Answer:
V2 = 15.53 [m/s]]
Explanation:
In order to solve this problem we must use the principle of energy conservation, where potential energy is transformed into kinetic energy. At the bottom is taken as a reference level of potential energy, where the value of this energy is equal to zero.
Above the inclined plane we have two energies, kinetics and potential. While when the sled is at the reference level all this energy will have been transformed into kinetic energy.
[tex]E_{1}=E_{2}\\ m*g*h+(\frac{1}{2} )*m*v_{1} ^{2}=\frac{1}{2}*m*v_{2} ^{2} \\(9.81*8.21)+(0.5*8.96^{2} )=(0.5*v_{2}^{2} )\\(0.5*v_{2}^{2} )=120.68\\v_{2} ^{2}=241.36\\v_{2} =\sqrt{241.36}\\ v_{2} =15.53[m/s][/tex]
PLEASE PROVIDE EXPLANATION.
THANK YOU!!
Answer:
11,000 kg
(a) 11.2 m/s
(b) 1.6 m/s
Explanation:
Momentum is conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(2200 kg) (60.0 km/h) + m (0 km/h) = (2200 kg) (10 km/h) + m (10 km/h)
132,000 kg km/h = 22,000 kg km/h + m (10 km/h)
110,000 kg km/h = m (10 km/h)
m = 11,000 kg
Momentum is conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(m) (-v) + (2m) (5v) = (m) (v₁) + (2m) (v₂)
-mv + 10mv = m v₁ + 2m v₂
9mv = m (v₁ + 2 v₂)
9v = v₁ + 2 v₂
Since the collision is elastic, kinetic energy is also conserved.
½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²
m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²
(m) (-v)² + (2m) (5v)² = m v₁² + (2m) v₂²
mv² + 50mv² = m v₁² + 2m v₂²
51mv² = m (v₁² + 2 v₂²)
51v² = v₁² + 2 v₂²
We know v = 1.60 m/s. So the two equations are:
14.4 = v₁ + 2 v₂
130.56 = v₁² + 2 v₂²
Solve the system of equations using substitution.
130.56 = (14.4 − 2 v₂)² + 2 v₂²
130.56 = 207.36 − 57.6 v₂ + 4 v₂² + 2 v₂²
0 = 6 v₂² − 57.6 v₂ + 76.8
0 = v₂² − 9.6 v₂ + 12.8
v₂ = [ 9.6 ± √(9.6² − 4(1)(12.8)) ] / 2(1)
v₂ = 1.6 or 8
If v₂ = 1.6 m/s, then v₁ = 14.4 − 2 v₂ = 11.2 m/s.
If v₂ = 8 m/s, then v₁ = 14.4 − 2 v₂ = -1.6 m/s.
We know v₁ can't be -1.6 m/s, since that would mean puck A didn't change speeds after the collision. Therefore, v₁ = 11.2 m/s and v₂ = 1.6 m/s.
In your research lab, a very thin, flat piece of glass with refractive index 2.30 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength λ0 in vacuum at normal incidence onto the surface of the glass. When λ0= 496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm.
Answer:
λ₀= 495.88 nm
Explanation:
To analyze this constructive interference interference experiment by reflection, let's look at two important aspects:
* when a ray of light passes from a medium with a lower index, they refact to another medium with a higher index, the reflected ray has a phase difference of pyres
* When a beam penetrates a material medium, the wavelength of the radiation changes according to the refractive index of the material.
λₙ = λ₀ / n
when we introduce these aspects in the expression of contributory interference, it remains
2 d sin θ = (m + ½) λ₀ / n
In general, reflection phenomena are measured at an almost normal angle, whereby θ = π/2 and sin θ = 1
2 d = (m +1/2) λ₀/ n
2n d = (m + ½) λ₀
Let's apply this expression to our case
d = (m + ½) λ₀ / 2n
Suppose we measure on the first interference, this is m = 0
d = ½ λ₀ / 2n
let's calculate
d = ½ 496 10⁻⁹ / (2 2.30)
d = 53.9 10-9 m
This is the thickness of the glass, the next wavelength that gives constructive interference is
λ₀ = 2 n d / (m + ½)
let's calculate
λ₀ = 2 2.3 5.39 10-8 / (1 + ½)
λ₀= 4.9588 10-7 m
λ₀= 495.88 nm
Think about the way your body works when you eat an ice cream cone. Provide an explanation of how organs are working together as a system to help you eat the cone. Describe one organ system and how its parts combine to get your body nutrition from the food.
PLZ HELP
Answer:
Explanation:
BAHAHAHAH karely omg HAHAHA un idke
A ball is launched directly upward and ultimately reaches a height of 40 ft on a day when the wind is gusting in different directions. From the time the ball is launched until it reaches a height of 20 ft off the ground the wind is blowing at constant 20 mph to the right. From that time to the time the ball has reached the top and traveled back down to a height of 20 ft the wind is blowing at constant 20 mph to the left. As it travels from 20 ft high back to the ground the wind again blows at a constant 20 mph to the right, where will the ball land?
Recall that
v² - u² = 2 a ∆x
where u and v are initial and final velocities, respectively; a is acceleration; and ∆x is the change in position.
We can use this formula to determine the launch speed of the ball. In the vertical direction, the ball has acceleration g in the downward direction, where g = 32.2 ft/s². At its maximum height, the ball has 0 vertical velocity. So we have
0² - u² = 2 (-32.2 ft/s²) (40 ft)
==> u ≈ 50.75 ft/s
Now, the ball's height y at time t is given by
y = u t - 1/2 g t²
Find at which time t the ball covers the first 20 ft of its trajectory:
20 ft = (50.75 ft/s) t - 1/2 g t²
==> t ≈ 0.462 s
(there is a second solution, t ≈ 2.69 s, which corresponds to the time it takes for the ball to return to this height as it's falling back down)
During this first interval, the ball's horizontal position x is
x = (20 mph) t ≈ (29.33 ft/s) t
so that at the moment the ball reaches a height of 20 ft, it will have moved
x = (29.33 ft/s) (0.462 s) ≈ 13.5 ft
to the right. (So we've take the right to be positive and the left to be negative.)
In the upper half of its trajectory, when the wind changes direction, the ball's horizontal position is given by
x = 13.5 ft - (29.33 ft/s) t
Note that t = 0 here means we now take the ball's current position to be the "initial" one. So the time we found earlier for when the ball has a height of 20 ft as it's falling back down is actually t = 2.69 s - 0.462 s ≈ 2.23 s. At this point, the ball's horizontal position is
x = 13.5 ft - (29.33 ft/s) (2.69 s - 0.462 s) ≈ -51.8 ft
or about 51.8 ft to the left of where it started.
The wind changes direction again, so that the ball's position at time t is now
x = -51.8 ft + (29.33 ft/s) t
Solving y = 0 for t gives the time when the ball reaches the ground:
0 = (50.75 ft/s) t - 1/2 g t²
==> t ≈ 3.15 s
Again, this is the time it takes for the ball to reach the ground since it was launched, while the position function takes t = 0 to refer to the moment the ball is 20 ft above the ground. So once it hits the ground, it will be
x = -51.8 ft + (29.33 ft/s) (3.15 s - 2.69 s) ≈ -38.3 ft
which means the ball lands on the ground about 38.3 ft to the left of where it started.
2.27 A gas is compressed from V1= 0.3 m3, p1=1 bar to V2= 0.1 m3, p2 =3 bar. The pressure and
the volume are related linearly during the process. For the gas, find the
work, in kJ.
2.29 A gas expands from an initial state where p1=500kPa and V1=0.1m3 to a state
end where p2= 100 kPa. The relationship between pressure and volume during the process is
pV=constant. Schematize the process in a p-V diagram and determine the work, in
kJ.
Answer:
-40 kJ
80 kJ
Explanation:
Work is equal to the area under the pressure vs volume graph.
W = ∫ᵥ₁ᵛ² P dV
2.27) Pressure and volume are linearly related. When we graph P vs V, the area under the line is a trapezoid. So the work is:
W = ½ (P₁ + P₂) (V₂ − V₁)
W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)
W = -40 kJ
2.29) Pressure and volume are inversely proportional:
pV = k
The initial pressure and volume are 500 kPa and 0.1 m³. So the constant is:
(500) (0.1) = k
k = 50
The final pressure is 100 kPa. So the final volume is:
(100) V = 50
V = 0.5
The work is therefore:
W = ∫ᵥ₁ᵛ² P dV
W = ∫₀₁⁰⁵ (50/V) dV
W = 50 ln(V) |₀₁⁰⁵
W = 50 (ln 0.5 − ln 0.1)
W ≈ 80 kJ
A cloth hat and large rock are dropped at at the same time on the moon
Answer:
What’s the question??
Explanation:
What is this car's instantaneous velocity ...
Wait for it ... Wait for it ... ok NOW!!! (in mi/h)
Answer:
it's a Bugatti
Explanation:
I seen pictures and videos of one for what they look like from the inside or the outside
1. Wind
A. Definition
B. What Heat Transfer Causes
c. Types
Answer:
B . a heat transfer caused
If a car with a mass of 4000 kg is accelerating at a rate of 2 ms and hits a tree what force does it have
That depends on the car's speed at the moment of impact, and how long it takes to stop after the impact.
It is NOT 8,000N. That answer would come from a totally incorrect application of " F = m a " . 8,000 Newtons is what it takes to accelerate that car in the first place, long before it hits the tree.
Force can be described as either a pull or a push
Now when a body with mass and acceleration hits an object, we say a collision has taken place, hence there will be transfer of energy within the system
Give data
mass = 4000kg
acceleration = 2m/s^2
We know that force is given as
Force = mass x acceleration
Substitute our given data into the expression for force
Force = 4000 x 2Force = 8000N
For more information on force kindly visit
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a strong inductive argument must have true premises
True
False
) The square plates of a 5000-pF parallel-plate capacitor measure 50 mm by 50 mm and are separated by a dielectric that is 0.23 mm thickand totally fills the region between the plates. The voltage rating (the maximum safe voltage) of the capacitor is 400 V. What is the maximum energy that can be stored in this capacitor without damaging it
Answer:
4 x 10⁻⁴ J
Explanation:
C = 5000 pF, V = 400 V
Energy = CV²/2 = 5000 x 10⁻¹² x 400²/2 = 4 x 10⁻⁴ J
Question 4
5 pts
What is microscopic and floats in the ocean?
Plankton
Benthos
Nekton
© Fish
Answer:
Plankton is the correct answer I think
The statement "All Romans are Italians" means the same as "If you are Roman, then you are Italian"
True
False
what is E=mc^2?
please explain..
Answer:
In physics, mass–energy equivalence is the principle that anything having mass has an equivalent amount of energy and vice ve
how does energy transfer between objects
Answer:
Object A runs into Object B. The force that object A has will transfer from Object A to Object B. But during this transfer you will loose some energy, therefore object B will not travel as fast.
Explanation:
I WILL MARK BRAINLYEST 40+ POINTS. PLEASE ANSWER!!!
The measure of the amount of atoms (matter) in something.
Group of answer choices
Weight
Mass
Volume
Law of Conservation of Matter (Mass) (LOCM)
Question 310 pts
The reference lines on a graph, i.e. X and Y
Group of answer choices
Coordinates
Line of best fit
Axis
Slope
Question 410 pts
How far apart objects or points are.
Group of answer choices
Inertia
Distance
Net Force
Question 510 pts
Moving from one position to another.
Group of answer choices
Balanced Forces
Traveling
Distance
Question 610 pts
The tendency of things to keep traveling in the same direction at the same speed, unless something stops them.
Group of answer choices
Velocity
Inertia
Gravity
Friction
Answer:
Mass
axis
distance
intertia
If any state tribunal decides a federal question and the litigant has no further remedy within the state court system, from which of the following remedies could the litigant possibly benefit?
A.
a hearing by a US district court
B.
a hearing by a federal court of appeals
C.
a hearing by the US Supreme Court
D.
a Congressional vote
Answer:
B. a hearing by a federal court of appeals
If any state tribunal decides a federal question and the litigant has no further remedy within the state court system, a hearing by a federal court of appeals.
What is appeal by federal courts ?Oral argument in the court of appeals is a prearranged conversation between the judges and the appellate attorneys that focuses on the legal principles at issue. Each party has a limited amount of time usually around 15 minutes to make their case to the judge.
If a litigant loses in a federal court of appeals or in the highest court of a state, they may ask the supreme court to reconsider the case by filing a petition.
Only when a case concerns an extremely significant legal concept or when two or more federal appellate courts have disagreed on how to interpret a statute will the court normally consent to consider it. Additionally, there are a few rare instances where the Supreme Court is mandated by the law to consider an appeal.
Find more on federal court :
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A large research balloon containing 2.00 × 10^3 m^3 of helium gas at 1.00 atm and a temperature of 15.0°C rises rapidly from ground level to an altitude at which the atmospheric pressure is only 0.900 atm. Assume the helium behaves like an ideal gas and the balloon’s ascent is too rapid to permit much heat exchange with the surrounding air.
Required:
a. Calculate the volume of the gas at the higher altitude.
b. Calculate the temperature of the gas at the higher altitude.
c. What is the change in internal energy of the helium as the balloon rises to the higher altitude?
Answer:
a
[tex]T_2 = 276.1 \ K[/tex]
b
[tex]V_2 = 2.13 *10^{3} \ m^3 [/tex]
c
[tex]\Delta U = -1.25 *10^{7} \ J [/tex]
Explanation:
From the question we are told that
The volume of the balloon is [tex]V = 2.00 * 10^3 \ m^3[/tex]
The pressure of helium is [tex]P_1 = 1.00 \ atm= 1.0 *10^{5} \ Pa [/tex]
The initial temperature is [tex]T_1 = 15.0^oC = 288 \ K [/tex]
The pressure of atmosphere is [tex]P_a = 0.900 \ atm[/tex]
Generally the equation representing the adiabatic process is mathematically represented as
[tex]P_1 V_1 ^{\gamma }= P_2 V_2 ^{\gamma }[/tex]
=> [tex]V_2 ^ {\gamma } = \frac{ V_1 ^{\gamma } * P_1 }{P_2}[/tex]
Generally [tex]\gamma[/tex] is a constant with value [tex]\gamma =\frac{5}{3}[/tex] for an ideal gas
So
[tex]V_2 ^ {\frac{5}{3} } = \frac{ ( 2.0 *10^{3}) ^{ \frac{5}{3} } * 1.00 }{0.900}[/tex]
[tex]V_2 = (\sqrt[5]{103.14641852} )^3[/tex]
=> [tex]V_2 = 2.13 *10^{3} \ m^3 [/tex]
Generally the adiabatic process can also be mathematically represented as
[tex]T_1 V_1 ^{\gamma -1 } = T_2 V_2^{\gamma -1 }[/tex]
=> [tex]T_2 = 288 * [\frac{2 * 10^{3}}{ 2.13 *10^{3}} ]^{ \frac{5}{3} -1 }[/tex]
=> [tex]T_2 = 276.1 \ K[/tex]
Generally the ideal gas equation is mathematically represented as
[tex]P_1 V_1 = nRT_1[/tex]
Here R is the gas constant with value [tex]R = 8.314\ J /mol \cdot K[/tex]
[tex]n = \frac{P_1 V_1 }{RT _1}[/tex]
=> [tex]n = \frac{1.0 *10^{5} * 2.0 *10^{3}}{8.314 * 288[/tex]
=> [tex]n = 84362 \ mol[/tex]
Generally change in internal energy i mathematically represented
[tex]\Delta U = n C_v \Delta T[/tex]
Here [tex]C_v[/tex] is the specific heats of gas at constant volume and the value is [tex]C_v = 12.47 J/mol \cdot K[/tex]
[tex]\Delta U = 84362 * 12.47 * [T_2 - T_1 ] [/tex]
[tex]\Delta U = 84362 * 12.47 * [276.1 - 288 ] [/tex]
[tex]\Delta U = -1.25 *10^{7} \ J [/tex]
I will give you branilest
Mrs. LaCross leaves school and accidentally leaves her coffee mug on the roof of her car as shown in the picture below.
She was traveling at a constant pace until a student rushes in front of her not using the sidewalks and crosswalks and she had to slam on her brakes. What will happen to her coffee mug?
[ Select ]
a. The mug will move side by side
b. The mug will move backwards
c. The mug will move forward
d. The mug will not move
Answer:
the mug will moveforwards
Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper, with a circular cross section of radius 1.53 mm. The other wire is composed of 19 strands of thin copper wire bundled together. Each strand has a circular cross section of radius 0.306 mm. The current density ???? in each wire is the same. ????=1750 A/m2 Two circles. One circle is solid, whereas the other contains 19 tightly packed smaller circles How much current does each wire carry? solid wire current: A stranded wire current: A The resistivity of copper is ????=1.69×10−8 Ω·m. What is the resistance of a 3.25 m length of each wire? solid wire resistance: Ω stranded wire resistance: Ω
Answer:
a
Solid Wire [tex]I = 0.01237 \ A [/tex]
Stranded Wire [tex]I_2 = 0.00978 \ A [/tex]
b
Solid Wire [tex]R = 0.0149 \ \Omega [/tex]
Stranded Wire [tex]R_1 = 0.0189 \ \Omega [/tex]
Explanation:
Considering the first question
From the question we are told that
The radius of the first wire is [tex]r_1 = 1.53 mm = 0.0015 \ m[/tex]
The radius of each strand is [tex]r_0 = 0.306 \ mm = 0.000306 \ m[/tex]
The current density in both wires is [tex]J = 1750 \ A/m^2[/tex]
Considering the first wire
The cross-sectional area of the first wire is
[tex]A = \pi r^2[/tex]
= > [tex]A = 3.142 * (0.0015)^2[/tex]
= > [tex]A = 7.0695 *10^{-6} \ m^2 [/tex]
Generally the current in the first wire is
[tex]I = J*A[/tex]
=> [tex]I = 1750*7.0695 *10^{-6}[/tex]
=> [tex]I = 0.01237 \ A [/tex]
Considering the second wire wire
The cross-sectional area of the second wire is
[tex]A_1 = 19 * \pi r^2[/tex]
=> [tex]A_1 = 19 *3.142 * (0.000306)^2[/tex]
=> [tex]A_1 = 5.5899 *10^{-6} \ m^2[/tex]
Generally the current is
[tex]I_2 = J * A_1[/tex]
=> [tex]I_2 = 1750 * 5.5899 *10^{-6} [/tex]
=> [tex]I_2 = 0.00978 \ A [/tex]
Considering question two
From the question we are told that
Resistivity is [tex]\rho = 1.69* 10^{-8} \Omega \cdot m[/tex]
The length of each wire is [tex]l = 6.25 \ m[/tex]
Generally the resistance of the first wire is mathematically represented as
[tex]R = \frac{\rho * l }{A}[/tex]
=> [tex]R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }[/tex]
=> [tex]R = 0.0149 \ \Omega [/tex]
Generally the resistance of the first wire is mathematically represented as
[tex]R_1 = \frac{\rho * l }{A_1}[/tex]
=> [tex]R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }[/tex]
=> [tex]R_1 = 0.0189 \ \Omega [/tex]
calculate the monentun pf 75 kg bicycle and boy who has ghe velocitg of 3m/s
Explanation:
n=75 kg
v=3m/s
m=n×v
m=75x3
m= 225n
In a photoelectric effect experiment, light from a blue LED (wavelength = 405 nm) is directed onto a piece of sodium. It is observed that when V = 0.52 volts, the current measured at the collector drops to zero. What is the work function of the sodium? (a) 0.85 eV (b) 2.5 eV (c) 1.6 eV (d) 3.6 eV (e) 0.088 eV 100% What is the highest velocity of an ejected electron, just above the surface of the sodium? (a) 680000 m/s (b) 300000 m/s (c) 430000 m/s 100% Now we decrease the power of the LED, keeping the wavelength fixed. Which of the following statements is true? (a) The maximum velocity of any ejected electrons will decrease. (b) The minimum possible time to eject an electron (relative to when the LED is turned on) will increase.
Answer:
1. (b) 2.5 eV
2. (c) 430000 m/s
3. (a) The maximum velocity of any ejected electrons will decrease.
Explanation:
1)
From Einstein's Photoelectric Equation:
hc/λ = K.E + ∅
∅ = hc/λ - K.E
∅ = hc/λ - eV
where,
e = charge on electron = 1.6 x 10⁻¹⁹ C
V = Stopping Potential = 0.52 volts
h = 6.625 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength = 405 nm = 4.05 x 10⁻⁷ m
∅ = Work Function = ?
Therefore,
∅ = (6.625 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(4.05 x 10⁻⁷ m) - (1.6 x 10⁻¹⁹ C)(0.52 volts)
∅ = 4.9 x 10⁻¹⁹ J - 0.832 x 10⁻¹⁹ J
∅ = (4.075 x 10⁻¹⁹ J)(1 eV)/(1.6 x 10⁻¹⁹ J)
∅ = 2.5 eV
therefore, correct answer is"
(b) 2.5 eV
2)
K.E = (1/2)mv² = eV
where,
m = mass of electron = 9.1 x 10⁻³¹ kg
v = speed = ?
therefore,
(1/2)(9.1 x 10⁻³¹ kg)v² = (1.6 x 10⁻¹⁹ C)(0.52 volts)
v = √(0.18 x 10¹² m²/s²)
v = 0.43 x 10⁶ m/s = 430000 m/s
Correct option is:
(c) 430000 m/s
3.
The decrease in power at constant wavelength means decrease in voltage, that results in the decrease of kinetic energy of electrons. So, correct option is:
(a) The maximum velocity of any ejected electrons will decrease.
Which of the following is a description of the Remote Associates Test (RAT)?
Answer:
The description is outlined throughout the clarification section following, and according to the given word.
Explanation:
Throughout the 1960s, Sarnoff Mednick created the RAT as a tool used for testing imaginative convergent thought. Through each RAT test query lists a set of terms, which demands that we have a single additional term that will tie any of the others around. Those other words may also be related in something like a variety of ways, such as through creating a compound word or perhaps a semantic connexon.A dog walks 30. m west, then 40. m east and then 20. m south. What is the
Answer:
90
Explanation:
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bkfjffy
22/22 is an even number greater than 20?
Answer:
yes
Explanation:
becuse it is an even that is 2 number then 20
Answer:
Yes
Explanation:
It is because 22 is greater than 20
1-4 help pls if u know it
Answer:
c,B,a,d i guess i m not sure it is the guessing
If electrons are added to an object the object becomes
charge.
negative
positive
neutral
nothing happens
Answer:
It can be either, depending on how much and how you add the electrons
Four students give their teacher identical apples. Each student sets his or her apple on a different stack of books. Which
apple has the most potential energy?
Answer:
Diagram B.
Explanation:
We'll begin by defining potential energy.
This is given below:
Potential energy is the energy stored in an object due to its position. Mathematically, it is expressed as:
P.E = m × g × h
Where:
P.E is the potential energy.
m is the mass of the object.
g is the acceleration due to gravity.
h is the height to which the object is located.
From:
P.E = m × g × h
We can say that the potential energy (P.E) is directly proportional to the height (h) to which the object is located. This implies that as the height increase, the potential energy of the object will also increase and as the height decrease, the potential energy of the object will also decrease.
Now, considering the question given above, we can see that the height of the object is greater in diagram B. Therefore, The potential energy of the object is greater in diagram B.
Answer:
The answer s B
Explanation:
what are the applications of measurement?
A cube with sides 12 cm is submerged in water to a depth of 30 cm. Given density of water is 1000 kg/m3. Calculate the pressure at the bottom surface of the cube due to water.
Answer:
P=2940 Pa
Explanation:
Given that,
Side of a cube, a = 12 cm
It is submerged to depth of 30 cm
The density of water, d = 1000 kg/m^3
We need to find the pressure at the bottom surface of the cube. We know that the pressure exerted is given by :
[tex]P=d gh\\\\P=1000\ kg/m^3\times 9.8\ m/s^2\times 0.3\ m\\\\P=2940\ Pa[/tex]
So, 2940 Pa of pressure is exerted at the bottom surface of the cube due to water.