Answer:
they will move away from each other
1. Odysseus traveled from Troy to Ithaca. What
was the acceleration of Odysseus' ship if its mass
was 900,000 kg and it moves across the water with
a force of 300,000 N?
Answer: 0.33 m/s^2
Explanation:
The acceleration of Odysseus' ship as it moves across the water from Troy to Ithaca is 0.33m/s²
Given the data in the question;
Mass of Odysseus' ship; [tex]m= 900000kg[/tex]Force with which Odysseus' ship moves across the water; [tex]F = 300000N[/tex]Acceleration; [tex]a = ?[/tex]To determine the acceleration of the ship, We the equation from Newton's Second Law of Motion:
[tex]F = m\ *\ a[/tex]
Where F is the force, m is the mass and a is the acceleration
Lets make acceleration ''a'', the subject of the formula
[tex]a = \frac{F}{m}[/tex]
Now, we substitute our given values into the equation
[tex]a = \frac{300000N}{900000kg}[/tex]
We know that, A newton is defined as [tex]1 kg.m/s^2[/tex]
[tex]a = \frac{300000 kg.m/s^2}{900000kg} \\\\a = 0.33m/s^2[/tex]
Therefore, the acceleration of Odysseus' ship as it moves across the water from Troy to Ithaca is 0.33m/s²
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Someone pls help!! Right now :(
___________is
objects total kinetic and potential energy.
Answer:
Bowling ball
Explanation:
How do you label a bar graph of work compared with power?
Answer:
You must label the horizontal axis with the names of the airlines and the vertical axis with the number of flights. The title must clearly state what data the bar chart is showing. With larger numbers, your scale may not go up by one.
Answer:
You must label the horizontal axis with the names of the airlines and the vertical axis with the number of flights. The title must clearly state what data the bar chart is showing. With larger numbers, your scale may not go up by one.
Explanation:
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19. In a turtle race, a 20 kg turtle moves with a velocity of 0.1 m/s. What is the kinetic energy of the turtle?
A. 20 J
B. 0.2 J
C. 0.1 J
Question 1 of 10
Which object has the most gravitational potential energy?
A. A 8 kg book at a height of 2 m
B. An 8 kg book at a height of 3 m
C. An 5 kg book at a height of 3 m
D. A5 kg book at a height of 2 m
SUBMIT
Answer:
B = An 8 kg book at a height of 3 m
PLEASE HELP ME WITH THIS ONE QUESTION
Which pendulum would have the highest frequency?
1) 1 kg mass and 0.5-meter length
2) 0.5 kg mass and 0.75-meter length
3) 0.5 kg mass and 1-meter length
4) 1 kg mass and 0.75-meter length
Answer:
B is the Answer
Follow me please
Mark brainliest
Answer:
The answer is A
Explanation:
This is because f = 1/T,
or frequency equals 1 over period, T
Since the highest T would be the answer with the greatest length and mass does not matter the answer would be c IF they were asking for the highest period.
F is found by 1 over T so the greatest F would be the smallest number, Hence answer A
Please Mark me brainiest
Object 1 has a momentum of 10 kg m/s and Object 2 has a momentum of 25 kg m/s. Will it be easier to change the direction of movement of Object 1 or of Object 2?
Answer:
I think its object 1
Explanation:
Because the object that has more weight has a greater momentum and the lightest object that has a less momentum will be easier to change because its lighter.
HELP PLEASE
60 POINTS
HAVE A GREAT REST OF YOUR DAY PEOPLE :>
Answer:
The sun and the stars
Explanation:
I hope this helps!
Answer:
Northstar & North Pole
Explanation:
yyggggggggggggggg
Define speed and what is it’s SI unit.
How does the current in a series circuit compare
The (BLANK) island were where Darwin made many important observations.
Answer:
I know theirs South American coast he was there a lot
PLS HELP
SHOW WORK
A car starting from rest accelerates at a rate of 1.5 m/s ^ 2 What is its final speed at the end of 18.0 seconds ?
Answer:
27 ms^-1
Explanation:
by using v= u + at
u = 0 ( because the object id starting from rest)
v= 0 + 1.5 x 18
v = 27 ms^-1
What us a magnetic domain?
magnetic domain is a region within a magnetic material in which the magnetization is in a uniform direction. This means that the individual magnetic moments of the atoms are aligned with one another and they point in the same direction.
When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how fast is the angle of elevation of the rocket increasing, as seen by an observer on the ground 5 kilometers from the launching pad
Answer:
The angle of elevation of the rocket is increasing at a rate of 48.780º per second.
Explanation:
Geometrically speaking, the distance between the rocket and the observer ([tex]r[/tex]), measured in kilometers, can be represented by a right triangle:
[tex]r = \sqrt{x^{2}+y^{2}}[/tex] (1)
Where:
[tex]x[/tex] - Horizontal distance between the rocket and the observer, measured in kilometers.
[tex]y[/tex] - Vertical distance between the rocket and the observer, measured in kilometers.
The angle of elevation of the rocket ([tex]\theta[/tex]), measured in sexagesimal degrees, is defined by the following trigonometric relation:
[tex]\tan \theta = \frac{y}{x}[/tex] (2)
If we know that [tex]x = 5\,km[/tex], then the expression is:
[tex]\tan \theta = \frac{y}{5}[/tex]
And the rate of change of this angle is determined by derivatives:
[tex]\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y[/tex]
[tex]\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}[/tex]
[tex]\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}[/tex]
[tex]\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}[/tex]
Where:
[tex]\dot \theta[/tex] - Rate of change of the angle of elevation, measured in sexagesimal degrees.
[tex]\dot y[/tex] - Vertical speed of the rocket, measured in kilometers per hour.
If we know that [tex]y = 4\,km[/tex] and [tex]\dot y = 400\,\frac{km}{h}[/tex], then the rate of change of the angle of elevation is:
[tex]\dot \theta = 48.780\,\frac{\circ}{s}[/tex]
The angle of elevation of the rocket is increasing at a rate of 48.780º per second.
A wave with a frequency of 5Hz travels a distance of 40mm in 2 seconds.What is the speed of the wave
Answer:
20mm per second
Explanation:
lol
During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of 39.0 kg and its horizontal component of velocity is 6.50 m/s just before the 65.0 kg performer catches it. If the performer is initially motionless on nearly frictionless roller skates, what is his speed immediately after catching the cannon ball
Answer:
[tex]2.4375\ \text{m/s}[/tex]
Explanation:
[tex]m_1[/tex] = Mass of cannon ball = 39 kg
[tex]m_2[/tex] = Mass of performer = 65 kg
[tex]u_1[/tex] = Initial horizontal component of cannon ball's velocity = 6.5 m/s
[tex]u_2[/tex] = Initial horizontal component of performer's velocity = 0
v = Velocity of combined mass
As the momentum of the system is conserved we have
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{39\times 6.5+0}{39+65}\\\Rightarrow v=2.4375\ \text{m/s}[/tex]
The speed of the performer immediately after catching the cannon ball is [tex]2.4375\ \text{m/s}[/tex].
Suppose that a uniform rope of length L resting on a frictionless horizontal surface, is accelerated along the direction of its length by means of a force F, pulling it at one end. A mass M is accelerated by the rope. Assuming the mass of the rope to be m and the acceleration is a. Stated in terms of the product ma, what is the tension in the rope at the position 0.3 L from the end where the force F is applied if the mass M is 1.5 times the mass of the rope m?
Answer:
2.2 ma
Explanation:
Given :
Length of the rope = L
Mass of the rope = m
Mass of the object pulled by the rope = M
M = 1.5 m
So, L [tex]$\rightarrow$[/tex] m
For unit length [tex]$\rightarrow \frac{m}{L}$[/tex]
∴ 0.3 L = [tex]$0.3 \ L \left(\frac{m}{L}\right)$[/tex]
= 0.3 m
And for remaining 0.7 L = [tex]$0.7 \ L \left(\frac{m}{L}\right)$[/tex]
= 0.7 m
By Newtons law of motion,
F - T = ( 0.3 m) a .........(1)
T = ( M + 0.7 m) a
T = ( 1.5 m + 0.7 m) a
T = ( 2.2 m ) a ..............(2)
So from equation (1) and (2), we have
Tension on the rope
T = 2.2 ma
Velocity tells us not only how fast something is going, but in what direction it is traveling.
True
False
Answer:
true
Explanation:
because I tride it
He-Ne Laser device emits photons of wave length 632.8 nm by rate 4.5 x 1020 photon/s, so
the power of the laser beam =
a. 3.14 W
b. 141.3 W
c. 314.1 w
d. 431.4 W
Answer: Option b.
Explanation:
We know:
Wavelength = 632.8 nm
Fluence = 4.5*10^20 photon/s
The energy of a single photon of wavelength λ is:
E = (h*c)/λ
where:
h = 6.6*10^(-34) J*s
c = 3*10^8 m/s
And we should rewrite the wave length in meters, so:
λ = 632.8 nm = 632.8*10^(-9) m
replacing these in the energy equation, we get:
E = (6.6*10^(-34)J*s)*(3*10^8 m/s)/(632.8*10^(-9) m) = 3.13*10^(-19) J
So each one of the 4.5x10^20 photon that flow each second have this energy, then the power is:
P = (3.13*10^(-19) J)*(4.5*10^20 /s) = 140.85 J/s
and 1 W = 140.85 J/S
Then the power is:
P = 140.85 W
Then the correct answer is the option b, where the units are a little bit different than mine because I used really simplified values for c and h.
Physics help, thank you guys so much!
Answer:
Δt = 5.85 s
Explanation:
For this exercise let's use Faraday's Law
emf = [tex]- \frac{d \phi}{dt}[/tex] - d fi / dt
[tex]\phi[/tex] = B. A
\phi = B A cos θ
The bold are vectors. It indicates that the area of the body is A = 0.046 m², the magnetic field B = 1.4 T, also iindicate that the normal to the area is parallel to the field, therefore the angle θ = 0 and cos 0 =1.
suppose a linear change of the magnetic field
emf = - A [tex]\frac{B_f - B_o}{ \Delta t}[/tex]
Dt = - A [tex]\frac{B_f - B_o}{emf}[/tex]
the final field before a fault is zero
let's calculate
Δt = - 0.046 (0- 1.4) / 0.011
Δt = 5.85 s
determine the value of a and K when 0.51 is written is scientific nation
Answer:
The value 0.51
Explanation:
A spring has a spring constant of 65.5 N/m and it is
stretched with a force of 15.3 N. How far will it stretch?
(iii) Gareth cycles 5 laps at this constant spoed. Each lap is 500m.
Calculate the distance he travels.
I
Distance
Answer:
2500 m
Explanation:
Given that,
Each lap is 500 m
Gareth takes 5 laps.
We need to find distance traveled by Gareth. The distance covered by him is given by :
d = 5×500
d = 2500 m
Hence, he will travel 2500 m.
Two automobiles, each of mass 1000 kg, are moving at the same speed, 20 m/s, when they collide and stick together. In what direction and at what speed does the wreckage move (a) if one car was driving north and one south (b) if one car was driving north and one east?
A. The wreckage after collision is moving at the speed 18 m/s to the south.
B. The wreckage after collision is moving at the speed 9.0 m/s to the north.
C. The wreckage after collision is moving at the speed 9.0 m/s to the south.
D. The wreckage after collision is moving at the speed 18 m/s to the north.
E. The wreckage after collision is motionless.
Answer:
The reckage after collision is motionless (E)
Explanation:
The first law of thermodynamics states that energy is neither created nor destroyed but is converted from one form to another.
The kind of collision described in the question above is known as a perfectly inelastic collision, and in this type of collision, the maximum kinetic energy is lost because the objects moving in opposite directions have a resultant momentum that is equal, but in opposite directions hence they cancel each other out.
The calculation is as follows:
m₁v₁ + m₂v₂
where:
m₁ = m₂ = 1000kg
v₁ = 20 m/s
v₂ = -20 m/s ( in the opposite vector direction)
∴ resultant momentum = (1000 × 20) + (1000 × -20)
= 20000 - 20000 = 0
∴ The reckage after collision is motionless
Answer:
The wreckage after collision is moving at the speed 18 m/s to the south.
Explanation:
Can someone plz explain :
Two cars have the same mass, the first car is moving towards the east and the second is
stationary. If the two cars merge together after a collision and head towards the east their speed
after the collision is equal to.......
A)1/4vi b)1/2vi c)vi d) 2vi
Answer: Option b, the final velocity is half of the initial velocity.
Explanation:
Here we will use the conservation of the total momentum of a system.
This means that the total momentum at the beginning must be the same as the final momentum.
Where momentum is:
P = M*v
Initially, we have two cars, both with the same mass M, and only one of them has a velocity v.
Then the initial momentum is:
P = M*v + M*0 = M*v
After the collision, the two cars move together. Then the total mass that is moving is equal to the sum of the masses of the cars, this is 2*M
and we can suppose that the two cars move at a final velocity v'
Then the final momentum is:
P' = (2*M)*v'
Now we use the conservation of momentum, then:
P = P'
M*v = (2*M)*v'
Now we need to solve this for v'
(M*v)/(2*M) = v'
v/2 = v'
This means that the final velocity is half of the initial velocity.
Then the correct option is option b.
You push a 1.5 kg ball across a desk for 2 seconds, so that 10 J of work have been done. How much power was produced?
Answer
Answer:
i need help with the same question
Explanation:
A ball is thrown straight up with an initial velocity of 128 ft/sec, so that its height (in feet) after t sec is given by s(t) = 128t-16t2. (a) What is the average velocity of the ball over the following time intervals? [3,4] ft/sec [3,3.5] ft/sec [3,3.1] ft/sec (b) What is the instantaneous velocity at time t = 3? ft/sec (c) What is the instantaneous velocity at time t = 6? ft/sec Is the ball rising or falling at this time? rising falling (d) When will the ball will hit the ground? t = sec
Answer:
one sec let me think
Explanation:
(a)The average velocity of the ball over the following time intervals will be [3,4] ft/sec.
(b)The instantaneous velocity at time t = 3 will be32 ft/sec.
(c)The instantaneous velocity at time t = 6 will be -64 ft/sec.
(d)The ball will hit the ground at 13.4 sec.
What is velocity?The change of displacement with respect to time is defined as the velocity. velocity is a vector quantity. it is a time-based component.
The given data in the question will be ,
u is the initial velocity by which ball thrown=128 ft/sec.
V₃ is the instantaneous velocity at time t=3 sec.
V₆ is the instantaneous velocity at time t=6 sec.
t is the time when ball hits the ground=?
(a) Given equation for the displacement
s(t) =128t-16t² (on differenting got the velocity )
v(t) = 128-32t
Time when velocity is zero will be
[tex]\rm{ t=\frac{128}{32}[/tex]
[tex]\rm{ t=4 sec[/tex]
If the velocity got in the equation is 128 and 32 ft /sec. it can be only when the average velocity is [3,4] ft/sec .
Hence the average velocity obtained from the problem will be [3,4] ft/sec
(b)
s(t) =128t-16t² (on differenting got the velocity )
v(t) = 128-32t
At time( t=3 sec)
v(3) = 128-32×3
v(3) =32 m/sec.
Hence the instantaneous velocity at time t = 3 will be32 ft/sec.
(c)
s(t) =128t-16t² (on differenting got the velocity )
v(t) = 128-32t
At time( t=6 sec)
v(6) = 128-32×6
v(6) = -64 m/sec.
Hence the instantaneous velocity at time t = 6 will be -64 ft/sec.
(d)
According to Newtons third equation of motion we got
v=u+gt
If the body returens from a certain height at max height its velocity must be zero; ( u=0)
[tex]\rm t=\frac{(v-u)}{g} \\\\\ \rm t=\frac{(128-0)}{9.81}\\\\\rm t=13.04 sec.[/tex]
Hence the ball will hit the ground at 13.4 sec.
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I need help!!!!!!!!!!!pleaseeeeeee
a rugby player passes the ball 5.34 m across the field, where it is caught at the same height as it left his hand. at what angle was the ball thrown if its initial speed was 7.7 m/s, assuming that the smaller of the two possible angles was used
Answer:
[tex]31.035^{\circ}[/tex]
Explanation:
x = Displacement in x direction = 5.34 m
t = Time taken to travel the displacement
y = Displacement in y direction = 0
u = Initial velocity of ball = 7.7 m/s
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Displacement in x direction is given by
[tex]x=u\cos\theta t\\\Rightarrow t=\dfrac{5.34}{7.7 \cos\theta}[/tex]
Displacement in y direction is given by
[tex]y=u\sin\theta t-\dfrac{1}{2}gt^2\\\Rightarrow 0=7.7\sin\theta \dfrac{5.34}{7.7\cos\theta}-\dfrac{1}{2}\times 9.81 (\dfrac{5.34}{7.7\cos\theta})^2\\\Rightarrow 0=7.7\sin\theta-4.905\times \dfrac{5.34}{7.7\cos\theta}\\\Rightarrow 0=7.7^2\sin\theta \cos\theta-4.905\times 5.34\\\Rightarrow 0=7.7^2\dfrac{\sin2\theta}{2}-4.905\times 5.34\\\Rightarrow 0=7.7^2\sin2\theta-4.905\times5.34\times 2\\\Rightarrow \sin2\theta=\dfrac{4.905\times 5.34\times 2}{7.7^2}\\\Rightarrow 2\theta=\sin^{-1}\dfrac{4.905\times 5.34\times 2}{7.7^2}[/tex]
[tex]\Rightarrow \theta=\dfrac{62.07}{2}\\\Rightarrow \theta=31.035^{\circ}[/tex]
The angle at which the ball was thrown is [tex]31.035^{\circ}[/tex].