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2 years ago

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An object weighs 7.84 N when it is in air and 6.86 N when it is immersed in water of density 1000 kg/m3. What is the density of

the object?

1 answer:

kirill [66]2 years ago

3 0

Answer:

8000 kg/m^3

Explanation:

Weight in air = 7.84 n

Weight in water = 6.86 N

density of water = 1000 kg/m^3

Let d be the density of object

According to the Archimedes principle, when a body is immersed in a liquid partly or wholly, it experiences an upward force which is called buoyant force. The buoyant force is equal to the loss in weight of the body.

Loss in weight of the object = Weight of object in air - weight of object in water

Loss in weight = 7.84 - 6.86 = 0.98 N

Volume of body x density of water x g = 0.98

Let V be the volume of body

V x 1000 x 9.8 = 0.98

V = 10^-4 m^3

Weight in air = Volume of body x density of body x g

7.84 = 10^-4 x d x 9.8

d = 8000 kg/m^3

Thus, the density of body is 8000 kg/m^3.

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kirill [66]

Answer:

U = 1 / r²

Explanation:

In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related

F = - dU / dr

this derivative is a gradient, that is, a directional derivative, so we must have

dU = - F. dr

the esxresion for strength is

F = B / r³

let's replace

∫ dU = - ∫ B / r³ dr

in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product

let's evaluate the integrals

U - Uo = -B (- / 2r² + 1 / 2r₀²)

To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)

U = B / 2r²

we substitute the value of B = 2

U = 1 / r²

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2 years ago

A 2.0-m-tall man is 5.0 m from the converging lens of a camera. His image appears on a detector that is 50 mm behind the lens. H

ladessa [460]

Answer:

20 cm

Explanation:

We can solve the problem by using the magnification equation:

$M=\frac{h_i}{h_o}=-\frac{q}{p}$

where

$h_i$ is the size of the image

$h_o = 2.0 m$ is the height of the real object (the man)

$q=50 mm =0.050 m$ is the distance of the image from the lens

$p = 5.0 m$ is the distance of the object (the man) from the lens

Solving the formula for $h_i$ , we find

$h_i = -\frac{q}{p}h_o=-\frac{0.050 m}{5.0 m}(2.0 m)=-0.02 m = -20 cm$

And the negative sign means the image is inverted.

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2 years ago

The gravity tractor is a proposed spacecraft that will fly close to an asteroid whose trajectory threatens to impact the Earth.

Talja [164]

Answer:

$F_g=461lb_f$

Explanation:

First calculate the mass of the asteroid. To do so, you need to find the volume and know the density of iron.

If r = d/2 = 645ft, then:

$V = \frac{4}{3} \pi r^3$

$V = 1.124*10^{9}ft^3$

So

$\delta_{iron}=m/V=491lb/ft^3$

$m=V*\delta=5.519*10^{11}lb$

Once you know both masses, you can calculate the force using Newton's universal law of gravitation:

$F_g=G\frac{m_1m_2}{d^2}$

Where G is the gravitational constant:

$G= 1.068846 * 10^{-9} ft^3 lb^{-1} s^{-2}$

$F_g=461lb_f$

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2 years ago

A Porsche challenges a Honda to a 200-m race.Because the Porsche's acceleration of 3.5 m/s2 is larger than the Honda's 3.0 m/s2,

Blizzard [7]

Answer:

Honda won by 0.14 s

Explanation:

We are given that

Distance =S=200 m

Initial velocity of Honda=u=0m/s

Initial velocity of Porsche=u'=0m/s

Acceleration of Honda= $3.0m/s^2$

Acceleration of Porsche's= $3.5m/s^2$

Time taken by Honda to start=1 s

$s=ut+\frac{1}{2}at^2$

Substitute the values

$200=0(t)+\frac{1}{2}(3)t^2$

$200=\frac{3}{2}t^2$

$t^2=\frac{200\times 2}{3}=\frac{400}{3}$

$t=\sqrt{\frac{400}{3}}=11.55s$

Time taken by Honda=11.55 s

Now, time taken by Porsche

$200=\frac{1}{2}(3.5)t^2$

$t^2=\frac{200\times 2}{3.5}$

$t=\sqrt{\frac{400}{3.5}}=10.69 s$

Total time taken by Porsche=10.69+1=11.69 s

Because it start 1 s late

Time taken by Honda is less than Porsche .Therefore, Honda won and

Time =11.69-11.55=0.14 s

Honda won by 0.14 s

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2 years ago

When jumping, a flea accelerates at an astounding 1000 m/s2 but over the very short distance of 0.50 mm. If a flea jumps straigh

Nadusha1986 [10]

Answer:

The flea reaches a height of 51 mm.

Explanation:

Hi there!

The equations of height and velocity of the flea are the following:

During the jump:

h = h0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

While in free fall:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the flea at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration of the flea due to the jump.

v = velocity of the flea at time t.

g = acceleration due to gravity.

First, let's calculate how much time it takes the flea to reach a height of 0.0005 m. With that time, we can calculate the speed reached by the flea during the jump:

h = h0 + v0 · t + 1/2 · a · t²

If we place the origin of the frame of reference on the ground, then, h0 = 0. Since the flea is initially at rest, v0 = 0. Then:

h = 1/2 · a · t²

We have to find the value of t for which h = 0.0005 m:

0.0005 m = 1/2 · 1000 m/s² · t²

0.0005 m / 500 m/s² = t²

t = 0.001 s

Now, let's find the velocity reached in that time:

v = v0 + a · t (v0 = 0)

v = a · t

v = 1000 m/s² · 0.001 s

v = 1.00 m/s

When the flea is at a height of 0.50 mm, its velocity is 1.00 m/s. This initial velocity will start to decrease due to the downward acceleration of gravity. When the velocity is zero, the flea will be at the maximum height. Using the equation of velocity, let's find the time at which the flea is at the maximum height (v = 0):

v = v0 + g · t

At the maximum height, v = 0:

0 m/s = 1.00 m/s - 9.81 m/s² · t

-1.00 m/s / -9.81 m/s² = t

t = 0.102 s

Now, let's find the height reached by the flea in that time:

h = h0 + v0 · t + 1/2 · g · t²

h = 0.0005 m + 1.00 m/s · 0.102 s - 1/2 · 9.81 m/s² · (0.102 s)²

h = 0.051 m

The flea reaches a height of 51 mm.

5 0

2 years ago