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Sedaia [141]
3 years ago
10

A shopping cart given an initial velocity of 2.0 m/s undergoes a constant acceleration of 3.0 m/s2. What is the magnitude of the

cart’s displacement after the first 4.0 s of its motion?
Physics
1 answer:
Tema [17]3 years ago
6 0

Answer:

32 m

Explanation:

initial velocity (u) = 2 m/s

accceleration (a) = 3 m/s^{2}

time (t) = 4 s

What is the magnitude of the cart’s displacement (s) after the first 4.0 s of its motion?

from the equation of motion, displacement (s) = ut + 0.5at^{2}

where

  • u is the initial velocity
  • t is the time
  • a is the acceleration

inserting all required values into  (s) = ut + 0.5at^{2}          

(s) = 2x4 + 0.5x3x4^{2}

s = 8 + 24

s = 32 m

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2. A rock is dropped off a bridge. How fast is the rock
zhenek [66]

Answer:

a) The velocity of rock at 1 second, v = 9.8 m/s

b) The velocity of rock at 3 second,  v = 29.4 m/s

c) The velocity of rock at 5.5 second,  v = 53.9 m/s

Explanation:

Given data,

The rock is dropped from a bridge.

The initial velocity of the rock, u = 0

a) The velocity of rock at 1 second,

   Using the first equation of motion

                         v = u + gt

                         v = 0 + 9.8 x 1

                          v = 9.8 m/s

b) The velocity of rock at 3 second,

                         v = u + gt

                         v = 0 + 9.8 x 3

                          v = 29.4 m/s

c) The velocity of rock at 5.5 second,

                         v = u + gt

                         v = 0 + 9.8 x 5.5

                         v = 53.9 m/s

5 0
3 years ago
Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect
Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a)  0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b)  r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 \pi r^{2}  = \frac{Q1}{E_{0} }

So,

Rearranging the above equation to get Electric field, we will get:

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   }

Multiply and divide by r1^{2}

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } x \frac{r1^{2} }{r1^{2} }

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x r1^{2}) /(E_{0} x r^{2})

c) r > r2 :

Electric Field = ?

E x 4 \pi r^{2}  = \frac{Q1 + Q2}{E_{0} }

Rearranging the above equation for E:

E = \frac{Q1+Q2}{E_{0} . 4 \pi. r^{2}   }

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

As we know from above, that:

\frac{Q1}{E_{0} . 4 \pi. r^{2}   } =  (σ1 x r1^{2}) /(E_{0} x r^{2})

Then, Similarly,

\frac{Q2}{E_{0} . 4 \pi. r^{2}   } = (σ2 x r2^{2}) /(E_{0} x r^{2})

So,

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

Replacing the above equations to get E:

E = (σ1 x r1^{2}) /(E_{0} x r^{2}) + (σ2 x r2^{2}) /(E_{0} x r^{2})

Now, for

d) Under what conditions,  E = 0, for r > r2?

For r > r2, E =0 if

σ1 x r1^{2} = - σ2 x r2^{2}

4 0
3 years ago
When atoms of an element are excited, they emit specific wavelengths of light. How is this similar to a fingerprint when Fraunho
Anika [276]

Answer:

As you may know, each element has a "fixed" number of protons and electrons.

These electrons live in elliptical orbits around the nucleus, called valence levels or energy levels.

We know that as further away are the orbits from the nucleus, the more energy has the electrons in it. (And those energies are fixed)

Now, when an electron jumps from a level to another, there is also a jump in energy, and that jump depends only on the levels, then the jump in energy is fixed.

Particularly, when an electron jumps from a more energetic level to a less energetic one, that change in energy must be compensated in some way, and that way is by radiating a photon whose energy is exactly the same as the energy of the jump.

And the energy of a photon is related to the wavelength of the photon, then we can conclude that for a given element, the possible jumps of energy levels are known, meaning that the possible "jumps in energy" are known, which means that the wavelengths of the radiated photons also are known. Then by looking at the colors of the bands (whose depend on the wavelength of the radiated photons) we can know almost exactly what elements are radiating them.

7 0
3 years ago
3. When a person is outside of the system and they add energy to the
NISA [10]

Answer:

We show added energy to a system as +Q or -W

Explanation:

The first law of thermodynamics states that, in an isolated system, energy can neither be created nor be destroyed;

Energy is added to the internal energy of a system as either work energy or heat energy as follows;

ΔU = Q - W

Therefore, when energy is added as heat energy to a system, we show the energy as positive Q (+Q), when energy is added to the system in the form of work, we show the energy as minus W (-W).

5 0
3 years ago
A student swings a container of water in a vertical circle of radius 1.0 m. Calculate the minimum speed of the container so that
12345 [234]

Answer:

Explanation:

The centripetal acceleration requirement must equal gravity at the top of the circle

mg = mv²/R

  v = √Rg

  v = √(1.0(9.8))

  v = 3.1304951...

  v = 3.1 m/s

6 0
3 years ago
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