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3 years ago

6

find the rate of positive acceleration of an automobile which went from a complete stop to a velocity of 30 meters per second in

10 seconds

1 answer:

maks197457 [2]3 years ago

3 0

Answer:

3 m/s^2

Explanation:

acceleration= Change in velocity/time

= 30-0 / 10

= 30/10

=3 m/s^2

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What type of elastic force is present in the strings on a guitar?

jeyben [28]

B) Tension , is the correct answer

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2 years ago

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts help.

GrogVix [38]

Answer:

$H = 532 m$

Explanation:

When teacher falls into the cliff then she shout for Help at the same time

so here we know that sound will go down and reflect back up

so here in 3 s distance traveled by the sound

$d = vt$

$d = 340 \times 3$

$d = 1020 m$

now in the same time the distance that teacher will fall down is given as

$d_1 = \frac{1}{2}gt^2$

$d_1 = \frac{1}{2}(9.81)(3^2)$

$d_1 = 44.1 m$

now total distance traveled by teacher and sound in 3 s

$d_{total} = d + d_1$

$d_{total} = 1020 + 44.1$

this total distance must be equal to twice the height of the cliff

$2H = 1064.1 m$

$H = 532 m$

7 0

3 years ago

Read 2 more answers

A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with

Komok [63]

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

$\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\ g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s$

$\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM$

Therefore the speed in RPM will be 62.64 RPM.

5 0

2 years ago

PROBLEMA DI FISICA:il volume di un attizzatoio da camino in ferro, a temperatura ambiente (20°C), è 12,2 cm cubi. A contatto con

lesantik [10]

Responder:

20.3 ° C

Explicación:

<u>Según la ley de Charles</u>: <em>cuando la presión sobre una muestra de gas seco se mantiene constante, la temperatura y el volumen estarán en proporción directa. </em>

Paso uno:

datos dados

Temperatura T1 = 20 ° C

Temperatura T2 =?

Volumen V1 = 12.2 cm ^ 3

Volumen V2 = 12.4 cm ^ 3

Aplicar la relación temperatura y volumen

$\frac {V_{1}}{T_{1}}=\frac {V_{2}}{T_{2}}$

sustituyendo tenemos

$\frac {12.2}{20}=\frac {12.4}{T_{2}}\\\\$

Cruz multiplicar tenemos

$T_2=\frac{20*12.4}{12.2} \\\\T_2=\frac{248}{12.2}\\\\T_2=20.3$

Temperatura delle braci 20.3°C

4 0

3 years ago

An object pulled to the right by two forces has an acceleration of 2.5m/s2. The free-body diagram shows the forces acting on the

Vanyuwa [196]

Answer:

490 N is the correct answer.

Explanation:

8 0

3 years ago