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Montano1993 [528]
3 years ago
7

What beat frequencies result if a piano hammer hits three strings that emit frequencies of 392.0, 587.3, and 146.8 hz? (enter yo

ur answers from smallest to largest.)?
Physics
1 answer:
FinnZ [79.3K]3 years ago
3 0
128.1-127.8= 0.3Hz 
<span>129.1-128.1= 1.0Hz </span>
<span>129.1-127.8= 1.3Hz</span>
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In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 8.13 km mark at a
SOVA2 [1]

Answer:0.084 m/s^2

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance d_1=8.13 km

time taken=25 min =1500 s

initial speed v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s

after 8.13 km mark steve started to accelerate

speed after 60 s

v_2=v_1+at

v_2=5.6+a\times 60

distance traveled in 60 sec

d_2=v_1\times 60+\frac{a60^2}{2}

d_2=336+1800 a

time taken in last part of journey

t_3=1663.6-1560=103.6 s

distance traveled in this time

d_3=v_2\times t_3

d_3=\left ( 5.6+a\times 60\right )103.6

and total distance=d_1+d_2+d_3

10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6

1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6

a=0.084 m/s^2

5 0
3 years ago
A farmer hitches her tractor to a sled loaded with firewood and pulls it a distance
Delvig [45]

(a) The work done by the force applied by the tractor is 79,968.47 J.

(b) The work done by the frictional force on the tractor is 55,977.93 J.

(c) The total work done by  all the forces is 23,990.54 J.

<h3>Work done by the applied force</h3>

The work done by the force applied by the tractor is calculated as follows;

W = Fd cosθ

W = (5000 x 20) x cos(36.9)

W = 79,968.47 J

<h3>Work done by frictional force</h3>

W = Ffd cosθ

W = (3500 x 20) x cos(36.9)

W = 55,977.93 J

<h3>Net work done by all the forces on the tractor</h3>

W(net) = work done by applied force  -  work done by friction force

W(net) = 79,968.47 J -  55,977.93 J

W(net) = 23,990.54 J

Learn more about work done here: brainly.com/question/25573309

#SPJ1

4 0
1 year ago
A 1.5 kg cart is attached to a spring with spring constant of 5 N/m. The cart &amp; spring is pulled to stretch the spring by 3
vaieri [72.5K]

22.5 J

Explanation:

Given:

x = 3 m

k = 5\:\text{N/m}

The spring potential energy PE_s is

PE_s = \frac{1}{2}kx^2 = \frac{1}{2}(5\:\text{N/m})(3\:\text{m})^2

\:\:\:\:\:\:\:=22.5\:\text{J}

3 0
2 years ago
In this first example of constant accelerated motion, we will simply consider a car that is initially traveling along a straight
algol13

Answer:

v_{f} =25m/s

Explanation:

Kinematics equation for constant acceleration:

v_{f}  =v_{o} + at=15+2*5=25m/s

4 0
3 years ago
A 67-kg base runner begins his slide into second base when he is moving at a speed of 3.6 m/s. The coefficient of friction betwe
USPshnik [31]

Answer:

434.16 Joules

Explanation:

u = Initial velocity

v = Final velocity

m = Mass of person

From work-energy theorem

KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow KE=\frac{1}{2}\times 67\times (3.6^2-0)\\\Rightarrow KE=434.16\ Joules

The runner loses 434.16 Joules of mechanical energy.

W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow \mu mg=\frac{W}{s}\\\Rightarrow s=\frac{W}{\mu mg}\\\Rightarrow s=\frac{434.16}{0.7\times 67\times 9.81}\\\Rightarrow s=0.94364\ m

He slides 0.94364 m

4 0
3 years ago
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