Answer:
A) E = 4.96 x 10³ eV
B) E = 4.19 x 10⁴ eV
C) E = 3.73 x 10⁹ eV
Explanation:
A)
For photon energy is given as:
\(E = hv\)
\(E = \frac{hc}{\lambda}\)
where,
E = energy of photon = ?
h = 6.625 x 10⁻³⁴ J.s
λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m
Therefore,
\(E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}\)
\(E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})\)
E = 4.96 x 10³ eV
B)
The energy of a particle at rest is given as:
\(E = m_{0}c^2\)
where,
E = Energy of electron = ?
m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg
c = speed of light = 3 x 10⁸ m/s
Therefore,
\(E = (9.1 x 10^{-31} kg)(3 x 10^8 m/s)^2\\\)
\(E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\\)
E = 4.19 x 10⁴ eV
C)
The energy of a particle at rest is given as:
\(E = m_{0}c^2\)
where,
E = Energy of alpha particle = ?
m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg
c = speed of light = 3 x 10⁸ m/s
Therefore,
\(E = (6.64 x 10^{-27} kg)(3 x 10^8 m/s)^2\\\)
\(E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\\)
E = 3.73 x 10⁹ eV
A) The energy in electron volts for a particle with this wavelength if the particle is a photon is; .E = 4969.5 eV or 4.9695 keV
B) The energy in electron volts for a particle with this wavelength if the particle is an electron is; E = 23.58 eV
C) E = 0.003306 eV
A) The formula for the energy here is;
E = hc/λ
where;
h is planck's constant = 6.626 × 10⁻³⁴ J.s
c is speed of light = 3 × 10⁸ m/s
λ is wavelength = 0.25 nm = 0.25 x 10⁻⁹ m
Thus;
E = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(0.25 x 10⁻⁹)
79.512 × 10⁻¹⁷ J
converting to eV gives;
E = (79.512 × 10⁻¹⁷)/(1.6 × 10⁻¹⁹)
E = 4969.5 eV or 4.9695 keV
B) Formula for the energy if the particle is an electron is;
E = h²/(2mλ²)
where m = 9.31 × 10⁻³¹ kg
E = (6.626 × 10⁻³⁴)²/(2 × 9.31 × 10⁻³¹ × (0.25 x 10⁻⁹)²)
E = 37.726 × 10⁻¹⁹ J
Converting to eV gives;
E = (37.726 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹)
E = 23.58 eV
C) Mass of alpha particle is; m = 6.64 × 10⁻²⁷ kg
E = h²/(2mλ²)
where m = 6.64 × 10⁻²⁷ kg
E = (6.626 × 10⁻³⁴)²/(2 × 6.64 × 10⁻²⁷ × (0.25 x 10⁻⁹)²)
E = 52.896 × 10⁻²³ J
Converting to eV gives;
E = (52.896 × 10⁻²³)/(1.6 × 10⁻¹⁹)
E = 0.003306 eV
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Students have four identical, hollow, uncharged conducting spheres, W, X, Y, and Z.
Sphere Z is given a positive charge of +40 C. Sphere Z is touched first to sphere W, then sphere X, and finally to sphere Y. What is the resulting charge on sphere Y?
a. +5 με
b. +10 μC
c. +20 μC
d. +40 με
d
Explanation:
because they made contact that means their new force will be the same
The resulting charge on sphere Y = ( A ) +5 με
Given that the spheres are hollow, uncharged and identical the charge on each of the sphere would be equal to each other.
Sphere Z having a positive charge of +40 C and in contact with other spheres ( w , x and y ), the charge on sphere z will be evenly distributed amongst the spheres. therefore the resulting charge on sphere Y will be +5με
Hence we can conclude that the resulting charge on sphere y is +5 με
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A 3kg cat falls out of a window that is 10m above the ground and lands
safely in the grass. Calculate the velocity at which the cat hit the ground.
Answer:
p = 30 kg·m/s
Explanation:
An object experiencing uniform circular motion is accelerating. t/f
Answer:
According to Newton's second law of motion, a net force causes the acceleration of mass according to Fnet = ma. For uniform circular motion, the acceleration is centripetal acceleration: a = ac. Therefore, the magnitude of centripetal force, Fc, is F c = m a c F c = m a c .
Explanation:
I need help with these questions please help
25 points
Answer:
A) 0.12 m
B) 11.5 degrees
A) Two convex lenses
D) Amplitude
A) 491.8 m
B) 2.92 m
C) 9.677 MHz
I couldn't see the options for number eight, so I'm unsure if this is correct, but I got 507.7 K in my calculations.
A candle is placed 50 cm from a diverging lens with a focal length of 28. What is the image distance in cm.
Answer:
v = -17.94 cm
Explanation:
Given that,
The candle is placed at a distance of 50 cm, u = -50 cm
The focal length of the diverging lens, f = -28 cm (negative in case of a concave lens)
We need to find the image distance. We know that the lens formula is as follows:
\(\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(-28)}+\dfrac{1}{(-50)}\\\\v=-17.94\ cm\)
So, the image distance is equal to 17.94 cm.
Suppose that scientists observe violent gas eruptions on a planet with an acceleration due to gravity of 3.8 m/s2. The jets throw sand and dust about 75 m above the surface.
What is the speed i of the material just as it leaves the surface?
Scientists estimate that the jets originate as high‑pressure gas speeds through vents just below ground at about 160 km/h. How much energy per kilogram of material does the jet lose to non conservative forces as the high‑speed matter forces its way to the surface and into the air?
Answer:
linarkuoouonpnpjjñihy
Explanation:
What is the effect of erosion?
A. New land forms at the mouth of a river.
B. New land forms at the top of a mountain.
C. A mountain forms.
D. A fossil is created.
find the velocity vof the block as a function of time.express your answer in terms of some or all of the variables: k, m, a, and t.
The expression "velocity as a function of time" refers to the change in an object's velocity over time, which can be observed by graphing the velocity against time on a graph.
Speed and time are related in the first equation of motion. V = u + at represents the first motion equation. Here, u denotes the starting velocity, a the acceleration, and t the duration, whereas v denotes the end velocity. The derivative of x with respect to time, or v(t)=ddtx, represents the instantaneous velocity of an object. It is the limit of average velocity as the time approaches zero (t). V (t) equals d d t x (t). Instantaneous velocity has a dimension of length per time, just like average velocity does.
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Express 6revolutions to radians
Answer:
About 37.70 radians.
Explanation:
1 revolution = 2\(\pi\) radians
∴ 6 revolutions = (6)(2\(\pi\) radians)
6 revolutions = 37.6991 or ≈ 37.70 radians
3. A car with a mass of 1600 kg has a kinetic energy of 125 000 J. How fast is it moving?
The car is moving at approximately 12.5 meters per second.
The kinetic energy (KE) of an object can be calculated using the formula:
KE = 1/2 * m * \(v^2\)
where
KE = kinetic energy,
m =Mass of the object, and
v = velocity.
In this case, we are given the mass (m) of the car as 1600 kg and the kinetic energy (KE) as 125,000 J. To find the velocity .
Substituting the values , we have:
125,000 J = 1/2 * 1600 kg *\(v^2\)
Now, we can solve for v by rearranging the equation:
\(v^2\) = (2 * 125,000 J) / 1600 kg
\(v^2\) = 156.25 \(m^2/s^2\)
Taking the square root, we find:
v = √156.25\(m^2/s^2\)
v ≈ 12.5 m/s
Therefore, the car is moving at approximately 12.5 meters per second.
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Which feature is created when a block of rock dropped down in relation to the block of rock beside it?
Answer:
Which feature is created when a block of rock dropped down in relation to the block of rock beside it?
Explanation:
Grabens drop down relative to adjacent blocks and create valleys. Horsts rise up relative to adjacent down-dropped blocks and become areas of higher topography.
An objects weight w is equal to the product of its mass and the acceleration due
to gravity. On Earth, this value is g = 9.8 m/s^2. If something weighs 400 N,
what is its mass?
Answer:hinwjdheiohuddwrr
87u7t
what causes diffferent air density?
Answer:
The temperature and gravity both affects the density
Each mass in the figure is 3 kg. Find the magnitude and direction of the net gravitational force on mass A due to the other masses.
A. 2.45 × 10–7 N toward B
B. 3.75 × 10–7 N toward C
C. 2.00 × 10–7 N toward D
D. 1.15 × 10–7 N toward D
Answer:
THE ANSER IS B
Explanation:
The magnitude and direction of the net gravitational force on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.
The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In this case, all of the masses are equal to 3 kg, and the distance between mass A and mass D is 3 m.
The gravitational force between mass A and mass D is therefore:
F = G * m_A * m_D / r²
= 6.674 × 10⁻¹¹ N m² / kg² * 3 kg * 3 kg / 3 m²
= 1.15 × 10⁻⁷ N
The direction of the gravitational force is towards mass D.
Therefore, the net gravitational force on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.
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I heat 29.292 g of an unknown metal up to 99.9 °C. While it is heating, I weigh out 27.777 g of water, and find its initial
temperature is 22.1 "C. When I mix the metal and water in an insulated container, the temperature of the mixture rises
to 29.3 °C.
What is the most likely specific heat of the metal?
Select one:
a. 4.8 (104) cal / (g *C)
b. 9.7 (102) cal/(g*C)
c. 7.9 (10³) cal/ (g*C)
Od: 0.13 cal / (g°C)
e. 9.8 (105) cal / (g°C)
The specific heat capacity of the metal, given that 27.777 g of water at 22.1 °C was mixed with the metal is 9.7×10⁻² Cal/gºC
How do I determine the specific heat capacity of the metal?Step 1: Obtain the heat absorbed by the water. This is shown below:
Mass of water (M) = 27.777 gInitial temperature (T₁) = 22.1 °CFinal temperature (T₂) = 29.3 °CTemperature change (ΔT) = 29.3 - 22.1 = 7.2 °CSpecific heat capacity of water (C) = 1 Cal/gºC Heat absorbed (Q) =?Q = MCΔT
= 27.777 × 1 × 7.2
= 199.9944 Cal
Step 2: Determine the specific heat capacity of the metal using the heat absorbed by the water. Details below:
Heat absorbed by water (Q) = 199.9944 CalHeat released by metal (Q) = -199.9944 CalMass of metal (M) = 29.292 gInitial temperature (T₁) = 99.9 °CFinal temperature (T₂) = 29.3 °CTemperature change (ΔT) = 29.3 - 99.9 = -70.6 °CSpecific heat capacity of metal (C) = ?Q = MCΔT
-199.9944 = 29.292 × C × -70.6
-199.9944 = -2068.0152 × C
Divide both sides by -2068.0152
C = -199.9944 / -2068.0152
= 9.7×10⁻² Cal/gºC
Thus, the specific heat capacity of the metal is 9.7×10⁻² Cal/gºC. None of the options are correct.
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Initial velocity: U = ?
Final velocity: v = 0
Gravity: g = 9,8 ms²
Height: h = 20m
v² = u² + 2gh
0 = u² + 2 (-9,8ms²) x 20m
The initial velocity of the object in motion is determined as 19.8 m/s.
What is the initial velocity of the object?The initial velocity of the object in motion is calculated by applying the third equation of motion as follows;
v² = u² + 2gh
where;
v is the final velocity of the objectu is the initial velocity of the objectg is acceleration due to gravityh is the height through which the object riseswhen the object reaches maximum height, the final velocity, v = 0
The initial velocity of the object in motion is calculated as;
0 = u² + 2 (-9,8ms²) x 20m
0 = u² - 392
u² = 392
u = √392
u = 19.8 m/s
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PLEASE HELP ME I NEED THIS ASAP
Investigating Energy Systems
Harnessing Human Energy Chapter 1 Lesson 1.2
Reasoning Tool
Possible subclaims:
The Battery System does have energy.
or
The Battery System does not have energy.
The Hand-Crank Generator System does have energy.
or
The Hand-Crank Generator System does not have energy.
The Solar Cell System does have energy.
or
The Solar Cell System does not have energy.
Evidence
(observations about whether the system does or does not have energy)
This matters because . . .
(How does this evidence support the subclaim?)
Therefore, . . .
(subclaim)
Considering the energy systems:
The Battery System does have energy because it can be used to light a bulb for example.The Hand-Crank Generator System does have energy because turning the lever generates electrical energy.The Solar Cell System does have energy because it can light a bulb.What is energy?Energy is the ability to do work.
There are several forms of energy such as:
chemical energy - energy stored in chemical bondsSolar energy - energy from the radiation of the sunLight energy - energy from the sensation of lightSound energy -Mechanical energy - there are two forms; kinetic energy and potential energyElectrical energy is the energy due to the movement of electrons.Generally, all matter possesses energy that can be used to do work.
The various forms of energy can be interconverted.
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Your lab partner tosses you a ball. As she throws the ball up in the air it follows this arc. Where is potential energy the lowest? Select all that apply.
A. point A
B. point B
C. point C
D. point D
E. They are all the same
Answer: c
Explanation:
Answer:
answer c
Explanation:
Yamel is heating up leftover mashed potatoes from Thanksgiving. She forgot to put gravy on them. So she puts the cold gravy on the hot mashed potatoes. Which statement best describes what will happen? A-The mashed potatoes will transfer heat into the gravy B-The gravy and mashed potatoes both transfer heat into the plate C-The gravy will transfer cold into the mashed potatoes D- The gravy transfers cold into the mashed potatoes and plate
Answer:
The correct answer is - option A. The mashed potatoes will transfer heat into the gravy.
Explanation:
In this case, where Yamel heats the mashed potatoes but forgets to heat the gravy and put the cold gravy on the hot mashed potatoes. Heat always transfers from the high-temperature object to the low-temperature object. So the hot mashed potatoes will transfer the heat to the gravy according to option A. Cold is not a form of heat but the condition of absence of heat or very low temperature.
Thus, the correct answer is - option A. The mashed potatoes will transfer heat into the gravy.
How is the biopsychological approach different from the medical model?
Answer:
Similar to a lock and key, the biomedical model seeks to find a solution that fits the respective problem. Biopsychosocial. The biopsychosocial model is aptly named because of its three essential components: biological, psychological and social ideologies. As a result, the concept of health is viewed as a balance between these three sectors.
Explanation:
hope it helps
have a great day
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Nitrogen makes up about what percent of a human's body weight?
Answer:
the answer is 3.3 %
Explanation:
1.at constant temperature a gas occupy 50cm³ at 10mmhg calculate the work done
2.if normal human body temperature is 34 °c, what is the equivalent value in fahrenheit
3. A stone is thrown upward from ground with a velocity of 40m/s. how long does it take to reach the highest point
Answer:
100= 4 (n+7) simplify with steps :)
Explanation:
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.
A vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. Determine the magnitude, F, of the force acting on the shaft, in N, required if the gas pressure is 3 bar. The masses of the piston and attached shaft are 24.5 kg and 0.5 kg, respectively. The piston diameter is 10 cm. The local atmospheric pressure is 1 bar. T
Answer:
The force acting on the shaft is 1324.75 N
Explanation:
Given that,
Cross sectional area of shaft \(A_{sh}=0.8\ cm^{2}\)
Gas pressure \(P_{gas}=3\ bar=3\times10^5\ Pa\)
Total mass M=24.5+0.5=25 kg
Diameter of piston, d=10 cm
We need to calculate the cross section area of piston
Using formula of area
\(A_{p}=\pi\times\dfrac{d^2}{4}\)
Put the value into the formula
\(A_{p}=\pi\times\dfrac{0.1^2}{4}\)
\(A_{p}=0.00785\ m^{2}\)
We need to calculate the weight of the piston and shaft
Using formula of weight
\(W=mg\)
Put the value into the formula
\(W=25\times9.81\)
\(W=245.25\ N\)
We need to calculate the force due to gas pressure
Using formula of force
\(F_{gas}=P_{gas}\times A_{p}\)
Put the value into the formula
\(F_{gas}=3\times10^{5}\times0.00785\)
\(F_{gas}=2355\ N\)
We need to calculate the force due to atmospheric pressure,
Using formula of force
\(F_{atm}=P_{atm}\times A_{p}\)
Put the value into the formula
\(F_{atm}=1\times10^5\times0.00785\)
\(F_{atm}=785\ N\)
We need to calculate the force acting on the shaft
from free body diagram
\(F_{gas}=F+F_{atm}+W\)
Put the value into the formula
\(2355=F+785+245.25\)
\(F=2355-785-245.25\)
\(F=1324.75\ N\)
Hence, The force acting on the shaft is 1324.75 N
The complete question is :
The figure (attached) shows a gas contained in a vertical piston-cylinder assembly. A vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. Determine the magnitude, F, of the force acting on the shaft, in N, required if the gas pressure is 3 bar. The masses of the piston and attached shaft are 24.5kg and 0.5kg respectively. The piston diameter is 10cm. The local atmospheric pressure is 1 bar. The piston moves smoothly in the cylinder and g=9.81 m/s2.
Solution :
Given
The cross sectional area of the shaft, \($A_{s} = 0.8 \ cm^2$\)
Gas pressure, \($P_{gas} = 3 \ bar = 3 \times 10^5 \ Pa$\)
The total mass, m = 24.5 + 0.5
= 25 kg
Diameter of the piston, d = 10 cm
The cross sectional area of the piston, \($A_{p} = \frac{\pi}{4} \times (0.1)^2$\)
Weight of the piston and shaft, W = mg
= 25 x 9.81
= 245.25 N
Force due to the gas pressure,
\($F_{gas} = P_{gas} \times A_{p}$\)
\($F_{gas} = 3 \times 10^5 \times 0.00785 $\)
= 2355 N
Force due to atmospheric pressure,
\($F_{atm} = P_{atm} \times A_{p}$\)
\($= 1 \times 10^5 \times 0.00785$\)
= 785 N
Now \($F_{gas} = F + F_{atm} + W$\)
⇒ 2355 = F + 785 + 245.25
⇒ F = 1324.75 N
Need help with questions 46 and 47. Thank you
46)
Since air resistance is ignored, both cannonballs will fall with the same acceleration due to gravity, which is approximately 9.8 m/s^2. The horizontal velocity of cannonball A does not affect its vertical motion, so it will fall at the same rate as cannonball B.
The time it takes for an object to fall to the ground from a certain height is given by the formula t = sqrt(2h/g), where t is the time, h is the initial height, and g is the acceleration due to gravity.
For cannonball B, which is dropped from a height of 20 meters, the time it takes to reach the ground is:
t = sqrt(2h/g) = sqrt(2*20/9.8) = 2.02 seconds
For cannonball A, which is fired horizontally with a velocity of 5 m/s, the time it takes to reach the ground is also 2.02 seconds, since the vertical motion is the same as cannonball B.
Therefore, both cannonballs will hit the ground at the same time.
47)
The recoil speed of the cannon can be calculated using the principle of conservation of momentum, which states that the total momentum of a system remains constant if no external forces act on it.
Before firing, the total momentum of the system is zero, since the cannon and cannonball A are at rest. After firing, the cannonball A has a momentum of 5 kg * 5 m/s = 25 kg m/s to the right, so the cannon must have an equal and opposite momentum to the left in order to conserve momentum.
The mass of the cannon is 500 kg, so its momentum after firing will be:
p = -25 kg m/s
The velocity of the cannon can be found using the equation:
p = mv
where p is the momentum, m is the mass, and v is the velocity. Solving for v, we get:
v = p/m = (-25 kg m/s) / (500 kg) = -0.05 m/s
Since the momentum of the cannon is negative, the recoil velocity is also negative, indicating that the cannon will move to the left after firing. The magnitude of the recoil velocity is 0.05 m/s, or approximately 0.18 km/h.
A 1,200 kg car accelerates at 3 m/s2. What net force is the car experiencing during this
acceleration?
0 3600 N
0 2000 N
2400 N
O ON
Answer:
3600 N
Explanation:
F=ma
1200*3 = 3600
4. A 40.0 kg child swings in a swing supported by two chains, each 3.00 m long. If the tension in each at the lowest point is 350N, find (i) The child’s speed at the lowest point ,
The child's speed at the lowest point is 5.42 m/s.
At the highest point of the swing, the child is momentarily at rest and has only potential energy. At the lowest point, the child has only kinetic energy.
Using the conservation of mechanical energy, we can write:
Potential energy at highest point = Kinetic energy at lowest point
mgh = (1/2)mv²
where m is the mass of the child, g is the acceleration due to gravity, h is the height of the swing at the highest point, and v is the speed of the child at the lowest point.
First, we need to find the height of the swing at the highest point. Since the swing is supported by two chains, the height of the swing at the highest point is half the length of the chains:
h = (1/2)3.00 m = 1.50 m
Next, we can solve for the child's speed at the lowest point:
mgh = (1/2)mv²
40.0 kg * 9.81 m/s² * 1.50 m = (1/2) * 40.0 kg * v²
588 J = 20.0 kg * v²
v² = 29.4 m²/s²
v = 5.42 m/s
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After a big snowfall, you take your favorite rocket‑powered sled out to a wide field. The field is 205 m across, and you know that your sled accelerates at a rate of 3.95 m/s2 when the rocket is on. How fast will the sled be moving when it reaches the other end of the field?
With the use of third equation of motion, the sled be moving as fast as 40.24 m/s when it reaches the other end of the field.
What is Acceleration ?Acceleration is the rate at which velocity is changing. It is a vector quantity.
Given that you accelerates at a rate of 3.95 m/s2 when the rocket is on across a field of 205 m. To know how fast the sled will move when it reaches the other end of the field, we will use third equation of motion. That is, v² = u² + 2as
Where
v = ?u = 0 ( assuming it is starting from rest )a = 3.95 m/s²s = 205 mSubstitute all the parameters into the formula
v² = 0 ² + 2 × 3.95 × 205
v² = 1619.5
v = √1619.5
v = 40.24 m/s
Therefore, the sled be moving as fast as 40.24 m/s when it reaches the other end of the field.
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A skateboarder, starting from rest, rolls down a 12.8-m ramp. When she arrives at the bottom of the ramp her speed is 8.89 m/s. (a) Determine the magnitude of her acceleration, assumed to be constant. (b) If the ramp is inclined at 32.6 ° with respect to the ground, what is the component of her acceleration that is parallel to the ground?
Answer:
a) a = 3.09 m/s²
b) aₓ = 2.60 m/s²
Explanation:
a) The magnitude of her acceleration can be calculated using the following equation:
\( V_{f}^{2} = V_{0}^{2} + 2ad \)
Where:
\(V_{f}\): is the final speed = 8.89 m/s
\(V_{0}\): is the initial speed = 0 (since she starts from rest)
a: is the acceleration
d: is the distance = 12.8 m
\( a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2} \)
Therefore, the magnitude of her acceleration is 3.09 m/s².
b) The component of her acceleration that is parallel to the ground is given by:
\( a_{x} = a*cos(\theta) \)
Where:
θ: is the angle respect to the ground = 32.6 °
\( a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2} \)
Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².
I hope it helps you!
A skateboarder, starting from rest, rolls down a 12.8-m ramp the magnitude of the skateboarder's acceleration is approximately 3.07 \(m/s^2\), the component of her acceleration that is parallel to the ground is approximately 1.66 \(m/s^2\).
(a) The following kinematic equation can be used to calculate the skateboarder's acceleration:
\(v^2 = u^2 + 2as\)
\((8.89)^2 = (0)^2 + 2a(12.8)\)
78.72 = 25.6a
a = 78.72 / 25.6
a = 3.07 \(m/s^2\)
(b) Trigonometry can be used to calculate the part of her acceleration that is parallel to the ground. We are aware that the ramp's angle with the ground is 32.6°.
\(a_{parallel }= a * sin(\theta)\)
Plugging in the values:
\(a_{parallel\) = 3.07 \(m/s^2\)* sin(32.6°)
\(a_{parallel\)≈ 1.66 \(m/s^2\)
Therefore, the component of her acceleration that is parallel to the ground is approximately 1.66 \(m/s^2\).
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Lab Report Sun, Earth, and Moon Models It’s time to complete your Lab Report. Save the lab to your computer with the correct unit number, lab name, and your name at the end of the file name (e.g., U5_ Lab_SunEarthAndMoonModels_Alice_Jones.doc). Introduction 1. What was the purpose of the investigation? Type your answer here: 2. What causes the bright part of the moon to appear bright? Type your answer here: Experimental Methods 1. What materials did you use to create your model? Type your answer here: 2. Describe how you created your model. Type your answer here: Develop a Model 1. Show your model and the relationships between the components. Include labeled pictures or diagrams that describe causal accounts for the phases of the moon and eclipses. Type your answer here: Use a Model 1. Use your model to predict the relative positions of the earth, sun, and moon when the moon is full. Type your answer here: 2. Use your model to explain why a lunar eclipse does not occur every month when there is a full moon. Type your answer here:
1. The purpose of the investigation was to study the relationship between the Sun, Earth, and Moon.
2. The bright part of the moon appears bright due to the reflection of sunlight.
Introduction: The purpose of the investigation was to study the relationship between the Sun, Earth, and Moon. By creating models of the three celestial bodies, we aimed to understand how their movements and positions influence the phases of the moon and eclipses.The bright part of the moon appears bright due to the reflection of sunlight. As sunlight hits the moon's surface, it bounces back and reflects into space. This reflected light is what we see as the bright part of the moon.
Experimental Methods: To create our model, we used a lamp to represent the Sun, a ball to represent the Earth, and a smaller ball to represent the Moon. We also used a ruler, tape, and a protractor to measure distances and angles.We created our model by placing the lamp at one end of a table, the Earth in the middle, and the Moon at the other end. We attached the Moon to a string and moved it around the Earth to simulate the Moon's orbit around the Earth.
Develop a Model: Our model consists of a lamp, a ball, and a smaller ball on a string. The lamp represents the Sun, the ball represents the Earth, and the smaller ball on the string represents the Moon. As the Moon moves around the Earth, it goes through phases, from a new moon to a full moon and back again.We used diagrams and pictures to label the components of our model and describe causal accounts for the phases of the moon and eclipses.
Use a Model: When the Moon is full, it is in a direct line with the Earth and the Sun. Using our model, we can predict that the Moon would be directly opposite the Sun in the sky during a full moon. A lunar eclipse does not occur every month when there is a full moon because the Moon's orbit around the Earth is tilted at an angle of about 5 degrees to the Earth's orbit around the Sun. Therefore, the Moon is not always in a direct line with the Earth and the Sun during a full moon, which is necessary for a lunar eclipse to occur.
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