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iVinArrow [24]
3 years ago
14

El muelle de un dinamómetro se alarga 6cm cuando aplicamos sobre él una fuerza de 10 N. Calcula el alargamiento del muelle al ap

licar una fuerza de 20 N.
Physics
1 answer:
Crank3 years ago
6 0

Answer:

x = 12cm

Explanation:

First, with the values given for the force and the displacement, you calculate the spring-constant of the dynamometer by using the following equation (primero, con los valores dados de la fuerza y el desplazamiento, calculas la constante del resorte aplicando la siguiente formula):

F=kx

F: force (fuerza) = 10N

k: spring constant (constante del resorte)

x: displacement of the spring (desplazamiento del resorte) = 6cm

by using these values you calculate k (usando estos valores calculas k):

k=\frac{F}{x}=\frac{10N}{6cm}=1.66\frac{N}{cm}

With this value you calculate the displacement for a force of 20N (con este valor de k calculas el desplazamiento del resorte para una fuerza de 20N):

x=\frac{F}{k}=\frac{20}{1.66N/cm}=12cm

hence, the displacement is 12cm

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Answer:

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Explanation:

Given request solutions

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3 years ago
A 210 Ohm resistor uses 9.28 W of
IgorLugansk [536]
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Current=I

\boxed{\sf P=I^2R}

\\ \sf\longmapsto I^2=\dfrac{P}{R}

\\ \sf\longmapsto I^2=\dfrac{9.28}{210}

\\ \sf\longmapsto I^2\approx0.04

\\ \sf\longmapsto I\approx\sqrt{0.04}

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4 0
3 years ago
In major league baseball, the pitcher's mound is 60 feet from the batter. If a pitcher throws a 85 mph fastball, how much time e
Nostrana [21]
The given from your problem are the following:
V = 85mph (This is miles per hour)
d = 60 feet

If you notice the units do not match. Before we can do anything else, we need to make the figures match. 

In this case, we will convert 85miles per hour to feet per hour.  There are 5,280 feet in 1 mile. 
\frac{85miles }{hr} x \frac{5,280feet}{1miles} = \frac{448,800feet}{hr}

But wait! If you think about the scenario, you are looking for how long it will take for the ball to reach the bat. The most applicable unit of time to use here is second. It would be very hard to really measure a short and instantaneous event in hours. So we convert it into feet per second: 

There are 3,600 seconds in 1 hour.

\frac{448,800feet }{hour} x \frac{1hour}{3,600seconds} = \frac{448,800feet}{3,600 seconds} = 124.67ft/s

So now we have our new given as:

v = 124.67ft/s
d = 60 ft

The formula for time can be derived from the formula from velocity, which is:
velocity = \frac{distance}{time}

The formula of time will then be:
time= \frac{distance}{velocity}

All you need to do is plug in what you know and solve for what you don't know. 

time= \frac{60feet}{124.67ft/s}

time= 0.48s

The answer then is 0.48s.

If you want this in hours, just divide the value in seconds by 3,600. The answer would then be 0.00013hr. (See how small it is? This is why seconds would be a more appropriate measure.)
8 0
3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

8 0
2 years ago
1. A car is slowing down at a constant rate from 40 m/s to come to a stop. The acceleration is - 5 m/s2. How
Oxana [17]

Answer:

           

Explanation:

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