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3 years ago

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A spring with a spring constant value of 2500 StartFraction N over m EndFraction is compressed 32 cm. A 1.5-kg rock is placed on

top of it, then the spring is released. Approximately how high will the rock rise? 9 m 17 m 27 m 85 m

2 answers:

PilotLPTM [1.2K]3 years ago

7 0

Answer:

9m

Explanation:

edge2o2o

Sedaia [141]3 years ago

5 0

Answer:

it's 9m

Explanation:

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a small asteroid of mass 125 kg is orbiting a planet that has a mass of 3.52x 10^13 what is the radial distance between the aste

asambeis [7]

Answer:

r = 2.031 x 10⁶ m = 2031 km

Explanation:

In order for the asteroid to orbit the planet, the centripetal force must be equal to the gravitational force between asteroid and planet:

Centripetal Force = Gravitational Force

mv²/r = GmM/r²

v² = GM/r

r = GM/v²

where,

r = radial distance = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Planet = 3.52 x 10¹³ kg

v = tangential speed = 0.034 m/s

Therefore,

r = (6.67 x 10⁻¹¹ N.m²/kg²)(3.52 x 10¹³ kg)/(0.034 m/s)²

<u>r = 2.031 x 10⁶ m = 2031 km</u>

7 0

3 years ago

Find τf, the torque about point p due to the force applied by the achilles' tendon.

Luba_88 [7]

The formula for the torque is
<span>τf = p F
where
</span><span>τf is the torque
p is the distance where the force is applied by the tendon
F is force applied by the tendon

If there are given values, substitute in the equation and solve for the torque.</span>

3 0

4 years ago

Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l

Masja [62]

Answer:

a) I2 = 3 (o-10) / (o- 30) , b) h ’/h= 3 (o-10) / o (o-30)

Explanation:

The builder's equation is

1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

1 / i1 = 1 / f - 1 / o

1 / i1 = 1/15 - 1 / o

i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

o2 = d-i1

We write the builder equation

1 / f2 = 1 / o2 + 1 / i2

1 / i2 = 1 / f2 -1 / o2

1 / i2 = 1 / f2 - 1 / (d-i1)

1 / i2 = 1/20 - 1 / (d-i1) (1)

Let's evaluate the last term

d-i1 = d - 15 o / (o-15)

d-i1 = (d (o-15) - 15 o) / (o-15)

d- i1 = (30 or -30 15 -15 o) / (o-15)

d-i1 = (15 or - 450) / (o- 15)

d-i1 = = (15 or -450) / (o-15)

replace in 1

1 / i2 = 1/20 - (or - 15) / (15 or -450)

1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

1 / i2 = (-5 or -150) / (15 or -150)

1 / i2 = (or -30) / (3 or - 30)

I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

m = h ’/ h = - i / o

h ’= - h i / o

In our case

h ’= h i2 / o

h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

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3 years ago

Your teacher gives you a piece of cardboard with two pinholes on it, saying that they are separated by 205 μm ± 3%. In order to

wlad13 [49]

Answer:

m,lkj,mkn,njn

Explanation:because she is telling you to do a project

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3 years ago

If 1495 j of heat is needed to raise the temperature of a 351 g sample of a metal from 55.0°c to 66.0°c, what is the specific he

forsale [732]

The amount of heat needed to increase the temperature of a substance by $\Delta T$ is given by
$Q= mC_s \Delta T$
where m is the mass of the substance, Cs is its specific heat capacity and $\Delta T$ is the increase of temperature.

If we re-arrange the formula, we get
$C_s = \frac{Q}{m \Delta T}$
And if we plug the data of the problem into the equation, we can find the specific heat capacity of the substance:
$C_s = \frac{1495 J}{(351 g)(66.0^{\circ}C-55.0^{\circ}C)}=0.39 J/g^{\circ}C$

6 0

3 years ago

Read 2 more answers