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2 years ago

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An electron and a positron are located 17 m away from each other and held fixed by some mechanism. The positron has the same mas

s and the same magnitude of charge as those of the electron, but its charge is positive. The electron and the positron are released at the same time by the mechanism. The electron and the positron begin to speed up towards each other. What velocities should they have when they are 1.3 m away from each other?

1 answer:

mariarad [96]2 years ago

6 0

Answer:

Velocity will be 13.9 m/s when they are 1.3 m away from each other.

Explanation:

Detailed steps are attached below.

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From the edge of a roof you throw a snowball downward that strikes the ground with 100J of kinetic energy. then you throw a seco

Vanyuwa [196]

Answer:

The second snowball hits the ground with a kinetic energy of 100 Joules

Explanation:

Given that,

From the edge of a roof you throw a snowball downward that strikes the ground with 100 J of kinetic energy. It is a case of conservation of energy.

At the highest point,

$mgh+\dfrac{1}{2}mu^2=mgh'+0$

$100=mgh'$

At lowest point,

$mgh'=K$

From above two equation, we get :

Kinetic energy, K = 100 J

So, the second snowball hits the ground with a kinetic energy of 100 Joules. So, the correct option is (A).

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2 years ago

The thrust of a certain boat’s engine generates a power of 10kW as the boat moves at constant speed 10ms through the water of a

Lunna [17]

Answer:

The change in power is 4400 W.

Explanation:

Given that,

Power = 10 kW

Speed = 10 m/s

Increases speed = 12 m/s

Given equation is,

$F=kv$

We know that,

The power is,

$P=Fv$

Put the value of F into the formula

$P=(kv)v$

$P=kv^2$

$P\propto v^2$

We need to calculate the new power

Using formula for power

$\dfrac{P}{P'}=\dfrac{v^2}{v'^2}$

Put the value into the formula

$\dfrac{10}{P'}=(\dfrac{10}{12})^2$

$P'=(\dfrac{12}{10})^2\times10$

$P'=14.4\ kW$

We need to calculate the change in power

Using formula of change in power

$\Delta P=P'-P$

Put the value into the formula

$\Delta P=14.4-10$

$\Delta P=4.4\ kW$

$\Delta P=4.4\times1000$

$\Delta P=4400\ W$

Hence, The change in power is 4400 W.

6 0

3 years ago

In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of you

olchik [2.2K]

PART A)

horizontal distance that will be moved = 14 m

Height of the fence = 5.0 m

height from which it is thrown = 1.60 m

angle of projection = 54 degree

So here we can say that stone will travel vertically up by distance

$\Delta y = 5 - 1.6 = 3.40 m$

now we will have displacement in horizontal direction

$\Delta x = 14 m$

now we know that

$v_x = vcos54$

$v_y = vsin54$

now we will have

$\Delta x = v_x t$

$14 = (vcos54)t$

also for y direction

$\Delta y = v_y t + \frac{1}{2}at^2$

$3.40 = (vsin54)t - \frac{1}{2}(9.8) t^2$

now from the two equations we will have

$3.40 = (vsin54)(\frac{14}{vcos54}) - 4.9 t^2$

$3.40 = 14 tan54 - 4.9 t^2$

$3.40 = 19.3 - 4.9 t^2$

$t = 1.8 s$

now from above equations

$14 = vcos54 (1.8)$

$v = 13.2 m/s$

So the minimum speed will be 13.2 m/s

Part B)

Total time of the motion after which it will land on the ground will be "t"

so its vertical displacement will be

$\Delta y = -1.60 m$

now we will have

$-1.60 = v_y t + \frac{1}{2}at^2$

$-1.60 = (13.2sin54)t - \frac{1}{2}(9.8)t^2$

$4.9 t^2 - 10.7t - 1.60 = 0$

$t = 2.3 s$

Now the time after which it will reach the fence will be t1 = 1.8 s

so total time after which it will fall on other side of fence

$t_2 = t - t_1$

$t_2 = 2.3 - 1.8 = 0.5 s$

now the displacement on the other side is given as

$\Delta x = (vcos54) t_2$

$\Delta x = (13.2 cos54)(0.5)$

$\Delta x = 3.88 m$

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2 years ago

A fisherman is fishing from a bridge and is using a "44.0-N test line." In other words, the line will sustain a maximum force of

vitfil [10]

Answer:

a) 4.485 kg b) 3.94 kg

Explanation:

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44 = m * 9.81m/s^2

m = 44/9.81 = 4.485kg

b) F(tension) = ma + mg ( where a is the acceleration of the body and g is the acceleration of the gravity)

44 = m (a +g)

44 = m (1.37 + 9.81)

44/11.18 = m

m = 3.94 kg

3 0

2 years ago

As a descending elevator approaches the correct floor, it slows to a stop with a constant acceleration of magnitude 1m/s*2. The

Nookie1986 [14]

Answer:

C

Explanation:

4 0

2 years ago