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3 years ago

5

Un coche de unos 500 kg viaja a 90 km/h. Percibe un obstáculo y debe frenar a tope. Por las marcas del suelo se sabe que el espa

cio de frenada fue de 125 m. ¿Cuánto valía la fuerza de rozamiento entre el coche y la carretera?

1 answer:

a_sh-v [17]3 years ago

3 0

Answer:

THERE'S A GIRL WHO WILL STEAL YOUR INFO, PLZ DON"T PRESS THE LINK IF SHE ASKS YOU TOO!!!!!

Explanation:

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Which phenomenon occurs when the moon and Earth are aligned with the sun? A. Retrograde motion B. Solstice C. Equinox D. Eclipse

muminat

Answer: Eclipse

Explanation: A lunar eclipse occurs when the full moon moves through the shadow of the Earth. This can only happen when the Earth is between the Moon and the Sun and all three are lined up in the same plane, called the ecliptic. The ecliptic is the plane of Earth's orbit around the Sun.

7 0

3 years ago

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} 
$
$2V = f_{2} *2L$
$V = \frac{ f_{2}*2L }{2}$
$V = \frac{42*2L}{2}$
$V = \frac{84L}{2}$
$V = 42L$

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

Which of the following is an impact of trampling of beach grass by humans and vehicles?

grigory [225]

Increase in sea water pollution

6 0

3 years ago

A cyclist traveling at constant speed of 12m/s when he passes a stationary bus.The bus starts moving just as the cyclist passes

Bogdan [553]

Answer:

A.) 8 seconds

B.) 16 seconds

C.) 48 m

Explanation:

Given that a cyclist traveling at constant speed of 12 m/s

and the bus accelerates uniformly at 1.5ms²

A.) The bus has the following parameters

Acceleration a = 1.5 m/s^2

Initial velocity U = 0. Since the bus is starting from rest.

Final velocity V = 12 m/s

Use equation one of linear motion.

V = U + at

Substitute V, U and a into the formula

12 = 0 + 1.5t

1.5t = 12

t = 12/1.5

t = 8 seconds

Therefore, the bus reach the same speed as the cyclist at 8 seconds.

B.) For the cyclist moving at constant speed, acceleration a = 0. Using second equation of motion

h = Ut + 1/2at^2

Since a = 0, the equation is reduced to:

h = Ut.

Also, for the bus,

h = Ut + 1/2at^2

Equate the two equations since the h is the same

Ut = Ut + 1/2at^2

Substitute all the parameters into the formula

12t = 0 + 1/2 × 1.5t^2

12t = 0.75t^2

0.75t = 12

t = 12/0.75

t = 16 seconds

Therefore, the bus takes 16 seconds to catch the cyclist

C.) Use third equation of linear motion.

V^2 = U^2 + 2as

Where s = distance

Substitute V, U and a into the formula

12^2 = 0 + 2 × 1.5 S

144 = 3S

S = 144/3

S = 48 m

8 0

3 years ago

Question 4 (1 point)

NNADVOKAT [17]

Answer: 0.42 Amperes

Explanation:

Given that:

Current, I = ?

Electric charge Q = 100 coulomb

Time, T = 4.0 minutes

(The SI unit of time is seconds. so, convert 4.0 minutes to seconds)

If 1 minute = 60 seconds

4.0 minutes = 4.0 x 60 = 240 seconds

Since electric charge, Q = current x time

i.e Q = I x T

100 coulomb = I x 240 seconds

I = 100 coulomb / 240 seconds

I = 0.4167 Amperes (round to the nearest hundredth which is 0.42 amperes)

Thus, 0.42 Amperes of current flows in the circuit.

6 0

3 years ago