Answer:
a) t = 6.62 s
b) x = 238.6 m
c) H = 53.7 m
Explanation:
a) We can find the time of flight as follows:
[tex] y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} [/tex]
Where:
[tex]y_{f}[/tex] is the final height = 0
[tex]y_{0}[/tex] is the initial height = 0
[tex]v_{0_{y}}[/tex] is the initial vertical velocity of the stone
t: is the time
g: is the gravity = 9.81 m/s²
[tex] v_{0}sin(42)t - \frac{1}{2}gt^{2} = 0 [/tex]
[tex] 48.5 m/s*sin(42)*t - \frac{1}{2}9.81 m/s^{2}*t^{2} = 0 [/tex]
By solving the above quadratic equation we have:
t = 6.62 s
b) The maximum range is:
[tex] x = v_{0_{x}}t = 48.5 m/s*cos(42)*6.62 s = 238.6 m [/tex]
c) The maximum height (H) can be found knowing that at this height the final vertical velocity of the stone is zero:
[tex] v_{f_{y}}^{2} = v_{0_{y}}^{2} - 2gH [/tex]
[tex] H = \frac{v_{0_{y}}^{2} - v_{f_{y}}^{2}}{2g} = \frac{(48.5 m/s*sin(42))^{2} - 0}{2*9.81 m/s^{2}} = 53.7 m [/tex]
I hope it helps you!
What is a possible equation fir an ionic compound with calcium
Answer:
CaO, CaMg, or CaF2
Ca2Cl, CaNa, or CaP
CaF, CaMg, or CaNa
CaO, CaF2, or CaCl2
Explanation: i think
Calculate the pressure in pascals if a 560N is applied to an area of 1.6m².
Answer:
P = 350 [Pa]
Explanation:
The pressure is calculated by knowing the force and area. By means of the following formula can be calculated by means of the following equation.
Units of 1 [Pa] = 1 [N/m²]
P = F/A
where:
P = pressure [Pa]
F = force = 560 [N]
A = area = 1.6 [m²]
P = 560/1.6
P = 350 [Pa]
Under what conditions the reaction rate of an enzymolysis that follows Michaelis-Menten kinetics is a quarter of its maximum value?
a. [S]=KM
b. [S]=KM/3
c. [S]=2KM
d. [S]=KM+3
e. [S]= (KM)^1/3
Solution :
Michaelis-Menten kinetics in the field of biochemistry is considered as one of the well known models for enzyme kinetics. The model represents an equation that describes the enzymatic reactions's rate by relating the reaction rate to the substrate's concentration. The equation is named after the two famous scientists, Leonor Michaelis and Maud Menten.
The formula is :
[tex]$v=\frac{V_{max}[S]}{K_M + [S]}$[/tex]
where v = velocity of reaction
[tex]$V_{max}$[/tex] = maximum rate achieved
[tex]$K_M$[/tex] = Michaelis constant
[S] = concentration of the substrate, S
According to the question, by putting the velocity of reaction, v as [tex]$\frac{V_{max}}{4}$[/tex], we get the above equation as
[tex]$[S]= \frac{K_M}{3}$[/tex]
Therefore the answer is [tex]$[S]= \frac{K_M}{3}$[/tex]
find the mass of an object with a density of 1.5 g/cm^3 and had a volume of 8cm^3
Answer:
12 gExplanation:
The mass of a substance when given the density and volume can be found by using the formula
mass = Density × volume
From the question we have
mass = 1.5 × 8
We have the final answer as
12 gHope this helps you
Which dwarf planet has a moon nearly as large as itself?
Answer:
The answer is the Pluto.
A group of students conduct an experiment with a block of wood sliding down an incline. They find that the final energy of the block is less than the initial energy of the block. Which statement best describes this situation?
A. The students made an error in deterring the final energy.
B. The students made an error in calculating the initial energy.
C. Some energy was transformed to other forms.
D. Some energy was destroyed or lost.
Light of wavelength 400 nm falls on a metal surface having a work function 1.70 eV. What is the maximum kinetic energy of the photoelectrons emitted from the metal?
a. 4.52 eV
b. 3.11 ev
c. 141 eV
d. 2.82 eV
e. 1.70 eV
Answer:
Ke = 1.41 eV
Explanation:
We know that
h = 6.626 x 10⁻²⁴
c = 3 x 10⁸
λ = 4 x 10⁻⁹ x 1.6 x 10⁻¹⁹
Computation:
Ke = [hc / λ] - w
Ke = [(6.626 x 10⁻²⁴)(3 x 10⁸) / (4 x 10⁻⁹ x 1.6 x 10⁻¹⁹)] - 1.70
Ke = 1.41 eV
If the hiker starts climbing at an elevation of 350 ft, what will their change in gravitational potential energy be, in joules, once they reach the top
Answer:
352,088.37888Joules
Explanation:
Complete question;
A hiker of mass 53 kg is going to climb a mountain with elevation 2,574 ft.
A) If the hiker starts climbing at an elevation of 350 ft., what will their change in gravitational potential energy be, in joules, once they reach the top? (Assume the zero of gravitational potential is at sea level)
Chane in potential energy is expressed as;
ΔGPH = mgΔH
m is the mass of the hiker
g is the acceleration due to gravity;
ΔH is the change in height
Given
m = 53kg
g = 9.8m/s²
ΔH = 2574-350 = 2224ft
since 1ft = 0.3048m
2224ft = (2224*0.3048)m = 677.8752m
Required
Gravitational potential energy
Substitute the values into the formula;
ΔGPH = mgΔH
ΔGPH = 53(9.8)(677.8752)
ΔGPH = 352,088.37888Joules
Hence the gravitational potential energy is 352,088.37888Joules
The change in gravitational potential energy be, once the hiker reach the top of the mountain is 352088 joules or 352.1 kJ.
What is gravitational potential energy?Gravitational potential energy is the energy which a body posses because of its position.
The gravitational potential energy of a body is given as,
[tex]U=mgh[/tex]
Here, (m) is the mass of the body, (g) is the gravitational force and (h) is the height of the body.
The mass of the hiker is 53 kg and the height of the climb is 2574 ft.
Now, the hiker starts climbing at an elevation of 350 ft. Thus, the net height of the hiker has to climb is,
[tex]h=2574-350\\h=2224\rm\; ft[/tex]
Convert this into the meter by multiplying 03048 as,
[tex]h=2224\times0.3048\\h=677.8752\rm\; m[/tex]
It is known that the value of g is 9.8 m/s². Plug in all the values as,
[tex]U=53\times9.8\times677.8752\\U=352088J\\U=352.1 \;\rm kJ[/tex]
Thus, the change in gravitational potential energy be, once the hiker reach the top of the mountain is 352088 joules or 352.1 kJ.
Learn more about the gravitational potential energy here;
https://brainly.com/question/15896499
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.330 mm wide. The diffraction pattern is observed on a screen 2.55 m away. Define the width of a bright fringe as the distance between the minima on either side.
a. What is the width of the central bright fringe?
b. What is the width of the first bright fringe on either side of the central one?
Answer:
a) y_total = 19.916 10⁻⁵ m , b) Δθ = 1.91 10-⁻³ rad
Explanation
This is a diffraction exercise that is described by the expression
a sin θ = m λ
the first minimum occurs for m = 1
a sin θ = λ
sin θ = λ / a
θ = sin⁻¹ (633 10⁻⁹ / 0.330 10⁻³)
θ = 1.918 10⁻³ rad
let's use trigonometry
tan θ = y / x
y = x tan θ
y = 2.55 tan (3.936 10-3)
y = 5.75 10⁻- m
this value is from the central maximum to one extreme of the value,
y_ total = 2 y
y_total = 2 (5.75e1)
y_total = 19.916 10⁻⁵ m
b) For the second point and constructive inference we have m = 2
sin θl = m λ
θ = sin⁻¹ (lat / a)
θ = sin⁻¹ (2 633 10-9 / 0.33010-3) = son-1 (3.836 10-3)
θ = 3.84 10-3 give
The width of this maximum is
Δθ = 1.3 10-3
Δθ = 3.84 10⁻³- 1.918 10⁻³
Δθ = 1.91 10-⁻³ rad
What is your worldview? Explain?
Answer:
My worldview is that the world is a beautiful place. IT has many wonders, and many kind people. There are also very nice places, governments, tech and more.
If the light ray moves from inside the glass toward the glass–air interface at an angle of 30.0° to the normal, determine the angle of refraction.
Answer:
This question is incomplete
Explanation:
This question is incomplete, however, the relationship to between the angle of incidence and angle of refraction is best explained by snell's law with the equation below
ni sinθi = nr sinθr
where ni is the refractive index of the incident medium (assuming it's plate glass then 1.52)
nr is the refractive index of the refractive medium (1.33)
θi is the angle of incidence (30°)
θr is the angle of refraction (X)
If we substitute the values, we have
1.52 × sin 30 = 1.33 × sin X
1.52 × 0.5 = 1.33 × sin X
0.75 = 1.33 × sin X
sin X = 0.75/1.33
sin X = 0.571
X = sin⁻¹ 0.571
X = 34.82°
Thus, the angle of refraction will be 34.82°
What is the best description for a transverse wave? *
A.The movement of the atoms around the wave is parallel to the movement of the ENERGY it is transferring.
B.The movement of the atoms around the wave is parallel to the movement of the MATTER it is transferring.
C.The movement of the atoms around the wave is perpendicular to the movement of the energy it is transferring.
D.The movement of the atoms around the wave is perpendicular to the movement of the matter it is transferring.
Answer:
Option - CHope it helps youA machinist with normal vision has a near point at 25 cm. This machinist wears +4.25-diopter eyeglasses in order to do very close work. With these eyeglasses, what is the near point of the machinist?
a. 12 cm
b. 10 cm
c. 7 cm
d. 15 cm
e. 17 cm
Answer: 12cm.
Explanation:
First we need to calculate the focal length which will be:
= 100 / 4.25
= 23.529
The near point of the machinist will be solved using the equation:
1/u + 1/v = 1/f
where,
v = -25
f = 23.529
1/u + 1/v = 1/f
1/u - 1/25 = 1/23.529
1/u = 1/23.529 + 1/25
u = 12.12cm
u = 12cm approximately
Therefore, the near point of the machinist is 12cm
what is simple harmonic motion
give two examples
Answer:
S.H.M:- If the acceleration of the vibrating body directly varies with the displacement of the body from the mean position and always directed to the mean position, the motion of that body is called simple harmonic motion.
Ex:- (i) The motion of a pendulum is an S.H.M.
(ii) The motion of vibrating mass attached to a spring is an S.H.M.
Artificial gravity is a must for any space station if humans are to live there for any extended length of time. Without artificial gravity, human growth is stunted and biological functions break down.
An effective way to create artificial gravity is through the use of a rotating enclosed cylinder, as shown in the figure. Humans walk on the inside of the outer edge of the cylinder, which has a diameter of =3335 m that is large enough such that its curvature is not readily noticeable to the inhabitants. (The space station in the figure is not drawn to scale.)
Once the space station is rotating at the necessary angular speed to create an artificial gravity of 1 , how many minutes would it take the space station to make one revolution?
I did [tex]2\pi \sqrt{\frac{1667.6}{9.8} } =81.96[/tex] then I converted 81.96 to minutes which was 1.4 but it still got marked wrong. 81.96 was wrong as well. Am I using the wrong equation for it? I'm not sure what to do.
The period of the enclosed cylinder is approximately 115.866 seconds.
The rotating enclosed cylinder is rotating at constant angular speed ([tex]\omega[/tex]), in radians per second, which means that experiments a constant radial angular acceleration ([tex]\alpha[/tex]), in radians per square second. Then, we derive an expression for the period of the cylinder, this is, the time needed by the cylinder to make one revolution:
[tex]g = \omega^{2}\cdot R[/tex] (1)
Where:
[tex]g[/tex] - Gravitational acceleration, in meters per square second. [tex]R[/tex] - Radius of the enclose cylinder, in meters.[tex]g = \frac{4\pi^{2}\cdot R}{T^{2}}[/tex]
[tex]T = 2\pi\cdot \sqrt{\frac{R}{g} }[/tex] (2)
Where [tex]T[/tex] is the period, in seconds.
If we know that [tex]R = 3335\,m[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the period of the enclosed cylinder is:
[tex]T = 2\pi\cdot \sqrt{\frac{3335\,m}{9.807\,\frac{m}{s^{2}} } }[/tex]
[tex]T \approx 115.866\,s\,(1.931\,min)[/tex]
The period of the enclosed cylinder is approximately 115.866 seconds.
We kindly invite to check this question on circular motion: https://brainly.com/question/2285236
For a 99 kg person standing at the equator, what is the magnitude of the angular momentum about Earth's center due to Earth's rotation?
Answer:
L = 4.58 x 10⁴ kg.m²/s
Explanation:
The angular momentum is given by the formula:
L = mvr
but, v = rω
Therefore,
L = mr²ω
where,
L = Angular Momentum of the person = ?
m = mass of person = 99 kg
r = radius of earth = 6.37 x 10⁶ m
ω = Angular Speed of Earth's Rotation = θ/t
Since, earth completes 1 rotation in 1 day. Hence,
ω = (2π rad/1 day)(1 day/24 h)(1 h/3600 s)
ω = 7.27 x 10⁻⁵ rad/s
Therefore,
L = (99 kg)(6.37 x 10⁶ m)²(7.27 x 10⁻⁵ rad/s)
L = 4.58 x 10⁴ kg.m²/s
1. The resistance of an electric device is 40,000 microhms. Convert that measurement to ohms
2. When an electric soldering iron is used in a 110 V circuit, the current flowing through the iron is
2 A. What is the resistance of the iron?
3. A current of 0.2 A flows through an electric bell having a resistance of 65 ohms. What must be
the applied voltage in the circuit?
Answer:
(1) 0.04 ohms (2) 55 ohms (3) 13 volt
Explanation:
(1) The resistance of an electric device is 40,000 microhms.
We need to convert it into ohms.
[tex]1\ \mu \Omega =10^{-6}\ \Omega[/tex]
To covert 40,000 microhms to ohms, multiply 40,000 and 10⁻⁶ as follows :
[tex]40000 \ \mu \Omega =40000 \times 10^{-6}\ \Omega\\\\=0.04\ \Omega[/tex]
(2) Voltage used, V = 110 V
Current, I = 2 A
We need to find the resistance of the iron. Using Ohms law to find it as follows :
V = IR, where R is resistance
[tex]R=\dfrac{V}{I}\\\\R=\dfrac{110}{2}\\\\R=55\ \Omega[/tex]
(3) Current, I = 0.2 A
Resistance, R = 65 ohms
We need to find the applied voltage in the circuit. Using Ohms law to find it as follows :
V=IR
V = 0.2 × 65
V = 13 volt
Answer:
1. 0.04 Ohms
2. 55 Ohms
3. 13 Volts
Explanation:
Penn Foster
This is a graph of a car speeding up and then reaching a constant speed of 105 m/s . After it finishes speeding up, how far does the car travel during the last 6 seconds of the graph, from 4 to 10 s?
60 m
Explanation:Concept Used:
We know that the area under a velocity-time graph represents the Displacement of the body
Displacement in the Last 6 seconds:
To find the Displacement in the last 6 seconds, we will find the area under the graph between x = 4 and x = 10
We can see that the shape formed is a rectangle also shown in the given graph. So, the area of the rectangle is the Displacement of the car in the last 6 seconds
Area of the Rectangle:
From the graph, we know that the rectangle is 10 (m/s) tall and 6 (s) wide
Area of Rectangle= length*Breadth
replacing the values
Area = 10 (m/s) * 6 (s)
Area = 60 m
Hence, the car travelled 60 m in the last 6 seconds of the graph
A student puts a besker of warm water next to a besker of cold water so that they fough which two statements are true? DA Thermal energy will move from the warm water to the cold water Thermal energy will move from the air to the cold water. Thermal energy utill move from the air to the warm water Thermal energy till move from the cold water to the warm water
Explanation:
so sorry
don't know but please mark me as brainliest please
n an experiment of a simple pendulum, measurements show that the pendulum has length m, mass kg, and period s. Take m/s2 . i. Use the measured length to predict the theoretical pendulum period with a range of error (use the error propagation method you learned in Lab 1). ii. Compute the percentage difference (as defined in Lab 1) between the measured value and the predicted value .
Answer:
The answer is "[tex](1.265 \pm 0.010) \ s \ and \ 0.709 \%[/tex]"
Explanation:
In point i:
[tex]T_{theo}= 2\pi \sqrt{\frac{l}{g}}[/tex]
[tex]=2\pi\sqrt{\frac{0.397}{9.8}}\\\\= 1.265 \ s[/tex]
If error in the theoretical time period :
[tex]\frac{\Delta T_{theo}}{T_theo} = \frac{1}{2} \frac{\Delta l }{l}\\\\\Delta T_{theo} = 1.265 \times \frac{1}{2} \times \frac{0.006}{0.397}[/tex]
[tex]= 0.010 \ s[/tex]
[tex]T_{theo} = (1.265 \pm 0.010) \ s[/tex]
In point ii:
[tex]\% \ difference = \frac{|T_{exp} -T_{theo}|}{\frac{T_{exp}+T_{theo}}{2}} \times 100[/tex]
[tex]= \frac{1.274 -1.265}{\frac{1.274+1.265}{2}} \times 100\\\\=0.709 \%[/tex]When a 20.0-ohm resistor is connected across the terminals of a 12.0-V battery, the voltage across the terminals of the battery falls by 0.300 V. What is the internal resistance of this battery?
a. 3.60 Ω
b. 0.51 Ω
c. 0.30 Ω
d. 1.56 Ω
e. 0.98 Ω
Answer:
B. 0.51 ohms
Explanation:
Our data is as follows:
Resistor R = 20.0ohms
V1 = 12 volts
V2 = 0.300 V
We are to get the internal resistance of this battery
Now,
I = change in voltage/R
Change in voltage = v1-v2
= 12-0.300
= 11.7
Then,
I = 11.7/20
= 0.585Ri
Now
O.300 = 0.585Ri
Ri = 0.300/0.585
= 0.51 ohms
This is therefore the internal resistance of the battery.
Thank you!
If an object has a mass of 47 kg and it is moved 27 meters in 60 seconds, how much power was used?
Answer: 207 W
Explanation: I assumed here that the object is moved vertically. If that is the case, the work done on the object is equal to its change in gravitational potential energy:
where
m = 47 kg is the mass of the object
g = 9.8 m/s^2 is the acceleration of gravity
is the change in height
Substituting,
Now we can calculate the power used, which is given by
Hope this helps I'm sorry if i'm wrong but I tried :(
A copper block rests 17.4 cm from the center of a steel turntable. The coefficient of static friction between block and surface is 0.465. The turntable starts from rest and rotates with a constant angular acceleration of 0.406 rad/s 2 . The acceleration of gravity is 9.8 m/s 2 . After what interval will the block start to slip on the turntable
Answer:
12.61 s
Explanation:
Given that
Distance from the center if the turntable, r = 17.4 cm = 0.174 m
Coefficient of static friction, μ = 0.465
Angular acceleration, α = 0.406 rad/s²
Acceleration due to gravity, g = 9.8 m/s²
We know that
F_max = μmg
Also, we know that
F = mω²r
Now, for slip to occur, both forces must be equal to one another, and thus
mω²r = μmg
ω²r = μg
ω² = μg/r
ω² = (0.465 * 9.8)/0.174
ω² = 4.557 / 0.174
ω² = 26.19
ω = √26.19
ω = 5.12 rad/s
t = ω/α
t = 5.12/0.406
t = 12.61 s.
Thus, after 12 seconds, the block will start to slip on the turntable
A metallic spherical shell has a charge density of 1 mC/m2 on its surface. The shell has radius of 7.5 cm. Determine the electric field magnitude (in kV/m) 10 meters away from the center of the shell.
Answer:
Explanation:
charge on the shell = 4πR² x charge density
= 4 x 3.14 x 7.5² x 10⁻⁴ x 10⁻³ C
= 706.5 x 10⁻⁷ C
electric field = k Q / d² , d is distance of point from centre
= 9 x 10⁹ x 706.5 x 10⁻⁷ / 10²
= 6358.5 N /C
= 6.358 kV / m
I’m testing if this app works.What is gravity?
Answer:
Gravity, or gravitation, is a natural phenomenon by which all things with mass or energy—including planets, stars, galaxies, and even light—are brought toward one another. On Earth, gravity gives weight to physical objects, and the Moon's gravity causes the ocean tides.
I need help ASAP
The water cycle functions because the matter in water changes:
a.forms
b.properties
c.minerals
Answer:
(a). forms.
Explanation:
Bro even i ain't sure.
Answer:
I think its A? Because in the water cycle it goes from a liquid to a gas?
Explanation:
Which vocabulary word would describe both a cell membrane and a screen door?
Answer:
flexible covering
Explanation:
frequency of a vibrating string with a period of 4 seconds
Answer: 2 Hz (8 cycles/4 s = 2 cycles/s).
Explanation:
If a coil of slinky makes 3 vibrational cycles in one second, then the frequency is 3 Hz. And if a coil makes 8 vibrational cycles in 4 seconds, then the frequency is 2 Hz (8 cycles/4 s = 2 cycles/s).
What form of energy is released into the atmosphere by the earth's surface
Answer:
Thermal Energy (Heat)
Answer:
Heat
Explanation:
the form that is raised from the atmosphere from earths surface is heat
the force of gravity acting on an object is known as