explanation:
first you find the original component of the force be a cause in 36 degrees then with ka sin 36 find the average interforce by multiplying it with hundred Newton then divide the original force by the mass 25 Kg is equals tto ma
Answer:
See image
Explanation:
Plato
(TCO 4) A signal consists of only two sinusoids, one of 65 Hz and one of 95 Hz. This signal is sampled at a rate of 245 Hz. Find the first six positive frequencies that will be present in the replicated spectrum.
Answer:
65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz
Explanation:
Given :
Frequencies of the sinusoids,
[tex]$f_{m_1}= 65 \ Hz$[/tex] , and
[tex]$f_{m_2}= 95 \ Hz$[/tex]
Sampling rate [tex]f_s = \ 245 \ Hz[/tex]
The positive frequencies at the output of the sampling system are :
[tex]$f_{o_1}=\pm f_{m_1} \pm nf_s, f_{o_2}=\pm f_{m_2} \pm nf_s $[/tex]
When n = 0,
[tex]$f_{o_1}= f_{m_1} = 65 \ Hz,\ \ f_{o_2}= f_{m_2} = 95 \ Hz $[/tex]
when n = 1,
[tex]$f_{o_1}=\pm f_{m_1} \pm f_s, \ \ f_{o_2}=\pm f_{m_2} \pm f_s $[/tex]
[tex]$f_{o_1}= \pm 65 \pm 245,\ \ f_{o_2}=\pm 95 \pm 245$[/tex]
[tex]$f_{o_1}= 180 \ Hz, 310 \ Hz,\ \ f_{o_2}= 150 \ Hz,340 \ Hz$[/tex]
When n = 2,
[tex]$f_{o_1}= \pm 65 \pm 2(245),\ \ f_{o_2}=\pm 95 \pm 2(245)$[/tex]
[tex]$f_{o_1}= 555 \ Hz, 425 \ Hz,\ \ f_{o_2}= 395 \ Hz,585 \ Hz$[/tex]
Therefore, the first six positive frequencies present in the replicated spectrum are :
65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz
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