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Cloud [144]
3 years ago
12

A "sun yacht" is a spacecraft with a large sail that is pushed by sunlight. Although such a push is tiny in everyday circumstanc

es, it can be large enough to send the spacecraft outward from the Sun on a cost-free but slow trip. Suppose that the spacecraft has a mass of 900 kg and receives a push of 20 N. (a) What is the magnitude of the resulting acceleration? If the craft starts from rest,(b) how far will it travel in 1 day and (c) how fast will it then be moving?
Physics
1 answer:
guajiro [1.7K]3 years ago
8 0

Answer:

acceleration = 0.022 m/s^2

distance = 8.3 x 10^7 m

speed = 1.9 x 10 ^3 m/s

Explanation:

the parameters given are:

mass = 900kg

force = 20N

  • from the formula force = mass x acceleration

acceleration = force / mass

acceleration = 20 / 900

acceleration = 0.022 m/s^2

  • distance travelled in 1 day (86,400 seconds) = (1/2) x a x t^2

   (1/2) x 0.022 x (86,400^2) = 8.3 x 10^7 m

  • speed of the sun yatch (v) = a x t

0.22 x 86400 = 1.9 x 10 ^3 m/s

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Compute the velocity of an electron that has been accelerated through a difference of potential of 100 volts. express your answe
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The velocity of an electron that has been accelerated through a difference of potential of 100 volts will be 5.93 * 10^{6} m/s

Electrons move because they get pushed by some external force. There are several energy sources that can force electrons to move. Voltage is the amount of push or pressure that is being applied to the electrons.

By conservation of energy, the kinetic energy has to equal the change in potential energy, so KE=q*V. The energy of the electron in electron-volts is numerically the same as the voltage between the plates.

given

charge of electron = 1.6 × 10^{-19} C

mass of electron  = 9.1 × 10^{-31} kg

Force in an electric field = q*E

potential energy is stored in the form of work done

potential energy = work done = Force * displacement

                                                   = q * (E * d)  

                                                   = q * (V) = 1.6 × 10^{-19} * 100

stored potential energy = kinetic energy in electric field

kinetic energy = 1/2 * m * v^{2}

                        = 1/2 *  9.1 × 10^{-31} *  v^{2}

equation both the equations

1/2 *  9.1 × 10^{-31} *  v^{2} = 1.6 × 10^{-17}

v^{2} = 0.352 * 10^{14} m/s

v^{2} = 35.2 * 10^{12}

    = 5.93 * 10^{6} m/s

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A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh
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(a) Zero

When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.

This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy

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

where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:

U=mgh

where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

a=\frac{v-u}{t}

where

a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

v = 0 is the final velocity (at the highest point)

u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

u=v-at = 0 -(-9.8 m/s^2)(1 s)= +9.8 m/s

and the positive sign means it points upward.

(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where

v = 0 is the final velocity (at the highest point)

u = 9.8 m/s is the initial velocity

Substituting, we find

\Delta v = 0 - (+9.8 m/s)=-9.8 m/s

(d) 9.8 m/s downward

We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

a=\frac{v-u}{t}

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v  is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

t = 1 s is the time interval

Solving for v, we find

v = u+at = 0 +(-9.8 m/s^2)(1 s)= -9.8 m/s

and the negative sign means it points downward.

(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

\Delta v = -9.8 m/s - 0=-9.8 m/s

(f) -19.6 m/s

The change in velocity during the overall 2-s interval is given by

\Delta v = v -u

where in this case we have:

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s

(g) -9.8 m/s^2

There is always one force acting on the ball during the motion: the force of gravity, which is given by

F=mg

where

m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so

mg = ma

which means that the acceleration is

a= g = -9.8 m/s^2

and the negative sign means it points downward.

7 0
3 years ago
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