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LekaFEV [45]
3 years ago
13

Please help! Calculate velocity. Show all work!

Physics
1 answer:
Eduardwww [97]3 years ago
4 0

Answer:

v = 23.66 m/s

Explanation:

recall that one of the equations of motion may be expressed:

v² = u² + 2as,

Where

v = final velocity (we are asked to find this)

u = initial velocity = 0 m/s since we are told that it starts from rest

a = acceleration = 0.56m/s²

s = distance traveled = given as 500m

Simply substitute the known values into the equation:

v² = u² + 2as

v² = 0 + 2(0.56)(500)

v² = 560

v = √560

v = 23.66 m/s

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If you use a force of 90 N to pick up a 10 pound bag of charcoal, what is the acceleration?
hjlf

Answer:

9ms^2

Explanation:

since ,Force=mass*acceleration

then, acceleration=force/mass

and, Force=90N

Mass=10pound

therefore, acceleration=90/10

=9ms^2

8 0
3 years ago
A 5 kg bowling ball with a velocity of +10 m/s collides with a stationary 2 kg bowling pin. If the ball's final velocity is +8 m
Whitepunk [10]

Answer:

<h2>The pin's final velocity is 5m/s</h2>

Explanation:

Step one:

given data

mass of ball m1=5kg

initial velocity of ball u1=10m/s

mass of pin m2=2kg

initial velocity of pin u2= 0m/s

final velocity of ball v2=8m/s

final velocity of pin v2=?

Step two:

The expression for elastic collision is given as

m1u1+m2u2=m1v1+m2v2

substituting we have

5*10+2*0=5*8+2*v2

50+0=40+2v2

50-40=2v2

10=2v2

divide both sides by 2

v2=10/2

v2=5m/s

The pin's final velocity is 5m/s

3 0
3 years ago
Consider a double slit experiment in the air. The wavelength of light is 500nm light, the screen is 1m away and two adjacent bri
Bess [88]

Answer: 0.75\ cm

Explanation:

Given

Wavelength of light \lambda=500\ nm

Screen is D=1\ m away

Distance between two adjacent bright fringe is \Delta y=\dfrac{\lambda D}{d}

When same experiment done in water, wavelength reduce to \dfrac{\lambda }{\mu}

So, the distance between the two adjacent bright fringe is \Delta y'=\dfrac{\lambda D}{\mu d}

Keeping other factor same, distance becomes

\Rightarrow \dfrac{1}{\frac{4}{3}}=\dfrac{3}{4}\quad \text{Refractive index of water is }\dfrac{4}{3}\\\\\Rightarrow 0.75\ cm

3 0
3 years ago
Three deer, a, b, and c, are grazing in a field. deer b is located 62.1 m from deer a at an angle of 54.3 ° north of west. deer
sveticcg [70]

by cosine law we know that

c^2 = a^2 + b^2 - 2 abcos\theta

\theta = 180 - 54.3 - 79.1 = 46.6 degree

now using above equation

93.8^2 = 62.1^2 + b^2 - 2*62.1*b * cos46.6

4942.03 = b^2 - 85.34 b

b^2 - 85.34b - 4942.03 = 0

by solving above quadratic equation we have

b = 124.9 m

so it is at distance 124.9 m from deer a

4 0
3 years ago
PLEASE HELP and actually help plz
frutty [35]

Answer:

iEvaluate  for \(x=2.\)Evaluate  for \(x=2.\)Evaluate  for \(x=2.\)Evaluate  for \(x=2.\)Evaluate  for \(x=2.\)Evaluate  for \(x=2.\)Evaluate  for \(x=2.\)

Explanation:

3 0
3 years ago
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