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Stolb23 [73]
3 years ago
13

Si pudieras viajar a la Luna:

Physics
1 answer:
Nostrana [21]3 years ago
4 0

Answer:

i) Distancia, ii) La cinta métrica es impracticable.

Explanation:

i) El concepto físico que se construye únicamente del punto de salida y el punto de llegada a la Luna es el concepto de desplazamiento, definido como la distancia en línea recta de un punto en el espacio con respecto a un punto de referencia (la Tierra en este caso).

La distancia puede involucrar trayectorias curvilíneas entre los puntos mencionados.

ii) Por último, el uso de una cinta métrica es impracticable debido a la cantidad de material a utilizar y los efectos gravitacionales, electromagnéticos y mecánicos que inducen a una deflexión o una ruptura de esa cinta debido a la magnitud de la distancia entre las superficies del planeta y el satélite, respectivamente.

En este caso, es mejor utilizar la medición con tecnología láser, basadas en el fenómeno del electromagnetismo.

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You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. (a) what fraction of its initial energy is los
Scilla [17]

a) At a position of 2.0m, the Initial energy is all made up of the potential energy=m*g*hi<span>
and meanwhile at 1.5 all its energy is also potential energy=m*g*hf 

The percentage of energy remaining is E=m*g*hi/m*g*hf x 100 

and since mass and gravity are constant so it leaves us with just E=hi/hf 
which 1.5/2.0 x100= 75% so we see that we lost 25% of the energy or 0.25 in fraction 

b) Here use the equation vf^2=vi^2+2gd 

<span>where g is gravity, vf is the final velocity and vi is the initial velocity while d is the distance travelled

so in here we are looking for the vi so let us isolate that variable 
we know that at maximum height or peak, the velocity is 0 so vf is 0 

therefore,</span></span>

vi =sqrt(-2gd) <span>
vi =sqrt(-2x-9.81x1.5) </span>
<span>vi =5.4 m/s

<span>c) The energy was converted to heat due to friction with the air and the ground.</span></span>

6 0
3 years ago
The ___ energy in a mechanical system is determined by adding the potential and kinetic enters together
Svetllana [295]

Answer:

A, total.

<em>The </em><em>total</em><em> energy in a mechanical system is determined by adding the potential and kinetic enters together.</em>

<em />

<u><em>i hope this helped at all.</em></u>

<em />

4 0
3 years ago
a runner covers the last straigjt stretch of a race in 4 s. during that time, he speeds up from 5m/s to 9m/s.
PtichkaEL [24]

During that final period of time,
his acceleration is
                                (9 m/s - 5 m/s) / (4 sec) = 1 m/s² .

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8 0
3 years ago
Find the mass and center of mass of the solid E with the given density function ρ. E lies under the plane z = 3 + x + y and abov
makvit [3.9K]

Answer:

The mass of the solid is 16 units.

The center of mass of the solid lies at (0.6875, 0.3542, 2.021)

Work:

Density function: ρ(x, y, z) = 8

x-bounds: [0, 1], y-bounds: [0, x], z-bounds: [0, x+y+3]

The mass M of the solid is given by:

M = ∫∫∫ρ(dV) = ∫∫∫ρ(dx)(dy)(dz) = ∫∫∫8(dx)(dy)(dz)

First integrate with respect to z:

∫∫8z(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8x+8y+24](dx)(dy)

Then integrate with respect to y:

∫[8xy+4y²+24y]dx, evaluate y from 0 to x

= ∫[8x²+4x²+24x]dx

Finally integrate with respect to x:

[8x³/3+4x³/3+12x²], evaluate x from 0 to 1

= 8/3+4/3+12

= 16

The mass of the solid is 16 units.

Now we have to find the center of mass of the solid which requires calculating the center of mass in the x, y, and z dimensions.

The z-coordinate of the center of mass Z is given by:

Z = (1/M)∫∫∫ρz(dV) = (1/16)∫∫∫8z(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫4z²(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[4(x+y+3)²](dx)(dy)

= ∫∫[4x²+24x+8xy+4y²+24y+36](dx)(dy)

Then integrate with respect to y:

∫[4x²y+24xy+4xy²+4y³/3+12y²+36y]dx, evaluate y from 0 to x

= ∫[28x³/3+36x²+36x]dx

Finally integrate with respect to x:

[7x⁴/3+12x³+18x²], evaluate x from 0 to 1

= 7/3+12+18

Z = (7/3+12+18)/16 = <u>2.021</u>

The y-coordinate of the center of mass Y is given by:

Y = (1/M)∫∫∫ρy(dV) = (1/16)∫∫∫8y(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫8yz(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8xy+8y²+24y](dx)(dy)

Then integrate with respect to y:

∫[4xy²+8y³/3+12y²]dx, evaluate y from 0 to x

= ∫[20x³/3+12x²]dx

Finally integrate with respect to x:

[5x⁴/3+4x³], evaluate x from 0 to 1

= 5/3+4

Y = (5/3+4)/16 = <u>0.3542</u>

<u />

The x-coordinate of the center of mass X is given by:

X = (1/M)∫∫∫ρx(dV) = (1/16)∫∫∫8x(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫8xz(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8x²+8xy+24x](dx)(dy)

Then integrate with respect to y:

∫[8x²y+4xy²+24xy]dx, evaluate y from 0 to x

= ∫[12x³+24x²]dx

Finally integrate with respect to x:

[3x⁴+8x³], evaluate x from 0 to 1

= 3+8 = 11

X = 11/16 = <u>0.6875</u>

<u />

The center of mass of the solid lies at (0.6875, 0.3542, 2.021)

4 0
3 years ago
A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule
ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

p_i = 0

The final total momentum is instead:

p_f = m_a v_a + m_c v_c

where

m_a = 125 kg is the mass of the astronaut

v_a = 2.50 m/s is the velocity of the astronaut

m_c = 1900 kg is the mass of the capsule

v_c is the velocity of the capsule

Since the total momentum must be conserved, we have

p_i = p_f = 0

so

m_a v_a + m_c v_c=0

Solving the equation for v_c, we find

v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

F \Delta t = \Delta p

The change in momentum of the astronaut is

\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s

And the duration of the push is

\Delta t = 0.600 s

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J

3 0
3 years ago
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