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SpyIntel [72]
3 years ago
7

A(n) 0.155 kg baseball is thrown with a speed of 18.4 m/s. It is hit straight back at the pitcher with a final speed of 18.4 m/s

. What is the magnitude of the impulse delivered to the ball? Answer in units of kg · m/s.
Physics
1 answer:
Ivan3 years ago
6 0

Answer:

5.704kg.m/s toward the pitcher

Explanation:

Impulse (Ft)= change in momentum

Impulse= m(v-u)

m=0.155 kg  ,       v=18.4 m/s  , u=18.4m/s

impulse=0.155[-(18.4) - (18.4)]

        =  0.155(-36.8)

        = -5.704 kg.m/s

the negative means towards the pitcher

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The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values
Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

to find out

distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level  β  = 10 log(I/I_{0})        ......................a

put here value (I/I_{0}) = 10^−12 W/m² and  β = 115

115  = 10 log(I/10^−12)

so

I = 0.316228 W/m²

and we know power = intensity × 4π r²    ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m    = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

5 0
3 years ago
An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10
s2008m [1.1K]

Answer:

<em>1.228 x </em>10^{-6}<em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm  

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x 10^{3} N

modulus of elasticity E = 85 GN/m^2  = 85 x 10^{9} Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = \frac{\pi D^{2} }{4} =  \frac{3.142* 40^{2} }{4} = 1256.8 mm^2

area of hole a = \frac{\pi(D^{2} - d^{2}) }{4} = \frac{3.142*(40^{2} - 30^{2})}{4} = 549.85 mm^2

Total contraction of the bar = \frac{F*L}{AE} + \frac{Fl}{aE}

total contraction = \frac{F}{E} * (\frac{L}{A} +\frac{l}{a})

==> \frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85}) = <em>1.228 x </em>10^{-6}<em> mm </em>

6 0
3 years ago
A 30 g bullet moving a horizontal velocity of 500 m/s comes to a stop 12 cm within a solid wall. (a) what is the change in the b
Sidana [21]
M = 30 g = 0.03 kg, the mass of the bullet
v = 500 m/s, the velocity of the bullet

By definition, the KE (kinetic energy) of the bullet is
KE = (1/2)*m*v²
      = 0.5*(0.03 kg)*(500 m/s)² = 3750 J
Because the bullet comes to rest, the change in mechanical energy is 3750 J.

The work done by the wall to stop the bullet in 12 cm is
W = (1/2)*(F N)*(0.12 m) = 0.06F J

If energy losses in the form of heat or sound waves are ignored, then
W = KE.
That is,
0.06F = 3750
F = 62500 N = 62.5 kN

Answer:
(a) 3750 J
(b) 62.5 kN

7 0
3 years ago
How will heat flow between objects?
alukav5142 [94]
Without an external agent doing work, heat will always flow from a hotter to a cooler object. Two objects of different temperature always interact. There are three different ways for heat to flow from one object to another. They are conduction, convection, and radiation.
5 0
3 years ago
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On his way to school, a student starts out walking quickly and then slows down as he gets closer to the school. Which graph best
Leni [432]

' A ' is the graph that shows it.

3 0
3 years ago
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