Answer:
effects of gravity on the Moon and Earth
Explanation: Newton found the Moon's inward acceleration in its orbit to be 0.0027 metre per second per second, the same as (1/60)2 of the acceleration of a falling object at the surface of Earth.
Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular acceleration of 0.311 rad/s2 for 4.13 s. What is the drill's angular displacement during that time interval?
Answer:
The drill's angular displacement during that time interval is 24.17 rad.
Explanation:
Given;
initial angular velocity of the electric drill, [tex]\omega _i[/tex] = 5.21 rad/s
angular acceleration of the electric drill, α = 0.311 rad/s²
time of motion of the electric drill, t = 4.13 s
The angular displacement of the electric drill at the given time interval is calculated as;
[tex]\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad[/tex]
Therefore, the drill's angular displacement during that time interval is 24.17 rad.
convert 100 Newton into dyne
Answer:10000000
Explanation:
A vertical tube one meter long is open at the top. It is filled with 50 cm of water. If the velocity of sound is 344 m/s, what will the fundamental resonant frequency be (in Hz)?
Answer:
The fundamental resonance frequency is 172 Hz.
Explanation:
Given;
velocity of sound, v = 344 m/s
total length of tube, Lt = 1 m = 100 cm
height of water, hw = 50 cm
length of air column, L = Lt - hw = 100 cm - 50 cm = 50 cm
For a tube open at the top (closed pipe), the fundamental wavelength is given as;
Node to anti-node (N ---- A) : L = λ / 4
λ = 4L
λ = 4 (50 cm)
λ = 200 cm = 2 m
The fundamental resonance frequency is given by;
[tex]f_0 = \frac{v}{\lambda}\\\\f_0 = \frac{344}{2}\\\\f_0 = 172 \ Hz\\\\[/tex]
Therefore, the fundamental resonance frequency is 172 Hz.
HELPP physics final will give brainliest
Hey guys this is Ap physics please help I need this to pass i will mark brainliest for a good attempt
Split up the forces into components acting parallel to and perpendicular to the slope. See the attached picture for the reference axes.
The box stays on the surface of the plane, so that the net force acting perpendicular to it is 0, and the only acceleration is applied in the parallel direction.
Let m be the mass of the box, θ the angle the plane makes with the ground, and a the acceleration of the box. By Newton's second law, we have
• net parallel force
∑ Force (//) = W (//) - F = m a
(that is, the net force in the parallel direction is the sum of the parallel component of the weight W and the friction F which acts in the negative direction)
• net perpendicular force
∑ Force (⟂) = W (⟂) + N = 0
Notice that
W (//) = W sin(θ) … … … which is positive since it points down the plane
W (⟂) = -W cos(θ) … … … which is negative since it points opposite the normal force N
So the equations become
W sin(θ) - F = m a
-W cos(θ) + N = 0
Solving for a gives
a = (W sin(θ) - F ) / m
which is good enough if you know the magnitude of the friction force.
If you don't, you can write F in terms of the coefficient of kinetic friction between the box and plane, µ, as
F = µ N
so that
a = (W sin(θ) - µ N ) / m
and the normal force itself has a magnitude of
N = W cos(θ)
so that
a = (W sin(θ) - µ W cos(θ) ) / m
The weight W has magnitude m g, where g is the magnitude of the acceleration due to gravity, so
a = (m g sin(θ) - µ m g cos(θ) ) / m
a = g (sin(θ) - µ cos(θ))
Determine the acceleration that results when a 12 N net force is applied to a 3 kg object.
Heya!!
For calculate aceleration, let's applicate second law of Newton:
[tex]\boxed{F=ma}[/tex]
⇒ Being:
→ F = Force = 12 N
→ m = Mass = 3 kg
→ a = aceleration = ?
Lets replace according formula and leave the "a" alone:
[tex]12\ N = 3\ kg * \textbf{a}[/tex]
[tex]\textbf{a} = 12\ N / 3\ kg[/tex]
[tex]\textbf{a} = 4\ m/s^{2}[/tex]
Result:
The aceleration of the object is of 4 m/s²
A. A piece of paper near a magnet
B. An aluminum nail near a magnet
C. An iron nail, not near a magnet
D. An iron nail near a magnet
Answer:
it’s c not d
Explanation:
took the test
Answer: D!!!
Explanation: jus got it wrong from the other answer.
in the case shown below, the 1 kg rock rides on a horizontal disk that rotates at constant speed 5m/s about its vertical axis. the radius of the disk is 1 meter. What is the magnitude of the friction?
Answer:
25
Explanation:
An unknown radioactive sample is observed to decrease in activity by a factor of two in a one hour period. What is its half-life?
Answer:
The half-life is [tex] t_{1/2} = 1.005 h[/tex]
Explanation:
Using the decay equation we have:
[tex]A=A_{0}e^{-\lambda t}[/tex]
Where:
λ is the decay constantA(0) the initial activityA is the activity at time tWe know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that [tex]A = \frac{A_{0}}{2}[/tex]
[tex]\frac{A_{0}}{2}=A_{0}e^{-\lambda*1 h}[/tex]
[tex]0.5=e^{-\lambda*1 h}[/tex]
Taking the natural logarithm on each side we have:
[tex]ln(0.5)=-\lambda[/tex]
[tex]\lambda=0.69 h^{-1}[/tex]
Now, the relationship between the decay constant λ and the half-life t(1/2) is:
[tex]\lambda = \frac{ln(2)}{t_{1/2}}[/tex]
[tex] t_{1/2} = \frac{ln(2)}{\lambda}[/tex]
[tex] t_{1/2} = \frac{ln(2)}{0.69}[/tex]
[tex] t_{1/2} = 1.005 h[/tex]
I hope it helps you!
d. If a dog has a mass of 12 kg, what is its weight on Neptune?
11.7N/kg
Answer:
133.8 N
Explanation:
Recall that the acceleration of gravity in Neptune is estimated as 11.15 m/s^2
Therefore, the weight of the dog on this planet would be:
Weight = mass x acceleration of gravity = 12 kg x 11.15 m/s^2 = 133.8 N
Allure of the seas is one of the most expensive cruise ships around the world with a length of 362 meters(1,187 ft) and a height of 72 meters(236 ft) above water line. On her first day of operation she moves with a uniform acceleration of 83.5 km/hr2 from rest has gone 10 nautical miles. How many seconds she is in motion? Note: 1 nautical mile = 1.852 km (help 3 mins left
Answer:
about 2398 seconds
Explanation:
The relation between time, distance, and acceleration is ...
d = (1/2)at²
t = √(2d/a) = √(2·10·1.852 km/(83.5 km/h²)) ≈ √0.4436 h ≈ 0.6660 h
That is about ...
(0.6660 h)(3600 s/h) ≈ 2397.7 s
The cruise ship takes about 2397.7 seconds to cruise 10 nautical miles, accelerating all the way.
what is the force of an egg that is thrown at a brick wall if the egg has a mass of 0.3 kg and an acceleration of 50 m/s/s
Answer:
15N
Explanation:
F=ma so F=.3*50 therefore F=15N
The force of an egg that is thrown at a brick wall is equal to 15 N.
What is force?Force can be defined as the influence or effect that changes the state of the body of from motion to rest or vice versa. The S.I. unit of force is Newton (N) as well as force is a vector quantity. Force can change the direction or the speed of the moving object.
The force acting on an object can be calculated from the multiplication of the mass(m) and acceleration(a). The mathematical form of the second law of motion for force can be written as follows:
F = ma
Given, the mass of the egg, m = 0.3 Kg
The acceleration of the egg with which it is thrown on the wall, a = 50 m/s²
The force of an egg that is thrown at a brick wall can be calculated as:
F = ma = 50 ×0.3 = 15 N
Learn more about force, here:
brainly.com/question/13191643
#SPJ2
Calculate the RMS speed of helium atoms near the surface of the Sun at a temperature of about 5300 K.
Answer:
RMS velocity, [tex]v_{rms}=5748.75\ m/s[/tex]
Explanation:
We need to find the RMS speed of helium atoms near the surface of the Sun at a temperature of about 5300 K.
The formula for RMS speed of a gas is given by :
[tex]v_{rms}=\sqrt{\dfrac{3RT}{m}}[/tex]
Where
R is ideal gas constant, R = 8.314 J /mol K
T = 5300 K
m is molar mass of Helium, [tex]m = 4\times 10^{-3}\ Kg/mol[/tex]
Substituting all the values in above formula :
[tex]v_{rms}=\sqrt{\dfrac{3\times 8.314\times 5300}{4\times 10^{-3}}}\\\\=5748.75\ m/s[/tex]
So, the RMS speed Helium atoms 5748.75 m/s.
n
Question 4
1 pts
A bus travels on an interstate highway at an average speed of 90 km/hrs. How far does it take to travel
in 30 mins? The distance equals speed times time, or d = st.
O 45 km
O 98 Km
O 56 km
O 432 Km
[tex]d = s \times t \\ d = 90 \times \frac{30}{60} \\ d = 90 \times \frac{1}{2 } \\ d = 45km[/tex]
A hare can run at a rate of 15 m/s, while a turbocharged tortoise can now crawl at a rate of 3 m/s, how much of a head-start (time-wise) does the tortoise need in order to tie the hare in a 250 meter race?
A.
16.7 seconds
B.
66.7 seconds
C.
83.3 seconds
D.
100 seconds
Answer:
t = 66.7 s
Explanation:
Given that,
Speed of a hare, v = 15 m/s
Speed of a turbocharged tortoise, v' = 3 m/s
The hare in a 250 meter race
Let the Hare takes time t. It can be calculated as follows :
[tex]t=\dfrac{250}{15}=16.67\ s[/tex]
Let a turbocharged tortoise takes t'. It can be calulated as follows :
[tex]t'=\dfrac{250}{3}= 83.33\ s[/tex]
To tie the race, required time is given by :
[tex]\Delta t = t'-t\\\\=83.33-16.67\\\\=66.66\ s\\\\\approx 66.7\ s[/tex]
Hence, the correct option is (b) i.e. 66.7 seconds.
Carter's favorite ride at playland amusement park is the rollercoaster. The roller coaster
car and passengers have a combined mass of 1620kg and they descend the first hill at
an angle of 45.0 degrees to the horizontal. With what force is the rollercoaster pulled
down the hill?
Answer:
F = 11226.02 N
Explanation:
The roller coaster car and passengers have a combined mass of 1620kg.
They descend the first hill at an angle of 45.0 degrees to the horizontal.
We need to find force is the rollercoaster pulled down the hill.
We firstly find the rectangular component of the downward. The force acting in the downward direction is mgsinθ such that,
F = mgsinθ
= 1620 × 9.8 × sin(45)
= 11226.02 N
So, the roller coaster is pulled down the hill with a force of 11226.02 N.
a mass of 2.00 kg rest on a rough horizontal table. The coefficient of static friction between the block and the table is 0.60. The block is attached to a hanging mass by a string that goes over a smooth pulley,as shown in the diagram. Determine the largest mass that can hang in this way without forcing the block to slide.
Answer:
1.2 kg
__________________________________________________________
We are given:
Mass of the block = 2 kg
Coefficient of Static Friction = 0.6
__________________________________________________________
Friction Force on the Block:
Finding the Normal Force:
We know that the normal force will be equal and opposite to the weight of the 2 kg block
So, Normal Force = mg
replacing the variables with the given values
Normal Force = (2)(9.8) [Taking g = 9.8]
Normal Force = 19.6 N
Friction force on the Block:
We know that:
Coefficient of Static Friction = Static Friction Force/Normal Force
replacing the variables
0.6 = Static Friction force / 19.6
Static Friction force = 0.6*19.6 N [Multiplying both sides by 19.6]
Static Friction force = 11.76 N
__________________________________________________________
Largest Mass that can Hang:
We know that the Static Friction force is 11.76 N, this means that a force of 11.76 N will be applied to keep the object at rest
So, if the weight of the second block is less than the static friction force, it will hang
Weight of the second block ≤ 11.76
We know that weight = mg
mg ≤ 11.76
m(9.8) ≤ 11.76 [since g = 9.8]
m ≤ 1.2 kg [dividing both sides by 9.8]
From this, we can say that the maximum mass of the second block is 1.2 Kg
3.00 kg block moving 2.09 m/s right hits a 2.22 kg block moving 3.92 m/s left. afterwards, the 3.00 kg block moves 1.71 m/s left. find the velocity of the 2.22 kg block afterwards
Momentum is conserved, so the total momentum before collision is equal to the total momentum after collision. Take the right direction to be positive. Then
(3.00 kg) (2.09 m/s) + (2.22 kg) (-3.92 m/s) = (3.00 kg) (-1.71 m/s) + (2.22 kg) v
where v is the velocity of the 2.22 kg block after collision. Solve for v :
6.27 kg•m/s - 8.70 kg•m/s = -5.13 kg•m/s + (2.22 kg) v
(2.22 kg) v = 2.70 kg•m/s
v = (2.70 kg•m/s) / (2.22 kg)
v ≈ 1.22 m/s
i.e. a velocity of about 1.22 m/s to the right.
A sled is pulled with a force of 540 N at an angle of 40° with the horizontal. What are the horizontal and vertical components of this force?
Answer:
Fx = 467.65N
Fy = 270N
Explanation:
Given
Force = 540N
angle of inclination = 40 degree
Horizontal component Fx = Fcos 30
Fx = 540cos30
Fx = 540(0.8660)
Fx = 467.65N
Hence the horizontal component is 467.65N
Vertical component Fy = Fsin 30
Fy = 540sin30
Fy = 540(0.5)
Fy = 270N
Hence the vertical component is 270N
There is a bell at the top of a tower that is 45 m high. The bell weighs 190 N. The bell has ____________
energy.
Answer:
The bell has a potential energy of 8550 [J]
Explanation:
Since the belt is 45 [m] above ground level, only potential energy is available. And this energy can be calculated by means of the following equation.
[tex]E_{p}= W*h\\E_{p} = 190*45\\E_{p}=8550[J][/tex]
A 100 kg. football player and a 20 kg. child sit on rolling carts and push off
from each other. Which person applies more force?*20 kg. child applies more force
100 kg. football player applies more force
Same force is applied
Answer:
the forces are the same
Explanation:
3) A 10kg object rests on a frictionless surface when it is struck by a 300N force. At what rate will it accelerate?
3m/s/s
30m/s/s
0.3m/s/s
300m/s/s
Answer: 0.3m/s/s
(i'm really sorry if i'm wrong)
:(
A car goes around a circular track at 30 m/s. If the radius of the curve is 90 m, what is the period of the car's revolution around the track?
Answer:
18.9s
Explanation:
Using the formula;
ω = v/r
Where;
ω = angular velocity (rad/s)
v = linear velocity (m/s)
r = radius of the circular track (m)
According to the given information, v = 30m/s, r = 90m
ω = v/r
ω = 30/90
ω = 3/9
ω = 0.3333 radians/seconds.
Since ω = 2π/T
Where;
π = 3.142
T = period (s)
ω = angular velocity
0.333 = 2 × 3.142/T
T = 2 × 3.142/0.333
T = 6.284/0.333
T = 18.87s
T = 18.9s
Describe the motion of an object as it accelerates. IN YOUR OWN WORD!! ASAP
Answer:
The aceleration of an object is in the direction of the net force. If you push or pull an object in a particular direction, it accelerates in that direction. The aceleration has a magnitude directly proportional to the magnitude of the net force.
Explanation:
Hope this helps Plz mark brainliest
A skier is moving 8.33 m/s when
he starts to slide UP a 8.44°
frictionless slope. How much time
does it take him to come to a stop?
(Unit = s)
Answer:
5.79
Explanation:
Vf=Vi+at
0=8.33+(-9.8)sin8.44t
t=8.33/(9.8sin8.44)=0.83/sin 8.44
=5.79
-I...................ok
Answer:
What?
Explanation:
(1-dimension) A fish has a mass of 6 kg and is moving at a speed of 4m/s to the right. What is its momentum?
Answer:
24 kg m/sExplanation:
The momentum of an object can be found by using the formula
momentum = mass × velocity
From the question we have
momentum = 6 × 4
We have the final answer as
24 kg m/sHope this helps you
A rocket will move upward as long as which condition applies?
If two exactly the same cars are driving down a road, which one would have the most kinetic energy. The one that is moving faster, the one that is moving downhill, the one that is moving uphill, or the one that is moving slower.
Answer: the car that is moving downhill
Explanation:
a car traveling at 28.4 m/s undergoes a constant deceleration of 1.92 m/s2 when the breaks are applied. How many revolutions does each tire make before the car comes to a stop
Complete Question
a car traveling at 28.4 m/s undergoes a constant deceleration of 1.92 m/s2 when the breaks are applied. How many revolutions does each tire make before the car comes to a stop? Assume that the car does not skid and that each tire has a radius of 0.307 m. Answer in units of rev.
Answer:
The value is [tex]N = 109 \ rev[/tex]
Explanation:
From the question we are told that
The speed of the car is [tex]u = 28.4 \ m/s[/tex]
The constant deceleration experienced is [tex]a = 1.92 \ m/s^2[/tex]
The radius of the tire is [tex]r = 0.307 \ m[/tex]
Generally from kinematic equation we have that
[tex]v^2 = u^2 + 2as[/tex]
Here v is the final velocity which is 0 m/s
So
[tex]0^2 = 28.4^2 + 2 * 1.92 * s[/tex]
=> [tex]s = 210.04 \ m[/tex]
Generally the circumference of the tire is mathematically represented as
[tex]C = 2 \pi r[/tex]
=> [tex]C = 2 * 3.142 * 0.307[/tex]
=> [tex]C = 1.929 \ m[/tex]
Generally the number of revolution is mathematically represented as
[tex]N = \frac{ s}{C}[/tex]
=> [tex]N = \frac{210.04}{1.929}[/tex]
=> [tex]N = 109 \ rev[/tex]