Answer: D. Heat transfer between the two things stops.
Anuja hit a golf ball on a level field at 70 degrees and 40 degrees with the same total speed as shown below.
70°
40°
Which launch angle causes the ball to be in the air for the longest time?
o not enough information
40 degrees
70 degress
times are the same
Answer:
At 40 deg Vy = V sin 40
at 70 deg Vy = V sin 70
So the ball launched at 70 deg has the greatest vertical velocity and will remain in the air the longest:
Since t = Vy / g time for to reach zero vertical velocity and also the time for the ball to reach velocity Vy on the downward path
Mathis kicked a ball on a level surface at 30∘ and 60∘ with the same total speed as shown below.
Which launch angle results in the greater maximum height for the ball?
Answer: CORRECT (SELECTED)
60
Proposed Exercise: Work-Energy Theorem
In the situation illustrated in the figure below, a 365 pile hammer is used to bury a beam. The hammer is raised to a height of 3.0 (point 1) above the beam (point 2) and released from rest, sinking the beam of 7.4 (point 3). The rails exert on the hammer a constant friction force equal to 54 . Using the work-energy theorem, calculate (a) the speed of the hammer at the exact instant it reaches point 2 and (b) the mean force exerted by the hammer on the beam when moving it from position 2 to 3.
Tip: the force requested in item (b) is equal to the normal force that the beam exerts on
the hammer.
Answer:
152,000 N
Explanation:
(a) Initial potential energy = final kinetic energy
mgh = ½ mv²
v = √(2gh)
v = √(2 × 10 m/s² × 3.00 m)
v = 7.75 m/s
(b) Work done on pile hammer = change in energy
FΔy = 0 − (mgh + ½ mv²)
F (-0.074 m) = -((365 kg) (10 m/s²) (0.074 m) + ½ (365 kg) (7.75 m/s)²)
F (-0.074 m) = -11220.1 J
F ≈ 152,000 N
NEED HELP DUE AT 11:59!! A ball is thrown horizontally from the top of
a building 130 m high. The ball strikes the
ground 53 m horizontally from the point of
release.
What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s.
Answer:
Since the ball was thrown horizontally, there was no vertical component in that force. and hence, the initial vertical velocity of the ball is 0 m/s and the initial horizontal velocity is r.
We are given:
initial velocity (u) = 0 m/s [vertical]
final velocity (v) = v m/s [vertical]
time taken to reach the ground (t) = t seconds
acceleration (a) = 10 m/s/s [vertical , due to gravity]
height from the ground (h) = 130 m
displacement (s) = 53 m [horizontal]
Solving for time taken:
From the third equation of motion:
s = ut + 1/2 at²
130 = (0)(t) + 1/2 * (10) * t²
130 = 5t²
t² = 26
t = √26 seconds or 5.1 seconds
Final Horizontal velocity of the ball
Since the horizontal velocity of the ball will remain constant:
the ball covered 53 m in 5.1 seconds [horizontally]
horizontal velocity of the ball = horizontal distance covered / time taken
Velocity of the ball = 53 / 5.1
Velocity of the ball = 10.4 m/s
Answer:
51.51519 m/s
Explanation:
Given: [tex]a_{x} =0[/tex] [tex]a_{y} -g[/tex] [tex]v_{yo} =0[/tex] [tex]x_{o} =0[/tex] [tex]x=53[/tex][tex]y_{o} =130[/tex]
X-direction | Y-direction
[tex]x=x_{o} +v_{xo}t[/tex] | [tex]y=y_{o} +v_{yo}t+\frac{1}{2}a_{y}t^2[/tex]
[tex]53=0v_{xo}(5.15078)[/tex] | [tex]0=130+\frac{1}{2}(-9.8)t^2[/tex]
[tex]53=v_{xo} (5.15078)[/tex] | [tex]-130=-4.9t^2[/tex]
[tex]\frac{53}{5.15078} =v_{xo}[/tex] | [tex]\sqrt{\frac{-130}{-4.9} }=\sqrt{t^2}[/tex]
[tex]10.2897=v_{xo}[/tex] | [tex]5.15078=t[/tex]
[tex]v=\sqrt{v_{y}^2+ v_{x}^2}[/tex] | [tex]v_{y}^2 =v_{yo}+2a_{y} d[/tex]
[tex]v=\sqrt{(50.27771)^2+(10.2897)^2}[/tex] | [tex]\sqrt{v_{y}^2} =\sqrt{2(-9.8)(0-130)}[/tex]
[tex]v=51.51519 m/s[/tex] | [tex]v_{y}=50.47771[/tex]
Which of the following is true as a ball that was thrown straight up into the air is traveling up?
A. The acceleration is upward until it stops and the acceleration becomes downward. The object is also slowing down as it rises.
B. The acceleration is downward as the object moves up. This causes the object to slow down until it stops and then speed up as it falls down.
Answer:
b
Explanation:
when a ball falls from the sky, it falls down fast due to no friction.
when water in a brook or system of pipes flows from a wide region to a narrow region, the speed of water in the narrow region is
Answer:
more
hope this helps
plz mark brainliest
What feature does not require a planet to have any particular characteristics?
Stream Beds
Dunes
Impact Craters
Volcanic Lava Flows
What is the correct answer?
Answer:
Impact Craters.
Explanation:
An impact crater can be defined as a circular depression that is caused by impact on any planet or asteroids or any other celestial body's surface. When smaller body in galaxy impacts these larger bodies, they form a circular depression on it's surface.
This is a major feature found in solid object such as the Moon, Mercury, etc.
Therefore, the feature that a planet does not require is an impact crater. As other features are important to define a planet. Thus correct option is C.
The suffix used to name a monatomic anion is
O A. ite
O B. ide
O C. anion
O D. ate
when an objects motion is not changing , the object is moving at a ___
A cannon is fired straight up into the air. If the cannon ball comes back down to the launch point in 5 seconds, what was the maximum height reached by the ball?
Answer:
30.63 m
Explanation:
From the question given above, the following data were obtained:
Total time (T) spent by the ball in air = 5 s
Maximum height (h) =.?
Next, we shall determine the time taken to reach the maximum height. This can be obtained as follow:
Total time (T) spent by the ball in air = 5 s
Time (t) taken to reach the maximum height =.?
T = 2t
5 = 2t
Divide both side by 2
t = 5/2
t = 2.5 s
Thus, the time (t) taken to reach the maximum height is 2.5 s
Finally, we shall determine the maximum height reached by the ball as follow:
Time (t) taken to reach the maximum height = 2.5 s
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) =.?
h = ½gt²
h = ½ × 9.8 × 2.5²
h = 4.9 × 6.25
h = 30.625 ≈ 30.63 m
Therefore, the maximum height reached by the cannon ball is 30.63 m
why did Iran experience almost 80 times more deaths in the 20th Century than California, despite having the same seismicity?
Answer: poor construction of houses
Explanation:
Majority of the people that died in Iran were as a result of poor building methods coupled with the fact that there was lack of proper regulation.
California experienced a similar earthquake but due to safer construction methods, about three people died.
Due to population boom in Iran and house shortage, this resulted in builders building cheap houses which were not strong enough.
what is power?
a- the magnitude of a force needed to move an object
b- how much work can be done in a given time
c- the distance over time that an object moves
d- the energy needed to create work
Answer:
b- how much work can be done in a given time
how much work can be done in a given time
Before the development of quantum theory, Ernest Rutherford's experiments with gold atoms led him to propose the so-called Rutherford Model of atomic structure. The basic idea is that the nucleus of the atom is a very dense concentration of positive charge, and that negatively charged electrons orbit the nucleus in much the same manner as planets orbit a star. His experiments appeared to show that the average radius of an electron orbit around the gold nucleus must be about 10−1010−10 m. Stable gold has 79 protons and 118 neutrons in its nucleus.
What is the strength of the nucleus' electric field at the orbital radius of the electrons?
What is the kinetic energy of an electron in a circular orbit around the gold nucleus?
Answer:
a) F = -1.82 10⁻¹⁵ N, b) K = 9.1 10⁻¹⁶ J
Explanation:
a) To calculate the force between the nucleus and the electrons, let's use the Coulomb equation
F = k q Q / r²
as the nucleus occupies a very small volume compared to electrons, we can suppose it as punctual
let's calculate
F = 9 10⁹ (-1.6 10⁻¹⁹) (79 1.6 10⁻¹⁹) / (10⁻¹⁰)²
F = -1.82 10⁻¹⁵ N
b) they ask us for kinetic energy
let's use Newton's second law
F = m a
acceleration is centripetal
a = v² / r
we substitute
F = m v² / r
v = √ (F r / m)
v = √ (1.82 10⁻¹⁵ 10⁻¹⁰ / 9.1 10⁻³¹)
v = √ (0.2 10⁻¹⁶)
v = 0.447 10⁸ m / s
kinetic energy is
K = ½ m v²
K = ½ 9.1 10⁻³¹ (0.447 10⁸)²
K = 0.91 10⁻¹⁵ J
K = 9.1 10⁻¹⁶ J
43. The particle consists of fast moving electrons is.
Answer:
Explanation: someone help me please you want free points right
You are piloting a General Aviation aircraft (e.g. Cessna 172) over the Georgia Tech campus in level flight at 3000 ft altitude. Your airspeed indicator shows that the true airspeed is 100 knots. You have trimmed the aircraft to have no sideslip and pitch angle at a 3 degree angle of attack. Your magnetic compass (which is uncorrected for magnetic variation) indicates that you are flying in the direction of local magnetic East. A local weather forecast predicted the winds to be steady at 10 knots coming exactly from the northeast direction at the time you took off.
a. Using the information provided above and any additional information you may require (state what it is), determine your aircraft's north and east velocity components, Vn and ve, in knots. Starting from Tech Tower at t = 0, you continue in this direction and you are planning to fly directly over a local VOR beacon exactly 10 minutes later. However, you note that you don't pass directly over the beacon and after 10 minutes and 15 seconds have elapsed you are exactly 300 meters away from the VOR beacon in the Easterly direction.
b. Assuming the discrepancy is attributable only to wind speed, estimate the actual average wind speed components, Vwind-n and Vwind-e in knots. Use a flat Earth approximation.
c. Based on the instruments you are using, describe 3 common causes of measurement error that could affect the output of either sensor.
Answer:
a) vₓ = 100 + 7.07 = 107.07 knot (East) , v_{y} = 7.07 knot (North)
b) v_wind_ North = 0 , v_wind_West = 61.12 knot (West)
Explanation:
a) This is a vector velocity addition problem, they tell us that the plane goes East at 100 Knots and the wind goes North East at 10 Knots
the components ask the total speed of the plane.
For this we decompose the wind speed
cos 45 = v₂ₓ / V₂
sin 45 = [tex]v_{2y}[/tex] / v₂
v₂ₓ = v₂ cos 45
v_{2y} = v₂ sin 45
v₂ₓ = 10 cos 45 = 7.07 m / s
v_{2y} = 10 sin45 = 7.07 m / s
the speed of the plane is
vₓ = v_plane + v₂ₓ
v_{y} = v_{2y}
vₓ = 100 + 7.07 = 107.07 knot (East)
v_{y} = 7.07 knot (North)
speed is
v = (107.07 i ^ + 7.07 j ^) knot
b) Estimated time for Target VOR t = 10 min
for a time of t = 10 min and 15 s, it is at a distance of d = 300 m from the VOR in an easterly direction
find the average speed of time on the ride
in the North direction there is no deviation so the average wind speed is zero
v_wind_ North = 0
In the east direction
as the estimated time was 10 min and the real time for the distance is 10 min 15 s
the time difference is t = 15 s to travel d = 300 m
with these data we can calculate the speed of the plane
v = x / t
v = 300/15
v = 20 m / s
let's reduce this speed to knot
v_real = 20 m / s (1knot / 0.5144 m / s) = 38.88 knots
therefore the wind speed is
v_tral = v_plane + v_wind_Este
v_wind_Este = v_real - v_avión
v_wind_Este = 38.88 - 100
v_wind_Este = -61.12 knot
the negative sign indicates that the wind is going west
v_wind_West = 61.12 knot (West)
3)
* The compass is not calibrated, to correct the magnetic deviation, therefore it gives an appreciable error in the direction
* The wind speed is taken when leaving, but there is no constant monitoring, so a change in direction or wind speed in the trajectory can significantly affect the results.
* The aircraft may noir throughout the trajectory at level, due to pitch errors
The day time temperature of Mercury is 800 F . What is Mercury’s temperature in Celsius and kelvin?
Answer:
In Celsius it is 426.667
In Kevin it is 699.817
Explanation:
A car rounds a banked curve as we will discuss in class on Tuesday. The radius of curvature of the road is R and the banking angle is θ. (a) In the absence of friction, what is the safe speed for the car to take this curve? (b) Now assume the coefficient of friction between the car’s tires and the road is µs. Determine the range of speeds the car can have without slipping up or down the road. (c) What is the minimum value of µs that makes the minimum speed zero? (d) If θ = 25.0 ◦ , for what values of µs can the curve be taken at any speed? Note: The upper limit of µs you will find is practically impossible to achieve for the car’s tires and the road.
Answer:
A) v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]
B)√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]
C) µ_s = tan θ
D) µ_s = 0.4663
Explanation:
A) The forces acting on the car will be;
Force due to friction; F_f
Force due to Gravity; F_g
Normal Force; F_n
Now, let us take the vertical direction to be j^ and the direction approaching the centre to be i^ downwards and parallel to the road surface by k^.
Also, we will assume that initially, F_n is in the negative k^ direction and that it will have a maximum possible value of; F_f = µ_s × F_n
Thus, sum of forces about the vertical j^ direction gives;
ΣF_j^ = F_n•cos θ − mg + F_f•sin θ = 0
Since F_f = µ_s × F_n ;
F_n•cos θ − mg + (µ_s × F_n × sin θ) =0
F_n = mg/[cos θ + (µ_s•sin θ)]
Also, sum of forces about the centre i^ direction gives;
ΣF_i^ = F_n(sin θ + (µ_s•cos θ)) = mv²/r
Plugging in formula for F_n gives;
ΣF_i^ = [mg/[cos θ + (µ_s•sin θ)]] × (sin θ + (µ_s•cos θ)) = mv²/r
Making v the subject gives;
v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]
B) What we got in a above is the minimum speed the car can have while going round the turn.
The maximum speed will be gotten by making the frictional force(F_f) to point in the positive k^ direction. This means that F_f will be negative.
Now, if we change the sign in front of F_f in the equation in part a that led to the minimum velocity, we will have the maximum as;
v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]
Thus the range is;
√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]
C) For the minimum speed to be 0, it implies that F_f will be in the negative k^ direction. Thus, Sum of the forces in the k^ direction gives;
ΣF_k^ = mg(sin θ - µ_s•cos θ) = 0
Thus;
mg(sin θ - µ_s•cos θ) = 0
Making µ_s the subject gives;
µ_s = sin θ/cos θ
µ_s = tan θ
D) If θ = 25.0°;
Thus;
µ_s = tan 25
µ_s = 0.4663
blank refers to the method of spreading fertilizer evenly over the entire field by hand it is done at the blank stage
Answer:
Broadcasting is the method, not sure about the stage it is done in
Explanation:
Answer:
Broadcasting is the first (blank) second (blank) is Cultivation.
Explanation:
I took the test & got this answer correct.
answer fast plz
....................
Answer:
114 m/s
Explanation:
see the image below
A 55-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 40.0o above the horizontal. If the tension in the rope is 125 N, how much work is done on the crate to move it 5.0 m?
Answer:
478.75 J
Explanation:
W=force* displacement
constant speed= (a=0) net F=0
Horizontal component of tension
Tcosx
125Ncos40= 95.76 N
W= (95.76 N)(5 m)
=478.75 J
The work done in moving the crate across the given distance is 478.75 J.
The given parameters;
Mass of the packing create, m = 55 kgAngle of inclination of the rope, Ф = 40°Tension on the rope, T = 125 NDistance through which the crate is the moved, d = 5 mThe work done in moving the crate is the product of the horizontal component of the tension and the distance through which the crate is moved.
The work-done in moving the crate is calculated as;
W = Tcos(Ф) x d
W = 125cos(40) x 5
W = 478.75 J.
Thus, the work done in moving the crate across the given distance is 478.75 J.
Learn more here: https://brainly.com/question/19498865
Which type of force occurs between two objects at a distance?
A. Applied force
B. Contact force
C. Non-contact force
D. Normal force
If your right you get 100 points & brainliest
Answer:
it has to be b contact forces
Explanation: because the two objects are next to each other but they still have force
Answer:
c
Explanation:
A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 3.54 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2230 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 5.25 V/m, (b) in the negative z direction and has a magnitude of 5.25 V/m, and (c) in the positive x direction and has a magnitude of 5.25 V/m
Answer:
(a). The magnitude of the net force is [tex](2.1\times10^{-18}\ N)k[/tex]
(b). The magnitude of the net force is [tex](4.23\times10^{-19}\ N)k[/tex]
(c). The magnitude of the net force is [tex](8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k[/tex]
Explanation:
Given that,
Magnetic field [tex]B=-3.54\times10^{-3}i\ T[/tex]
Velocity = 2230j m/s
We know that,
The net force acting on the proton is equal to the sum of electric and magnetic force.
[tex]F=F_{e}+F_{B}[/tex]
(a). If the electric field is in the positive z direction and has a magnitude of 5.25 V/m,
We need to calculate the magnitude of the net force acting on the proton
Using formula of net force
[tex]F_{net}=e(E+v\times B)[/tex]
Put the value into the formula
[tex]F_{net}=1.6\times10^{-19}(5.25k+2230\times-3.54\times10^{-3}(j\times i))[/tex]
[tex]F_{net}=1.6\times10^{-19}(5.25k+2230\times-3.54\times10^{-3}(-k))[/tex]
[tex]F_{net}=(2.1\times10^{-18}\ N)k[/tex]
(b). If the electric field is in the negative z direction and has a magnitude of 5.25 V/m,
We need to calculate the magnitude of the net force acting on the proton
Using formula of net force
[tex]F_{net}=e(E+v\times B)[/tex]
Put the value into the formula
[tex]F_{net}=1.6\times10^{-19}(-5.25k+2230\times-3.54\times10^{-3}(j\times i))[/tex]
[tex]F_{net}=1.6\times10^{-19}(-5.25k+2230\times-3.54\times10^{-3}(-k))[/tex]
[tex]F_{net}=(4.23\times10^{-19}\ N)k[/tex]
(c). If the electric field is in the positive x direction and has a magnitude of 5.25 V/m
We need to calculate the magnitude of the net force acting on the proton
Using formula of net force
[tex]F_{net}=e(E+v\times B)[/tex]
Put the value into the formula
[tex]F_{net}=1.6\times10^{-19}(5.25i+2230\times-3.54\times10^{-3}(j\times i))[/tex]
[tex]F_{net}=(8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k[/tex]
Hence, (a). The magnitude of the net force is [tex](2.1\times10^{-18}\ N)k[/tex]
(b). The magnitude of the net force is [tex](4.23\times10^{-19}\ N)k[/tex]
(c). The magnitude of the net force is [tex](8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k[/tex]
a force of magnitude 30 n stretches a spring 0.83 m from equilibrium. what is the value of the spring constant?
Answer:
36 N/m
Explanation: I got the answer right on the test ight noo cap
Explain why a ping pong ball and bouncy ball of the same size have different weights
Answer:
the material
Explanation:
weight is defined as the amount of force on the object because of gravity. ping pong balls and bouncy balls are made out of different materials that are different weights. most bouncy balls are also not hollow, unlike ping pong balls. these factors affect the weight of these objects.
Electrically neutral objects become___when they or lose electrons.
Answer: positively charged
Explanation: basically when a neutral atom gets some electrons it becomes negatively charged and when a neutral atom loses elections it's positively charged Ik science is weird lol.
I hope this was the answer you were looking for.
please help me. i have 2 hours
the missing word is clockwise moment. I hope this helps good luck
A magician performs in a room with a ceiling, which is 2.70 m above his hands. He throws a ball upwards such that it reaches the ceiling with zero speed. Calculate the initial speed of the ball and the time it takes for it to reach the ceiling. A magician performs in a room with a ceiling, which is 2.70 m above his hands. He throws a ball upwards such that it reaches the ceiling with zero speed. Calculate the initial speed of the ball and the time it takes for it to reach the ceiling.
Answer:
a) v = 7.28 m/s
b) t = 0.74 s
Explanation:
a) The initial speed of the ball can be calculated using the following equation:
[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]
Where:
[tex]V_{f}[/tex] is the final speed = 0
[tex]V_{0}[/tex] is the initial speed =?
g: is the gravity = 9.81 m/s²
h: is the height = 2.70 m
[tex] V_{0} = \sqrt{2gh} = \sqrt{2*9.81 m/s^{2}*2.70 m} = 7.28 m/s [/tex]
Hence, the initial speed of the ball is 7.28 m/s.
b) To find the time that takes the balls to reach the ceiling we can use the next equation:
[tex] V_{f} = V_{0} - gt [/tex]
[tex] t = \frac{V_{0} - V_{f}}{g} = \frac{7.28 m/s}{9.81 m/s^{2}} = 0.74 s [/tex]
Therefore, the time it takes for the ball to reach the ceiling is 0.74 s.
I hope it helps you!
Light refracts when traveling from air into glass because light
O A Travels at the same speed in air and in glass
B Frequency is greater in air than in glass
OC Frequency is greater in glass than in air
D Travels slower in glass than in air
Answer:
the last one
Explanation:
ii took the quiz and it was right... i think
what will happen if a low massive main sequence star runs out of hydrogen fuel?
Answer:
Low mass stars are: hydrogen burning in the core while on the Main Sequence. As the hydrogen fuel runs out, extreme pressure raises the temperature to 100 million degrees, where helium burning becomes possible.
A black widow spider hangs motionless from a web that extends vertically from the ceiling above. If the spider has a mass of 1.81 g, what is the tension in the web?
N=
Answer:
The tension in the web is 0.017738 N
Explanation:
Net Force
The net force exerted on an object is the sum of the vectors of each individual force applied to an object.
If the net force equals 0, then the object is at rest or moving at a constant speed.
The spider described in the question is hanging at rest. It means the sum of the forces it's receiving is 0.
A hanging object has only two forces: The tension of the supporting string (in our case, the web) and its weight. If the object is in equilibrium, the tension is numerically equal to the weight:
T=W=m.g
The mass of the spider is m=1.81 gr = 0.00181 Kg, thus the tension is:
[tex]T = 0.00181\ Kg\cdot 9.8\ m/s^2[/tex]
[tex]T=0.017738\ N[/tex]
The tension in the web is 0.017738 N
The displacement of an object moving 330 km North for 2 hours and an additional 220 km North for 5 hours is?
Answer:
Usbe
Explanation:
