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borishaifa [10]
3 years ago
11

A car with an initial speed of 6.5 m/s accelerates at a uniform rate of 0.92 m/s2 for 3.6 s. Find the final speed of the car dur

ing this time.
Physics
1 answer:
geniusboy [140]3 years ago
8 0
U = 6.5 m/s, initial speed
t = 3.6 s, time
a = 0.92 m/s², acceleration

Let v = the final velocity.
Then
v = u +at
v = (6.5 m/s) + (0.92 m/s²)*(3.6 s) = 9.812 m/s

Answer: 9.81 m/s 
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Elaborate on the classification of aluminum. A) Aluminum is a noble gas and has a full valence electron shell, is chemically non
scZoUnD [109]
The correct option is (B) <span>Aluminum is a metal and is shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.

Since Aluminium is in group 13, and all the elements in group 13 are either metals or metalloids(Boron). Hence we are left with option (B) and (D). Boron is the only metalloid in group 13 and aluminium is a metal(not a metalloid); therefore, we are left with only one option which is Option (B). And Aluminium is </span>shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.<span>
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4 0
3 years ago
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A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they pr
Westkost [7]

A) Order of the first laser: 3, order of the second laser: 2

B) The overlap occurs at an angle of 34.9^{\circ}

Explanation:

A)

The formula that gives the position of the maxima (bright fringes) for a diffraction grating is

d sin \theta = m \lambda

where

d is spacing between the lines in the grating

\theta is the angle of the maximum

m is the order of diffraction

\lambda is the wavelength of the light

For laser 1,

d sin \theta = m_1 \lambda_1

For laser 2,

d sin \theta = m_2 \lambda_2

where

\lambda_1 = 420 nm\\\lambda_2 = 630 nm

Since the position of the maxima in the two cases overlaps, then the term d sin \theta on the left is the same for the two cases, therefore we can write:

m_1 \lambda_1 = m_2 \lambda_2\\\frac{m_1}{m_2}=\frac{\lambda_2}{\lambda_1}=\frac{630}{420}=\frac{3}{2}

Therefore:

m_1 = 3

m_2 = 2

B)

In order to find the angle at which the overlap occurs, we use the 1st laser situation:

d sin \theta = m_1 \lambda_1

where:

N = 450 lines/mm = 450,000 lines/m is the number of lines per unit length, so the spacing between the lines is

d=\frac{1}{N}=\frac{1}{450,000}=2.2\cdot 10^{-6} m

m_1 = 3 is the order of the maximum

\lambda_1 = 420 nm = 420\cdot 10^{-9} m is the wavelength of the laser light

Solving for \theta, we find the angle of the maximum:

sin \theta = \frac{m_1 \lambda_1}{d}=\frac{(3)(420\cdot 10^{-9})}{2.2\cdot 10^{-6}}=0.572

So the angle is

\theta=sin^{-1}(0.572)=34.9^{\circ}

Learn more about diffraction:

brainly.com/question/3183125

#LearnwithBrainly

5 0
3 years ago
A football punter accelerates a .55 kg football
Ronch [10]

Answer:

17.6 N

Explanation:

The force exerted by the punter on the football is equal to the rate of change of momentum of the football:

F=\frac{\Delta p}{\Delta t}

where

\Delta p is the change in momentum of the football

\Delta t=0.25 s is the time elapsed

The change in momentum can be written as

\Delta p = m(v-u)

where

m = 0.55 kg is the mass of the football

u = 0 is the initial  velocity (the ball starts from rest)

v = 8.0 m/s is the final velocity

Combining the two equations and substituting the values, we  find the force exerted on the ball:

F=\frac{m(v-u)}{\Delta t}=\frac{(0.55)(8.0-0)}{0.25}=17.6 N

5 0
4 years ago
We can catch a rolling ball but not a flying bullet?? give reason
Triss [41]

Answer:

yeah this statement is tru

Explanation:

it is because the speed of the bullet is more than the speed of rolling ball .so from this reason we cannot catch a bullet.

3 0
3 years ago
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Acceleration is zero if
tester [92]
It’s doesn’t change meaning it’s 0
4 0
3 years ago
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