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12345 [234]
3 years ago
15

I got 13 min to do 20 problems helpppp

Physics
1 answer:
Alina [70]3 years ago
3 0

Answer:

I believe it's 200 miles...

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In Which senecio is an animal doing work
Arisa [49]

Answer:

A horse pulls a wagon along a road

4 0
3 years ago
A daredevil is shot out of a cannon at 45.0° to the horizontal with an initial speed of 31.0 m/s. A net is positioned a horizont
hodyreva [135]

Answer:

s = vcos(x)t

50 = 25cos(45)t

cos(45)t = 2

t = 2/cos(45) = 2sqrt(2)

h = vsin(x)t + gt^2/2

h = 25sin(45)*2sqrt(2) - 4.9*8

h = 10.8 metres

Explanation:

5 0
3 years ago
Read 2 more answers
A 1,100 kg car is traveling initially 20 m/s when the brakes are applied. The brakes apply a constant force while bringing the c
Natalka [10]

Answer:

Work done = -220,000 Joules.

Explanation:

<u>Given the following data;</u>

Mass = 1100kg

Initial velocity = 20m/s

To find workdone, we would calculate the kinetic energy possessed by the car.

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

K.E = \frac{1}{2}MV^{2}

Where,

  • K.E represents kinetic energy measured in Joules.
  • M represents mass measured in kilograms.
  • V represents velocity measured in metres per seconds square.

Substituting into the equation, we have;

K.E = \frac{1}{2}*1100*20^{2}

K.E = 550*400

K.E = 220,000J

Therefore, the workdone to bring the car to rest would be -220,000 Joules because the braking force is working to oppose the motion of the car.

4 0
2 years ago
Kristina demonstrates a toy for her younger brother. The steps to operate the toy are listed below. Step 1. Push the toy down. S
IRINA_888 [86]
The object has potential and kinetic energy when it jumps into the air
0 0
3 years ago
Read 2 more answers
If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?
sdas [7]
Data:
f_{2} = 42 Hz
n (Wave node)
V (Wave belly) 
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
f_{n} =  \frac{nV}{2L}

Wire 2 → 2º Harmonic → n = 2

f_{n} = \frac{nV}{2L}
f_{2} = \frac{2V}{2L} &#10;
2V =  f_{2} *2L
V =  \frac{ f_{2}*2L }{2}
V =  \frac{42*2L}{2}
V =  \frac{84L}{2}
V = 42L

Wire 1 → 1º Harmonic or Fundamental rope → n = 1


f_{n} = \frac{nV}{2L}
f_{1} = \frac{1V}{2L}
f_{1} =  \frac{V}{2L}

If, We have:
V = 42L
Soon:
f_{1} = \frac{V}{2L}
f_{1} = \frac{42L}{2L}
\boxed{f_{1} = 21 Hz}

Answer:

<span>The fundamental frequency of the string:
</span>21 Hz

7 0
3 years ago
Read 2 more answers
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