Answer:
Explanation:
Take base of the ground as origin .
component of initial velocity along i and j direction is 145 con23 and 145 sin23 . Along j , gravity acts but along i , no force acts .
The path of ball in vector form
s = (145 cos23- 14 )t i + ( 2.5 + 145sin23 t - 1/2 g t² ) j
t is time period .
b )
vertical component of initial velocity = 145 sin 23 =
for vertical displacement
v² = u² - 2gH
For maximum height , v = 0
0 = (145 sin 23 )² - 2 g H , H is maximum height attained .
H = 3209.56 / 2 x 9.8
= 163.75 m
Total height attained = 163.75 + 2.5 = 166.25 m
if time be t for reaching maximum height
v = u -gt
0 = 145 sin 23 - gt
t = 145 sin23 / g
= 5.78 s
c )
For time of flight , vertical displacement = 2.5 m
2.5 = - 145 sin 23 t + 1/2 g t²
2.5 = -56.65 t + 4.9 t²
4.9 t² - 56.65 t - 2.5 = 0
t = 11.60s
horizontal displacement during this period = 145 cos23 x 11.60 = 1548.28 m
Range = 1548.28 m.
A star's emission line of 400 nm appears shifted to 404 nm in the spectrum. What can you conclude from this shift?
A. The star is approaching you with the speed of 3000 km/s.
B. The star is approaching you with the speed of 30300 km/s.
C. The star is receding from you with the speed of 3000 km/s.
D. The star is receding from you with the speed of 30300 km/s.
Answer:
C. The star is receding from you with the speed of 3000 km/s
Explanation:
To get this answer we use the doppler effect equation . The formula for a receding emissor is given in the attachment.
We solve for V
V = 3x10⁶m/s
V = 3000km/s
We have the wavelength to be shifting towards red. Therefore we conclude that it is receding. We say the star is receding with speed of 3000km/s towards you.
Thank you!
someone help me with this exercise ?
1. if a body with a mass of 350kg is subjected to a fare of 90n what will be its mass
?
Mass remains mass no matter what you do to it.
how much heat energy is needed to raise the temperature of 2.0 kg of concrete from 10c to 30c
A skier pushes off the top of a hill with an initial speed of 3.30 m/s. How fast will she be moving after dropping 5.00 meters in elevation if friction is negligible?
Answer:
eeeeeeeeeeeeeeeeeeeee
Define position
i am not sure?
The equation r(t)= (3t+9)i+(sqrt(2)t)j+(t^2)k is the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time t=0. What is the angle?
Answer:
θ = 90º
Explanation:
The velocity is given by
v = [tex]\frac{dr}{dt}[/tex]
calculate
v = 3 i ^ + √2 j ^ + 2t k ^
acceleration is defined by
a = dv / dt
a = 2 k ^
one way to find the angle is with the dot product
v. a = | v | | a | cos θ
cos θ= v.a / | v | | a |
Let's look for the value of each term
v. a = 4 t
| v | = [tex]\sqrt{3^2 + 2 + (2t)^2 }[/tex] = [tex]\sqrt{ 11 + 4t^2}[/tex]
| a | = 2
they ask us for the angle for time t = 0
v. a = 0
| v | = √11 = 3.317
we substitute
cos θ = 0 /√11
cos θ = 0
therefore the angles must be θ = 90º
Which of the following is not a true statement?
A
B
C
D
Answer:
I think A
Explanation:
because I don't think a unknowned number can be a division problem
Which of the following statements is true? A. Both warming up and cooling down or important. B. It is more important warm up then it is to cool down. C. Is more important to cool down then is to warm up. D. Both warming up and cooling down are not important
Answer:
No. A is correct because both warming up and cooling down are important
Both warming up and cooling down are important.
This is based on aerobics and human body balance regulation as regards exercising.
Warming up and cooling down in exercising are just based on the level of intensity at which the exercise is carried out.
Now, warming up when exercising involves activities like jogging. Warming up is a very vital part of exercising as it helps to get a person's cardiovascular system ready for the subsequent exercises and physical activities to be engaged. This will help in making sure there is enough blood flowinh to your muscles as well as increasing your body temperature.In another sense, cooling down is also very vital in activity because after the blood pressure and heart rates have been raised after exercising, they will need to be restored to their normal levels at which they were before commencement of the exercise. It also helps to regulate the blood flow.Thus, Both warming up and cooling down are important.
Read more at; https://brainly.com/question/18984273
if the forces on an object are balanced the resultant force is equal to zero true false
Answer:
If the forces are balanced, the resultant force is zero. If the forces on an object are unbalanced, this is what happens: a stationary object starts to move in the direction of the resultant force. a moving object changes speed and/or direction in the direction of the resultant force.
Explanation:
A certain green light bulb emits at a single wavelength of 550 nm. It consumes 55 W of electrical power and is 75% efficient in converting electrical energy into light. (a) How many photons does the bulb emit in one hour? (b) Assuming the emitted photons to be distributed uniformly in space, how many photons per second strike a 10 cm by 10 cm paper held facing the bulb at a distance of 1.0 m?
Answer:
a) #_total = 4 10²³ photons / h, b) # _photon_area = 3 10²² photons
Explanation:
a) Let's start by calculating the energy of an emitted photon
E₀ = h f
c = λ f
substituting
E₀ = h c /λ
E₀ = [tex]\frac{6.62 \ 10^{-34} \ 3 \ 10^{8} }{550 \ 10^{9} }[/tex]
E₀ = 3.6 10⁻¹⁹ J
Let's use a direct proportion rule (res rule) if a photon has Eo, how many photons are there in 55W
#_foton = 55 / E₀
# _photon = 55 / 3.6 10⁻¹⁹ = 15.27 10¹⁹ photons
This version is with 100% if the conversion is 75%, how many footnes are there
#_foton_real = # _foton 75/100
#real_photon = 15.27 10¹⁹ 0.75
# _real_photon = 11.45 10¹⁹ photons
this is the broadcast in a second
#_total = # _real_photon t
#_total = 11.45 10¹⁹ 3600
#_total = 4.1 10²³ photons / h
#_total = 4 10²³ photons / h
b) This number of photons is constant, so after being emitted they are distributed on the surface of a sphere, in this case of radius r = 1.0 m
the volume of a sphere is
A = 4π r²
A = 4π 1²
A = 12,566 m²
the area of the plate is
A₁ = l₁ l₂
A₁ = 0.10 0.10
A₁ = 1 10⁻² m
Let's use a direct proportion rule, if there are 4.1 10²³ photons in an area A, how much are there in an area A₁
# _photon_area = #_total A₁ / A
# _photon_area = [tex]4.1 \ 10^{23} \ \frac{1 \ 10^{-2} }{12.566}[/tex]
# _photon_area = 3.26 10²²
as the number of photons must be a whole number
# _photon_area = 3 10²² photons
A 20 kg box has an initial velocity of 2 m/s starting at the bottom of a 30-degree inclined plane. A person pushes on the box directly up the frictionless inclined plane so that it travels up the inclined plane at a constant velocity of 2 m/s. Calculate the how much is done by the person after 5 seconds have past.
Answer:
Explanation:
The box is moving with constant velocity so acceleration of box is zero . That means net force on the box is zero .
The weight component acting on box parallel to incline plane
= mg sin 30⁰ = 20 x 9.8 x sin 30 = 98 N
This force is acting down the plane , hence to make the net force zero acting on box , force exerted by person will also be 98 N up the incline .
Force exerted by person = 98 N
distance travelled in 5 s
= velocity x time
= 2 x 5 = 10 m
Work done by person
= 98 x 10
= 980 J .
For this assignment, you should mathematically solve and record a video testing your solution for the following prompt: Two rolls of toilet paper, of equal mass and radius, are dropped from different heights so that they hit the ground at the same time. One roll of toilet paper is dropped normally while the other is dropped while a person holds onto a sheet of toilet paper such that the roll unravels as it descends. Determine the ratio of heights h1/h2, where h1 represents the height of the toilet paper dropped normally and h2 represents the height of the toilet paper that unravels, so that both rolls hit the ground at the same time.
Answer:
h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]
Explanation:
Using two rolls of tissue paper : One roll dropped normally while the other drops as some holds onto a sheet of the toilet paper ( I.e. the tissue paper drops rotating about its axis )
Determine the ratio of heights h1/h2
mass of tissues = same
radius of tissues = same
h1 = height of tissue 1
h2 = height of tissue 2
For the first tissue ( Tissue that dropped manually )
potential energy = kinetic energy
mgh = 1/2 mv^2
therefore the final velocity ( v^2 ) = 2gH ----- ( 1 )
second tissue ( Tissue that dropped while rotating )
gh = [tex]\frac{v^2}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] ) ------ ( 2 )
To determine the ratio of heights we will equate equations 1 and 2
hence :
gh = [tex]\frac{2gH}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] )
∴ h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]
why no tempature can be lower than 0 kelvin
Answer:
At zero kelvin (minus 273 degrees Celsius) the particles stop moving and all disorder disappears. Thus, nothing can be colder than absolute zero on the Kelvin scale. Physicists have now created an atomic gas in the laboratory that nonetheless has negative Kelvin values.
Explanation:
A spring has a spring constant of 25 Newtons per meter. The minimum force required to
stretch the spring 0.20 meter from its equilibrium position is approximately
Answer:
6.3N
Explanation:
Guessed it right on castle learning
Answer:
6.3 N
Explanation:
F=kx
F=(25N/m)(0.25m)
6.3 N
The all-digital touch-tone phones use the summation of two sine waves for signaling. Frequencies of these sine waves are defined as 697, 770, 852, 941, 1209, 1336, 1477, and 1633 Hz. Since the sampling rate used by the telecommunications is 8000 Hz, convert those eight analog frequencies into digital frequencies of radians and cycles.
A ball 12 m in 4 seconds and then 2.5 seconds later it rolls 8 m in 2 seconds what is its acceleration
Answer:
If it accelerates at 20 m/s2 for a period of 22 seconds, what is its final velocity? ... How fast is the ball falling after 5 seconds? v = v0 + gt v = 0 + 10(5) v = 50 m/s. 4. ... + ½ 2.5(15)2 x = 281 m. 5. What is the total displacement of the car in question 2? ... 8. A base jumper falls until he reaches a speed of 200 m/s
Explanation:
Balance the following equation:
H3B03 →_B203 +_H20
a. 1, 3,2
b. 2,4,6
C. 4, 2, 6
d. 6, 4,2
It's c I think ( 4 , 2 , 6 , )
A 1325 kg car and a 2050 kg pickup truck approach a curve on a highway that has a radius of 255 m. At what angle should the highway engineer bank this curve so that vehicles traveling at 75.0 mi/h can safely round it regardless of the condition of their tires
Answer:
the banking angle of the road is 24.2⁰
Explanation:
Given;
speed of the vehicles considered, v = 75 mi/h
Speed in m/s ⇒ 1 mi/h --------> 0.44704 m/s
75 mi/h --------> ?
= 75 x 0.44704 m/s = 33.528 m/s
radius of the curve, r = 255 m
The banking angle of the road is calculated as;
[tex]\theta = tan^{-1} (\frac{v^2}{rg} )\\\\\theta = tan^{-1} (\frac{33.528^2}{255\times 9.8} )\\\\\theta = tan^{-1}(0.44983)\\\\\theta =24.2^0[/tex]
Therefore, the banking angle of the road is 24.2⁰
The angle of banking is 24 degrees.
What is the angle of banking?As a driver approaches a bend two equal and opposite forces act on him which are the centripetal force and the centrifugal force. The driver will have to ben through a certain angle called the angle of banking to avoid falling off.
The angle of banking depends on the speed of the vehicle and the radius of the curve.
θ = v^2/rg
speed = 75.0 mi/h or 33.5 m/s
r = 255 m
g = 9.8 ms-1
θ = tan-1 (33.5 m/s)^2/ 255 m × 9.8 ms-1
θ = tan-1(1122.3/2499)
θ = 24 degrees
Learn more about banking: https://brainly.com/question/12134554
As the distance between the sun and earth decreases, the speed of the planet
a
increases
b
decreases
c
stays the same
Answer:
Explanation:
Increases. The force of gravity is distance dependent. Therefore, a smaller 'r' value will result in a larger force. Net force is proportional to the acceleration, so the planet will increase its speed.
before:
2m
What is the change in momentum if the v before = 1m/s and the velocity after = 1m/s?
(Change in momentum - momentum after-
after:
2m
momentum before) p - mv
Answer:
22
Explanation:
before:
2m
What is the change in momentum if the v before = 1m/s and the velocity after = 1m/s?
(Change in momentum - momentum after-
after:
2m
momentum before) p - mv
An object with an initial horizontal velocity of 20 ft/s experiences a constant horizontal acceleration due to the action of a resultant force applied for 10 s. The work of the resultant force is 10 Btu. The mass of the object is 55 lb. Determine the constant horizontal acceleration, in ft/s2.
Answer:
a = 7.749 ft/s²
Explanation:
First to all, we need to convert all units, so we can work better in the calculations.
The horizontal acceleration is asked in ft/s² so the units of speed will be the same. The Work is in BTU and we need to convert it in ft.lbf in order to get the acceleration and final speed in ft/s:
W = 10 BTU * 778.15 Lbf.ft / BTU = 7781.5 lbf.ft
Now, to get the acceleration we need to get the final speed of the object first. This can be done, by using the following expression:
W = ΔKe (1)
And Ke = 1/2mV²
So Work would be:
W = 1/2 mV₂² - 1/2mV₁²
W = 1/2m(V₂² - V₁²) (2)
Finally, we need to convert the mass in lbf too, because Work is in lbf, so:
m = 55 lb * 1 lbf.s²/ft / 32.174 lb = 1.7095 lbf.s²/ft
Now, we can calculate the final speed by solving V₂ from (2):
7781.5 = (1/2) * (1.7095) * (V₂² - 20²)
7781.5 = 0.85475 * (V₂² - 441)
7781.5/0.85475 = (V₂² - 400)
9103.83 + 400 = V₂²
V₂ = √9503.83
V₂ = 97.49 ft/s
Now that we have the speed we can calculate the acceleration:
a = V₂ - V₁ / t
Replacing we have:
a = 97.49 - 20 / 10
a = 7.749 ft/s²Hope this helps
An aircraft flies 800km due East and then 600km due north. Determine the magnitude of it's displacement. please it's urgent
Answer:
Magnitude of it's displacement = 1,000 km
Explanation:
Given:
Distance towards east = 800 km
Distance towards north = 600 km
Find:
Magnitude of it's displacement
Computation:
Magnitude of it's displacement = √800² + 600²
Magnitude of it's displacement = √640000 + 360000
Magnitude of it's displacement = √10000000
Magnitude of it's displacement = 1,000 km
John attaches a ball to a spring. The diagram below shows what happens. Which option shows the direction of the force of the ball on the spring?
Option C shows the direction of the force of the ball on the spring. The direction of the force of the ball on the spring will be downwards.
What is force?Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.
Force is defined as the product of mass and acceleration. Its unit is Newton.
The spring is extended downward because the weight is always act downwards. The direction of the force of the ball on the spring will be downwards.
Hence, option C shows the direction of the force of the ball on the spring
To learn more about the force, refer to the link;
https://brainly.com/question/26115859
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A fairgrounds ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has a 9.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 1.80 times that due to gravity
Answer:
13.37 rev/min
Explanation:
acceleration due to gravity (g) = 9.8 m/s², centripetal acceleration ([tex]a_c[/tex]) = 1.8 * g = 1.8 * 9.8 m/s² = 17.64 m/s².
r = 9 m
Centripetal acceleration ([tex]a_c[/tex]) is given by:
[tex]a_c=\frac{v^2}{r} \\\\v=\sqrt{a_c*r} \\\\v=\sqrt{17.64\ m/s^2*9\ m}\\\\v=12.6\ m/s[/tex]
The velocity (v) is given by:
v = ωr; where ω is the angular velocity
Hence:
ω = v/r = 12.6 / 9
ω = 1.4 rad/s
ω = 2πN
N = ω/2π = 1.4 / 2π
N = 0.2228 rev/s
N = 13.37 rev/min
Earth has seasons because _____.
it rotates on its axis as it moves around the sun
the temperature of the sun changes
its axis is tilted
the distance between Earth and the sun changes
Answer:
c, its axis is tilted
maybe
As it works its way around the sun, its tilted axis exposes different parts of earth.
C would be it because the roation of Earth on its axis doesn't have anything to do with the exposer of the revolution on its axis
A sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 35 km/h at the 67-m mark. He then maintains this speed for the next 88 meters before uniformly slowing to a final speed of 32 km/h at the finish line. Determine the maximum horizontal acceleration which the sprinter experiences during the run. Where does this maximum acceleration value occur
Answer:
0.705 m/s²
Explanation:
a) The sprinter accelerates uniformly from rest and reaches a top speed of 35 km/h at the 67-m mark.
Using newton's law of motion:
v² = u² + 2as
v = final velocity = 35 km/h = 9.72 m/s, u = initial velocity = 0 km/h, s = distance = 67 m
9.72² = 0² + 2a(67)
134a = 94.484
a = 0.705 m/s²
b) The sprinter maintains this speed of 35 km/h for the next 88 meters. Therefore:
v = 35 km/h = 9.72 m/s, u = 35 km/h = 9.72 m/s, s = 88 m
v² = u² + 2as
9.72² = 9.72² + 2a(88)
176a = 9.72² - 9.72²
a = 0
c) During the last distance, the speed slows down from 35 km/h to 32 km/h.
u = 35 km/h = 9.72 m/s, v = 32 km/h = 8.89 m/s, s = 200 - (67 + 88) = 45 m
v² = u² + 2as
8.89² = 9.72² + 2a(45)
90a = 8.89² - 9.72²
90a = -15.4463
a = -0.1716 m/s²
The maximum acceleration is 0.705 m/s² which is from 0 to 67 m mark.
what is the acceleration of a satellite moving in a circular orbit around the earth of radius 2r
Explanation:
You do the radius times the circumference of the earth
HELP ! ILL MARK BRAINLIEST HELP ASAP
Answer:
A
Explanation:
In 5 minutes, they went 10 miles at both 2, 3, and 4 checkpoints. The bus then starts to speed up.
Hope this helps!
calculate the force needed to push the ball up a 4 m ramp if the work is equal to 16 joules.
A long, straight wire has a uniform constant charge with linear charge density, - 3.60 nC/m. The wire is surrounded by a long nonconducting, thin-walled cylindrical shell that is charged on its outside surface, such that the electric field outside the shell is zero. The shell has a radius of 1.50 cm.
Required:
What uniform area charge density rho is needed on the shell for the electric field to be zero outside the shell?
Answer:
Uniform area charge density rho is needed is 3.82*10^-8 C.m^-2
Explanation:
See the attached files.
To find the rho, I used Gauss law for cylindrical shell which is equation 1 and Gauss law for the rod which is equation 4.
Note that in equation 4, Lamda is the charge per length while L is the length if the rod. Also R is the radius of the shell.
The final answer is 3.82*10^-8 C.m^-2 which is the uniform area charge density rho is needed.