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MrRa [10]
2 years ago
15

Romeo lanza suavemente guijarros a la ventana de julieta y quiere que los guijarros golpeen la ventana solo con con un component

e horizontal de velocidad el esta parado en el extremo de un jardin de rosas 4.5m por abajo de la ventana y a 5.0m de la base de la pared cual es la rapidez de los guijarros cuando golpean la ventana
Physics
1 answer:
Yuki888 [10]2 years ago
6 0

Answer:

5.219\,\frac{m}{s}

Explanation:

Las condiciones del problema requieren el cálculo de la rapidez inicial de los guijarros. Se sabe que el componente vertical de la rapidez final es cero. Por tanto, el tiempo se determina a continuación: (The conditions of this problems require the calculation of the initial speed of the peebles. It is known that vertical component of the final speed is zero. Therefore, the time is determined herein:).

(0\,\frac{m}{s})^{2} = v_{o,y}^{2} - 2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.5\,m)

v_{o,y} = 9.395\,\frac{m}{s}

0\,\frac{m}{s} = 9.395\,\frac{m}{s} - \left(9.807\,\frac{m}{s^{2}} \right)\cdot \Delta t

\Delta t = 0.958\,s

Además, se determina el componente horizontal de la rapidez inicial (Likewise, the horizontal component of the initial speed is determined):

v_{o,x} = \frac{5\,m}{0.958\,s}

v_{o,x} = 5.219\,\frac{m}{s}

El guijarro tiene una rapidez de 5.219\,\frac{m}{s} cuando golpea la ventana (The peeble has a speed of  5.219\,\frac{m}{s} when it hits the window).

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jeyben [28]

Answer:

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It now becomes necessary to compute the diffusion coefficient at 750°C (1023 K) given that D0= 8.5 ×10-5m2/s and Qd= 202,100 J/mol.

we have D= D0exp( -Qd/RT)

=(8.5×105m2/s)exp(-202,100/8.31×1023)

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4 0
2 years ago
A student attaches a block to a vertical spring so that the block-spring system will oscillate if the block-spring system is rel
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Answer:

Time period of the motion will remain the same while the amplitude of the motion will change

Explanation:

As we know that time period of oscillation of spring block system is given as

T= 2\pi\sqrt{\frac{M}{k}}

now we know that

M = mass of the object

k = spring constant

So here we know that the time period is independent of the gravity

while the maximum displacement of the spring from its mean position will depends on the gravity as

mg = kx

x = \frac{mg}{k}

so we can say that

Time period of the motion will remain the same while the amplitude of the motion will change

4 0
2 years ago
A 10. g cube of copper at a temperature T1 is placed in an insulated cup containing 10. g of water at a temperature T2. If T1 &g
Anna35 [415]

Answer:

a. The temperature of the copper changed more than the temperature of the water.

Explanation:

Because we're only considering the isolated system cube-water, the heat of the system should be constant, that implies the heat the cube loses is equal the heat the water gains (because by zero law of thermodynamics heat (Q) flows from hot body to cold body until reach thermal equilibrium and T1>T2). So:

Q_{cube}=Q_{water} (1)

But Q is related with mass (m), specific heat (c) and changes in temperature (\varDelta T)in the next way:

Q=cm\varDelta T(2)

Using (2) on (1):

c_{cooper}*m_{cooper}*\varDelta T_{cooper}=c_{water}*m_{waterer}*\varDelta T_{water}

(10g)(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(10g)(4.186 \frac{J}{g\,C})(\varDelta T_{water})

(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(4.186 \frac{J}{g\,C})(\varDelta T_{water})

Because we have an equality and 0.385 < 4.186 then \varDelta T_{cooper}>\varDelta T_{waterer} to conserve the equality

4 0
2 years ago
12*8A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 13 feet. The ball is started in
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Answer:

See attached pictures.

Explanation:

See attachments for explanation.

6 0
2 years ago
In a 1.25-T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50μC and initially moving north
Goshia [24]

Answer:

a) The sign of the charge is positive.

b) The magnetic force on the particle is 0.050 newtons.

Explanation:

The magnetic force F on a moving charge with velocity v passing through a magnetic field B is:

\overrightarrow{F}=q\overrightarrow{v}\times\overrightarrow{B}(1)

a)

Because it is a cross product, we can find the direction of the force using the right-hand rule, that is too the direction of the movement. We have two possibilities here because the velocity vector and magnetic field are perpendicular: the particle deflects towards east or toward west, which depends on the charge of the particle. Note that if you put your right hand fingers, except thumb, pointing towards north (direction of velocity) and later close them in the direction of the magnetic field, if you maintain your thumb perpendicular to this movement it will point towards east (See figure), so that will be de direction of the force if the charge is positive, but if the charge is negative, the direction will be opposite (towards west). So the charge has to be positive to deflects towards east.

b)

Now by 1:

F=qvB\sin\theta=(8.50\times10^{-6})(4750)(1.25)\sin90\simeq\mathbf{0.050\,N}

5 0
2 years ago
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