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Elena-2011 [213]
3 years ago
6

A bullet is fired through a board 13.0 cm thick in such a way that the bullet's line of motion is perpendicular to the face of t

he board. The initial speed of the bullet is 560 m/s and it emerges from the other side of the board with a speed of 460 m/s. (a) Find the acceleration of the bullet as it passes through the board.
Physics
1 answer:
GREYUIT [131]3 years ago
5 0

Answer:

392307.6923 m/s²

Explanation:

t = Time taken

u = Initial velocity = 560 m/s

v = Final velocity = 460 m/s

s = Displacement = 13 cm

a = Acceleration

From the equation of motion we have

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{460^2-560^2}{2\times 13\times 10^{-2}}\\\Rightarrow a=-392307.6923\ m/s^2

The acceleration of the bullet as it passes through the board is -392307.6923 m/s²

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If an object accelerates from rest, with a constant acceleration of 5.4 m/s2, what will its velocity be after 28s?
aleksklad [387]
Vf = Vi + at
Vf = 0 + 5.4•28
= 151.2m/s..
not sure if its right
6 0
2 years ago
Coherent light of frequency 6.37×1014 Hz passes through two thin slits and falls on a screen 88.0 cm away. You observe that the
IgorC [24]

Answer:

The distance between the two slits is 40.11 μm.

Explanation:

Given that,

Frequency f= 6.37\times10^{14}\ Hz

Distance of the screen l = 88.0 cm

Position of the third order y =3.10 cm

We need to calculate the wavelength

Using formula of wavelength

\lambda=\dfrac{c}{f}

where, c = speed of light

f = frequency

Put the value into the formula

\lambda=\dfrac{3\times10^{8}}{6.37\times10^{14}}

\lambda=471\ nm

We need to calculate the distance between the two slits

m\times \lambda=d\sin\theta

d =\dfrac{m\times\lambda}{\sin\theta}

Where, m = number of fringe

d = distance between the two slits

Here, \sin\theta =\dfrac{y}{l}

Put the value into the formula

d=\dfrac{3\times471\times10^{-9}\times88.0\times10^{-2}}{3.10\times10^{-2}}

d=40.11\times10^{-6}\ m

d = 40.11\ \mu m

Hence, The distance between the two slits is 40.11 μm.

7 0
2 years ago
PLEASE HELP QUICKLY THANKS
Dmitriy789 [7]

Answer: the answer is d

Explanation:

7 0
3 years ago
A 75 kg ball carrier is running to the right at 6.5 m/s. An 80 kg defender is chasing the ball carrier running at 7.0 m/s. The d
Sophie [7]

Answer:

3,544.375Joules

Explanation:

Kinetic energy is the energy possessed by a body by virtue of its motion, It is expressed as;

Kinetic energy = 1/2mv²

m is the mass of the body

v is the velocity

For the ball carrier;

KE = 1/2(75)(6.5)²

KE = 3168.75/2

KE = 1584.375Joules

For the defender;

KE = 1/2(80)(7)²

KE = 3920/2

KE = 1960Joules

The kinetic energy of the ball carrier/defender system BEFORE the tackle = KE for the carrier + KE for the defender

kinetic energy of the ball carrier/defender system BEFORE the tackle= 1584.375+1960 = 3,544.375Joules

7 0
2 years ago
A car travels along a straight line at a constant speed of 60.0 mi/h for a distance d and then another distance d in the same di
Makovka662 [10]

Answer:

velocity during second d = 20.0 mi/h

Explanation:

Total distance travelled is 2d, with an average velocity of 30.0 mi/h you can express the time travelled in terms of d:

distance = velocity * time

time = distance / velocity

time = 2d/30.0

The time needed for the first d at 60.0 is:

time = d/60.0

The time in the second d you can get it by substracting both times (total time - time for the first d)

second d time = 2d/30.0 - d/60.0

= 4d/60.0 - d/60.0

= 3d/60.0

and with the time (3d/60.0) and the distance travelled (d) you can get the velocity:

velocity = distance / time

velocity = d / (3d/60.0)

= 60.0/3 = 20.0 mi/h

8 0
2 years ago
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