Answer:
v = 10.46 m/s
x = 30.134 m from house
Explanation:
given data
speed = 3.05 m/s
time = 7 s
height = 40.8 mm
solution
we get here first time required to fall that is t
t = [tex]\sqrt{\frac{2\times 40.8}{9.8}}[/tex] ..................1
t = 2.88 s
now we take here initial speed that is v
so v × t = 3.05 × ( t+ 7)
v = [tex]\frac{3.05\times (2.88 + 7)}{2.88}[/tex]
v = 10.46 m/s
and
when she catch the bagel henrietta was at
x = 3.05 × ( 2.88 + 7)
x = 30.134 m from house
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.
A 0.14 kg baseball moving at 23.00 m s is slowed to a stop by a catcher who exerts a constant force of 370 N How long does it take this force to stop the ball Incorrect Your answer is incorrect s How far does the ball travel before stopping
Answer:
(A) The time taken for the ball to stop is 8.7 x 10⁻³ s
(B) The distance traveled by the baseball before stopping is 0.3 m
Explanation:
Given;
mass of the baseball, m = 0.14 kg
velocity of the baseball, v = 23 m/s
force exerted on the baseball by the catcher, F = 370 N
(A) The time taken for the ball to stop;
[tex]F = ma = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{0.14 \times 23}{370} \\\\t = 8.7\times 10^{-3} \ s\\\\t = 8.7 \ ms[/tex]
(B) The distance traveled by the baseball before stopping is calculated as;
acceleration of the ball, [tex]a = \frac{v}{t} = \frac{23}{8.7\times 10^{-3}} = 2643.678 \ m/s^2[/tex]
Distance traveled, s;
s = ut + ¹/₂at²
s = (23)(8.7 x 10⁻³) + ¹/₂(2643.678)(8.7 x 10⁻³)²
s = 0.2001 + 0.1001
s = 0.3 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!
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 .
calculate the force needed to push the ball up a 4 m ramp if the work is equal to 16 joules.
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
A baseball is hit when it is 2.5 ft above the ground. It leaves the bat with an initial velocity of 145 ft/sec at a launch angle of 23°. At the instant the ball is hit, an instantaneous gust of wind blows against the ball, adding a component of -14i (ft/sec) to the ball’s initial velocity. A 15-ft-high fence lies 300 ft from home plate in the direction of the flight.
a. Find a vector equation for the path of the baseball.
b. How high does the baseball go, and when does it reach maxi-mum height?
c. Find the range and flight time of the baseball, assuming that the ball is not caught.
d. When is the baseball 20 ft high? How far (ground distance) is the baseball from home plate at that height?
e. Has the batter hit a home run? Explain.
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.
Two motorcycles are traveling due east with different velocities. However, 5.68 seconds later, they have the same velocity. During this 5.68-second interval, motorcycle A has an average acceleration of 3.87 m/s2 due east, while motorcycle B has an average acceleration of 18.2 m/s2 due east. (a) By how much did the speeds differ at the beginning of the 5.68-second interval, and (b) which motorcycle was moving faster?
Answer:
The answer is below
Explanation:
Let a be the initial velocity of motorcycle A and b be the initial velocity of motorcycle B.
After 5.68 seconds, both motorcycle had the same velocity (v), therefore for motorcycle A:
(a - v) / 5.68 = 3.87
a - v = 21.9816
v = a - 21.9816
For motorcycle B:
(b - v) / 5.68 = 18.2
b - v = 103.376
v = b - 103.376
Therefore:
a - 21.9816 = b - 103.376
b - a = -21.9816 + 103.376
b - a = 81.3944
a) The difference between their speeds at the beginning was 81.3944 m/s
b) Since b - a = 81.3944. This means that the initial velocity of motorcycle B is greater than that of motorcycle A by 81.3944 m/s.
Therefore motorcycle B was moving faster
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]
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
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.
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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.
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.
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
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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:
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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 lake has a surface area of 410 m2 and a volume of 1140 m3 . Suppose that during the day, sunlight with a power averaging 820 W/m2 shines on the lake, and that about 10% of this power is absorbed in the lake, producing heat. Assume that the temperature of the lake water stays constant because the absorbed solar power is exactly balanced by heat lost due to evaporation of water from the lake surface. What is the evaporation rate, in g/s
Answer:
Explanation:
Total solar energy falling on total surface per second
= 410 x 820 W
= 336200 W
10 % of 336200 = 33620 J is converted into heat which is absorbed by lake water . But its temperature does not rise because heat is used up in evaporating water in the form of vapor .
Total heat released during evaporation = 33620 J
Let evaporation rate be m gram /s
heat absorbed by m gram water = m x latent heat of evaporation
= m x 2260 J .
Given ,
m x 2260 = 33620
m = 14.87 g /s .
How does energy move in relation to the medium in a transverse wave?
Answer:
Only the energy of the wave travels through the medium. In a transverse wave, particles of the medium vibrate up and down perpendicular to the direction of the wave. ... In a surface wave, particles of the medium vibrate both up and down and back and forth, so they end up moving in a circle.
The Earth’s orbit _____.
is an ellipse
goes around the moon
is a circle
causes day and night
Cara is building a model of the solar system, which includes the Sun. She plans to include a written description to provide details about each piece in her model. In order for her model to be realistic, which of the following should she include in her representation of the Sun?
Answer:
she should write about how big is it and what the sun looks and how far away is it from earth.
Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.400 m and the length of the copper section is 0.800 m . Each segment has cross-sectional area 0.00700 m2 . The free end of the brass segment is in boiling water and the free end of the copper segment is in an ice-water mixture, in both cases under normal atmospheric pressure. The sides of the rods are insulated so there is no heat loss to the surroundings.
(a) What is the temperature of the point where the brass and copper segments are joined?
(b) What mass of ice is melted in 5.00 min by the heat conducted by the composite rod?
Answer:
a) 36°
b) 0.109 kg
Explanation:
Heat flows from brass to copper with the brass having its temperature
Length of brass = 0.4
Length of copper = 0.8
Temperature of = 36.15
See attachment for calculation
The temperature at the joint is 36.15°C
The amount of ice melted is 1.086 kg
The rate of transfer of thermal energy,
H = Q/t = KAΔT/L
where, K is the thermal conductivity of the substance, A is cross-sectional area, ΔT is temperature difference at the ends and L is the length
As given in the question,
the length of the brass section [tex]L_{1}[/tex] = 0.4 m
it's thermal conductivity [tex]K_{b}[/tex] = 109 J[tex]s^{-1}m^{-1}K^{-1}[/tex]
the temperature at the brass end [tex]T_{1}[/tex] = 373K
the length of the copper section [tex]L_{2}[/tex] = 0.8 m
it's thermal conductivity [tex]K_{c}[/tex] = 385 J[tex]s^{-1}m^{-1}K^{-1}[/tex]
the temperature at the brass end [tex]T_{1}[/tex] = 273K
cross-sectional area of both the substance is same A = 0.007 [tex]m^{2}[/tex]
Let the temperature at the joint be T
The rate of heat flow must be constant across the whole length of the setup.
Hence at the joint,
[tex]\frac{K_{b}A(T_{1}-T) }{L_{1} } =\frac{K_{c}A(T-T_{2}) }{L_{2} }[/tex]
⇒ [tex]\frac{ 109*A*(373-T)}{0.4} =\frac{385*A(T-273)}{0.8}[/tex] ⇒ T=309.15 K
⇒ T = 36.15°C is the temperature at the joint.
Now we have to calculate the equivalent thermal conductivity K of the setup in order to calculate the amount of heat transfer.
considering equivalent thermal conductivity K throughout the setup we can form the following equation to calculate its value
[tex]\frac{KA(T_{1}-T_{2} ) }{L_{1}+L_{2} } =\frac{K_{c}A(T-T_{2}) }{L_{2} }[/tex]
⇒ [tex]\frac{ K*A*(100)}{1.2} =\frac{385*A(36.15)}{0.8}[/tex]
⇒ K = 208.76 J[tex]s^{-1}m^{-1}K^{-1}[/tex]
the amount of heat transferred at the copper end in ice-water mixture in 5 minutes(300 seconds) :
Q = [tex]\frac{KA(T_{1}-T_{2} ) }{L_{1}+L_{2} }[/tex] × t = [tex]\frac{208.76*0.007*100}{1.2}[/tex] × 300 = 36533 J
latent heat of fusion of ice [tex]L_{f}[/tex] = 33600 J/kg
[tex]Q=mL_{f}[/tex]
[tex]m=\frac{Q}{L_{f} }[/tex]
[tex]m=\frac{36533}{33600}[/tex] ⇒ m = 1.086 kg of ice is melted in 5 minutes
Learn more about heat transfer:
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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.
Define position
i am not sure?
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.
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 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
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:
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:
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