1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
zhuklara [117]
3 years ago
8

Two objects were lifted by a machine.One object had a mass of 2 kilograms and was lifted at a speed of 2 m/sec.The other had a m

ass of 4 kilograms and was lifted at a rate of 3 m/sec
Physics
1 answer:
NikAS [45]3 years ago
7 0

Answer:

While lifting two object the machine needs the different momentum for different mass object.

Explanation:

  • Momentum is the quantity of motion contained in an object. Usually it is measured by the product of mass and velocity.
  • Momentum of first mass = 2 kg × 2 m/sec = 4 kg m/sec
  • Momentum of second mass = 4 kg × 3 m/sec = 12 kg m/sec
  • So the machine requires higher mass in motion for second object ( i.e. momentum) than the first one while lifting.
You might be interested in
In Hooke’s law, what does the x represent?
Arte-miy333 [17]
X Represents the distance the spring is stretched or compressed away from its equilibrium or rest position.
7 0
3 years ago
Read 2 more answers
Two traveling sinusoidal waves are described by the wave functions y1 = 4.85 sin [(4.35x − 1270t)] y2 = 4.85 sin [(4.35x − 1270t
Tamiku [17]

Answer:

Approximately 9.62.

Explanation:

y_1 = 4.85\, \sin[(4.35\, x - 1270\, t) + 0].

y_2 = 4.85\, \sin[(4.35\, x - 1270\, t) + (-0.250)].

Notice that sine waves y_1 and y_2 share the same frequency and wavelength. The only distinction between these two waves is the (-0.250) in y_2\!.

Therefore, the sum (y_1 + y_2) would still be a sine wave. The amplitude of (y_1 + y_2)\! could be found without using calculus.

Consider the sum-of-angle identity for sine:

\sin(a + b) = \sin(a) \cdot \cos(b) + \cos(a) \cdot \sin(b).

Compare the expression \sin(a + b) to y_2. Let a = (4.35\, x - 1270) and b = (-0.250). Apply the sum-of-angle identity of sine to rewrite y_2\!.

\begin{aligned}y_2 &= 4.85\, \sin[(\underbrace{4.35\, x - 1270\, t}_{a}) + (\underbrace{-0.250}_{b})]\\ &= 4.85 \, [\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}.

Therefore, the sum (y_1 + y_2) would become:

\begin{aligned}& y_1 + y_2\\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t) \\ &\quad \quad \quad\;+\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}.

Consider: would it be possible to find m and c that satisfy the following hypothetical equation?

\begin{aligned}& (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)\\&= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}.

Simplify this hypothetical equation:

\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\&=\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)\end{aligned}.

Apply the sum-of-angle identity of sine to rewrite the left-hand side:

\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\[0.5em]&=m\, \sin(4.35\, x - 1270\, t)\cdot \cos(c) \\ &\quad\quad + m\, \cos(4.35\, x - 1270\, t)\cdot \sin(c) \\[0.5em] &=\sin(4.35\, x - 1270\, t)\cdot (m\, \cos(c)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot (m\, \sin(c)) \end{aligned}.

Compare this expression with the right-hand side. For this hypothetical equation to hold for all real x and t, the following should be satisfied:

\displaystyle 1 + \cos(-0.250) = m\, \cos(c), and

\displaystyle \sin(-0.250) = m\, \sin(c).

Consider the Pythagorean identity. For any real number a:

{\left(\sin(a)\right)}^{2} + {\left(\cos(a)\right)}^{2} = 1^2.

Make use of the Pythagorean identity to solve this system of equations for m. Square both sides of both equations:

\displaystyle 1 + 2\, \cos(-0.250) +  {\left(\cos(-0.250)\right)}^2= m^2\, {\left(\cos(c)\right)}^2.

\displaystyle {\left(\sin(-0.250)\right)}^{2} = m^2\, {\left(\sin(c)\right)}^2.

Take the sum of these two equations.

Left-hand side:

\begin{aligned}& 1 + 2\, \cos(-0.250) + \underbrace{{\left(\cos(-0.250)\right)}^2 + {\left(\sin(-0.250)\right)}^2}_{1}\\ &= 1 + 2\, \cos(-0.250) + 1 \\ &= 2 + 2\, \cos(-0.250) \end{aligned}.

Right-hand side:

\begin{aligned} &m^2\, {\left(\cos(c)\right)}^2 + m^2\, {\left(\sin(c)\right)}^2 \\ &= m^2\, \left( {\left(\sin(c)\right)}^2 +  {\left(\cos(c)\right)}^2\right)\\ &= m^2\end{aligned}.

Therefore:

m^2 = 2 + 2\, \cos(-0.250).

m = \sqrt{2 + 2\, \cos(-0.250)} \approx 1.98.

Substitute m = \sqrt{2 + 2\, \cos(-0.250)} back to the system to find c. However, notice that the exact value of c\! isn't required for finding the amplitude of (y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c).

(Side note: one possible value of c is \displaystyle \arccos\left(\frac{1 + \cos(0.250)}{\sqrt{2 \times (1 + \cos(0.250))}}\right) \approx 0.125 radians.)

As long as \! c is a real number, the amplitude of (y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c) would be equal to the absolute value of (4.85\, m).

Therefore, the amplitude of (y_1 + y_2) would be:

\begin{aligned}|4.85\, m| &= 4.85 \times \sqrt{2 + 2\, \cos(-0.250)} \\&\approx 9.62 \end{aligned}.

8 0
3 years ago
5. What happens to the inductor current as time passes? 6. What happened to the inductor current when the supply voltage was tri
lawyer [7]

Answer:

i ran out of time

Explanation:

6 0
3 years ago
A car starts from rest with an acceleration of 2.84 m/s2 at the instant when a second car moving with a velocity of 25.7 m/s pas
vredina [299]

Answer:

464.69 m

Explanation:

First car

s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}2.84t^2\\\Rightarrow s=1.42t^2

Second car

Distance = Speed × Time

\text{Distance}=25.7t

Here, the time taken and the distance traveled will be the same

Equating the two equations

1.42t^2=25.7t\\\Rightarrow t=\frac{25.7}{1.42}\\\Rightarrow t=18.09\ s

So, the first would have to move 1.42t^2=1.42\times 18.09^2=464.69\ m in order to overtake the second car.

7 0
3 years ago
Which light packs the highest energy per photon? 1. infrared 2. red 3. blue 4. ultraviolet 5. green?
WINSTONCH [101]
A single photon carries an energy equal to
E=hf
where h is the Planck's constant and f is the frequency of the photon.
This means that the higher the frequency of the light, the higher the energy. Among the 5 different options mentioned by the problem, the light with highest frequency is ultraviolet, which has frequencies in the range [3-30] PHz, while visible light (red, blue, green) and infrared have lower frequency, so ultraviolet light has the highest energy per photon.
3 0
3 years ago
Other questions:
  • Sadi Carnot came up with a hypothetical heat engine that had the maximum possible efficiency. What discovery did Carnot make tha
    10·1 answer
  • What are two functions of the respiratory system
    9·2 answers
  • How to get a + b on a graph
    8·1 answer
  • Fill in the blank-There is very little precipitation in the Taiga, however when moisture does fall, it is typically in the form
    13·1 answer
  • Carlos gives a grocery cart a 60-N push. The cart has a mass of 40 kg. What is the cart's acceleration?
    15·1 answer
  • In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal:__
    11·1 answer
  • Where is the most of the liquid freshwater on earth located?
    15·2 answers
  • Average speed is the total distance divided by the
    8·1 answer
  • PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP
    12·2 answers
  • Why is the efficiency of a machine always less than 100%?
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!