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Effectus [21]
3 years ago
14

During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.148 kg baseball crashing through the pane

of a second-floor window in a nearby building. The ball strikes the glass at 15.5 m/s , shatters the glass as it passes through, and leaves the window at 10.1 m/s with no change of direction. What is the direction of the impulse that the glass imparts to the baseball
Physics
1 answer:
LiRa [457]3 years ago
3 0

Answer:

Explanation:

mass of baseball, m = 0.148 kg

initial velocity, u = 15.5 m/s

final velocity, v = 10.1 m/s

Impulse is defined as the change in momentum of the body.

Impulse = change in momentum

I = m (v - u)

I = 0.148 (10.1 - 15.5)

I = - 0.8 Ns

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What does the universe look like on very large scales?
Anestetic [448]

Answer:

it looks like dots and just black space on a large scale

Explanation:

on a large scale the universe especially our milky way looks small

hope this helps  

3 0
2 years ago
A car traveling 85 km/h is 250 m behind a truck<br> traveling 73 km/h.
amm1812

Time needed = t = 20.83 s

<h3>Further explanation</h3>

Given

car speed = 85 km/h

truck speed = 73 km/h

Required

the time it takes for the car to reach the truck

Solution

When the car reaches the truck, the distance between them will be the same

x car - 250 m = x truck

General formula for distance (d) :

d = v.t

So the equation becomes :

85t-250 = 73t

12t=250

t = 20.83 s

8 0
3 years ago
What type of energy increases as air expands
Vinvika [58]
Ummm..Kinetic Engery??
3 0
3 years ago
Read 2 more answers
An LED operation at 850 nm center wavelength has a spectral width of 45 nm. What is the pulse spreading in ns/km
Papessa [141]

Answer:

\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}

Explanation:

From the given information, the LED is operating with a given wavelength of 850 nm or 0.85 μm.

Hence, the material dispersion is \dfrac {d \tau _{mat}}{d \lambda } \simeq (80 \ ps / (nm.km) \ )

Now, using the pulse spread formula:

\dfrac{\sigma_{mat}}{L} = \dfrac{d \tau _{mat} }{d \lambda} \sigma \lambda

\dfrac{\sigma_{mat}}{L} = (80 \ ps/ ( m.km) \ )  \times (45 \ nm)

Thus, the pulse spreading as a result of  material dispersion is:\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}

3 0
3 years ago
Dos cargas puntuales q1 = −50μC y q2 = +30μC se encuentran
alina1380 [7]

Answer:

Datos:

q1 = -50 μC = 50*10^{-6}

q2 = +30 μC = 30*10^{-6}

F = 10 N

a) x si la <em>F = 10N</em>

Aplicando la Ley de Coulomb:

x = \sqrt\frac{k0 * q1 *q2}{F} = \sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{10} = 1,162m

b) x si la <em>F = 20 N</em>

x=<em> </em>\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{20}<em> </em>= 0,822m

c)x si la <em>F = 50 N</em>

x = \sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{50} = 0,520m

3 0
3 years ago
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