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2 years ago

Se tiene un objeto a una altura de 8 m generando 450 J de energía potencial cuál es la masa de dicho objeto

Physics

1 answer:

fgiga [73]2 years ago

8 0

Answer:

5.73kg

Explanation:

Given data

Height = 8m

Potential Energy = 450J

Required

The mass of the object

From the following expression

PE=mgh

we can substitute g= 9.81m/s^2 and the other parameters to get mass m

450= m*9.81*8

450= m* 78.48

m= 450/78.48

m= 5.73kg

Hence the mass is 5.73kg

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A roller coaster is moving at 25m/s at the bottom of a hill. Three seconds later it reaches the top of the hill moving at 10m/s.

marishachu [46]

Answer:

$5 m/s^{2}$

Explanation:

acceleration is defined as the rate of change of velocity per unit time. Therefore, acceleration, $a=\frac {\triangle v}{\triangle t}$

where $\triangle v$ is change in velocity and t is time

Substituting initial and final velocities with 25 m/s and 10 m/s then using time as given in the question of 3 s then

$a=\frac {25-10}{3}=\frac {15}{3}=5 m/s^{2}$

Therefore, the acceleration is $5 m/s^{2}$

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2 years ago

What's an easy way to create an interference pattern of waves?

igor_vitrenko [27]

Answer:

Explanation:

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2 years ago

You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30

12345 [234]

Answer:

r₂ = 0.2 m

Explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

distance moved closer = ?

new frequency = 80 dB

using formula

$\beta = 10 log(\dfrac{I_1}{I_0})$

I₀ = 10⁻¹² W/m²

now,

$30 = 10 log(\dfrac{I_1}{10^{-12}})$

$\dfrac{I_1}{10^{-12}}= 10^3$

$I_1= 10^{-8}\ W/m^2$

to hear the whisper sound = 80 dB

$80 = 10 log(\dfrac{I_2}{10^{-12}})$

$\dfrac{I_2}{10^{-12}}= 10^8$

$I_2= 10^{-4}\ W/m^2$

we know intensity of sound is inversely proportional to square of distances

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}$

$\dfrac{10^{-8}}{10^{-4}}=\dfrac{r_2^2}{20^2}$

$10^{-4}=\dfrac{r_2^2}{20^2}$

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6 0

2 years ago

The High Speed Industrial Drill With Diameter Of 98 Cm Develops 5.85hp At 1900 Rpm. What Torque And Force Is Applied To The Dril

STatiana [176]

Answer:

The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.

Explanation:

Given:-

- The diameter of the drill bit, d = 98 cm

- The power at which drill works, P = 5.85 hp

- The rotational speed of drill, N = 1900 rpm

Find:-

What Torque And Force Is Applied To The Drill Bit?

Solution:-

- The amount of torque (T) generated at the periphery of the cutting edges of the drilling bit when it is driven at a power of (P) horsepower at some rotational speed (N).

- The relation between these quantities is given:

T = 5252*P / N

T = 5252*5.85 / 1900

T = 16.171 Nm

- The force (F) applied at the periphery of the drill bit cutting edge at a distance of radius from the center of drill bit can be determined from the definition of Torque (T) being a cross product of the Force (F) and a moment arm (r):

T = F*r

Where, r = d / 2

F = 2T / d

F = 2*16.171 / 0.98

F = 33 N

Answer: The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.

4 0

2 years ago

Read 2 more answers

A particle moves according to the equation x = 11t^2, where x is in meters and t is in seconds.

Savatey [412]

We are given the equation:

x = 11t^2

We use that equation to calculate for the distance traveled.
For (a)

At t=2.20 sec,
x =53.24 meters

At t=2.95 sec,
x =95.73 meters

Velocity = (95.73 meters - 53.24 meters) / (2.95 s - 2.20 s ) = 56.65 m/s

For (b)

At t=2.20 sec,
x =53.24 meters

At t=2.40 sec,
x =63.36 meters

Velocity = (63.36 meters - 53.24 meters) / (2.40 s - 2.20 s ) = 50.6 m/s

4 0

2 years ago