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Ad libitum [116K]
3 years ago
13

Tenemos un muelle sobre el que se ejerce una fuerza de 20 N produciéndose una deformación de 5 cm. Determinar: a) la constante r

ecuperadora del muelle b) la deformación que sufriría si se le aplicara una fuerza de 50 N- c) la fuerza aplicada para que se deformara 2 cm
Physics
1 answer:
zzz [600]3 years ago
4 0

Answer:

I have no clue on what to do I am trying my hardest to figure out what to do but this is frustrating I do not know I am trying I'm trying I'm trying I'm trying I'm trying but does it supposed to rain the advocacy try to answer questions have a question on top of questions what I'm trying to do what I can sorry I do not have an answer for you if I could I would give you an answer but I do not have an answer so I do not know hopefully you can find somebody they can figure out the answer for you but I do not know the answer sorry

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How much work does a student (m = 60 kg) do when he Climb a tower 80 m high?
gtnhenbr [62]

Answer:

W = 47040  J

Explanation:

Given that,

The mass of a student, m = 60 kg

Height of the tower, h = 80 m

We need to find the work done in climbing the tower. The work done is given by :

W = mgh

So,

W = 60 × 9.8 × 80

W = 47040  J

So, the required work done is 47040  J.

7 0
3 years ago
A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters
Artist 52 [7]

Answer:

Part a)

F = 7.76 N

Part b)

\theta = -137.7 degree

Part c)

\theta = -127.7 degree

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

x = -19 + t - 3t^3

y = 20 + 7t - 9t^2

Part a)

differentiate x and y two times with respect to time to find the acceleration

a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)

a_x = \frac{d}{dt}(0 +1 - 9t^2)

a_x = -18t

a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)

a_y = \frac{d}{dt}(0 +7 - 18t)

a_y = -18

Now the acceleration of the object is given as

\vec a = (-18t)\hat i + (-18)\hat j

at t= 1.1 s we have

\vec a = -19.8 \hat i - 18 \hat j

now the net force of the object is given as

\vec F = m\vec a

\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)

\vec F = -5.74 \hat i - 5.22 \hat j

now magnitude of the force will be

F = \sqrt{5.74^2 + 5.22^2} = 7.76 N

Part b)

Direction of the force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{-5.22}{-5.74}

\theta = -137.7 degree

Part c)

For velocity of the particle we have

v_x = \frac{dx}[dt}

v_x = (0 +1 - 9t^2)

v_y = \frac{dy}{dt}

v_y = (0 +7 - 18t)

now at t = 1.1 s

\vec v = -9.89\hat i - 12.8 \hat j

now the direction of the velocity is given as

\theta = tan^{-1}(\frac{v_y}{v_x})

\theta = tan^{-1}(\frac{-12.8}{-9.89})

\theta = -127.7 degree

7 0
3 years ago
A 0.50 kg toy is attached to the end of a 1.0 m very light string. The toy is whirled in a horizontal circular path on a frictio
xenn [34]

Answer:

The maximum speed will be 26.475 m/sec

Explanation:

We have given mass of the toy m = 0.50 kg

radius of the light string r = 1 m

Tension on the string T = 350 N

We have to find the maximum speed without breaking the string  

For without breaking the string tension must be equal to the centripetal force

So T=\frac{mv^2}{r}

So 350=\frac{0.5\times v^2}{1}

v^2=700

v = 26.475 m /sec

So the maximum speed will be 26.475 m/sec

6 0
3 years ago
A scooter is accelerated from rest at the rate of 8m/s
algol13

Explanation:

time=Distance/speed

t=32/8

t=4 seconds

8 0
3 years ago
If a 20 g cannonball is shot from a 5 kg cannon with a velocity of 100
balu736 [363]

Strange as it may seem, the statement in the question appears to be <em>TRUE</em>.  

-- Before the shot, neither the cannon nor the ball is moving, so their combined momentum is zero.  

-- Since momentum is conserved, we know immediately that their combined momentum AFTER the shot also has to be zero.

-- (20g is rather puny for a "cannonball" ... about the same weight as four nickels. But we'll take your word for it and just do the Math and the Physics.)

-- Momentum = (mass) x (velocity)

After the shot, the momentum of the cannonball is

(0.02 kg) x (100 m/s ==> that way)

Momentum of the ball = 2 kg-m/s ==> that way.

-- In order for both of them to add up to zero, the momentum of the cannon must be (2 kg-m/s this way <==) .

Momentum of cannon = (5 kg) x (V m/s this way <==)

2 kg-m/s this way <== = (5 kg) x (V m/s this way <==)

Divide each side by (5 kg):

V m/s  =  (2/5) m/s this way <==

Speed of recoil of the cannon = <em>-- 0.4 m/s</em>

3 0
3 years ago
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