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kolbaska11 [484]

3 years ago

11

If a sled has at the top of 10 m hill had 1000 J of potential energy what would happen to the PE if the sled were to moved to a

5 m hill?

1 answer:

nataly862011 [7]3 years ago

7 0

Answer:

500J

Explanation:

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there are a wats group where you can get your answer by trusted experts.

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A rocket firing its engine and accelerating in outer space (no gravity, no air resistance) suddenly runs out of fuel. Which best

notsponge [240]

the correct choice is

D. It immediately stops accelerating, and continues moving at the velocity it had when burnout occurred.

as soon as the rocket runs out of fuel, its engine stops firing gases which is mainly responsible for the acceleration of the rocket. hence the rocket stops accelerating. also, since there is no gravity or air resistance , there is no net force on the rocket after it runs out of the fuel. hence

a = acceleration = 0

we know that , at no acceleration , object moves at constant velocity. hence the rocket keeps moving at the velocity it had when burnout occured.

4 0

4 years ago

What tripositive ion has the electron configuration [kr] 4d3 ? what neutral atom has the electron configuration [kr] 5s 2 4d2 ?

tigry1 [53]

(a)

Electronic configuration is given as follows:

$[Kr]4d^{3}$

Since, this is the electronic configuration of ion with+3 that means 3 electrons are removed. On adding the 3 electrons, the electronic configuration of neutral atom can be obtained.

Thus, electronic configuration of neutral atom is $[Kr]4d^{5}5s^{1}$ .

The atomic number of Kr is 36, thus, total number of electrons become 36+6=42.

This corresponds to element: molybdenum. Thus, the tripositive atom will be $Mo^{3+}$ .

(b) The given electronic configuration is $[Kr]5s^{2}4d^{2}$ .

The atomic number of Kr is 36, thus, total number of electrons become 36+4=40.

This corresponds to element zirconium, represented by symbol Zr.

6 0

3 years ago

Points a, b, and c are at the corners of an equilateral triangle of side 1 m. equal positive charges of 5 c are at a and

soldier1979 [14.2K]

Solution:

we know that the potential of any point is given by

v= kq/r^2............(1)

According to the question we have point charges is applied only two point so we must multiply the above equation by 2.

Then we get

V=2kq/r^2 [since charge is unity].

6 0

3 years ago

A mass of 0.273 kg is placed on an incline of 38.382 degrees. It is attached by a rope over a pulley to a mass of 0.254 kg. Find

Natali5045456 [20]

here tension in the string is counter balanced by weight of block of mass m1

so we can say

$T = m_1g$

$T = 0.254 \times 9.8 = 2.49 N$

now on the other side the block which is placed on the inclined plane

we can say that component of weight of the block and friction force is counter balanced by tension force

$m_2g sin\theta + F_f = T$

now we can plug in all values to find the friction force

$0.273 \times 9.8 sin38.382 + F_f = 2.49$

$1.66 + F_f = 2.49$

$F_f = 0.83 N$

so it will have 0.83 N force on it due to friction

now to find the friction coefficient

$F_f = \mu \times F_n$

here we know that

$F_n = m_2gcos\theta$

$F_n = 0.273 \times 9.8 \times cos38.382 = 2.1 N$

now from above equation

$0.83 = \mu \times 2.1$

$\mu = 0.38$

so friction coefficient will be 0.38

8 0

3 years ago

4.72 A full-wave bridge-rectifier circuit with a 1-k load operates from a 120-V (rms) 60-Hz household supply through a 12-to-1 s

melisa1 [442]

Answer:

a) 12.74 V

b) Two pairs of diode will work only half of the cycle

c) 8.11 V

d) 8.11 mA

Explanation:

The voltage after the transformer is relationated with the transformer relationshinp:

$V_o=Vrms*\frac{1}{12}\\V_o=10Vrms$

the peak voltage before the bridge rectifier is given by:

$V_{op}=Vo*\sqrt{2}\\V_{op}=14.14V$

The diodes drop 0.7v, when we use a bridge rectifier only two diodes are working when the signal is positive and the other two when it's negative, so the peak voltage of the load is:

$V_l=V_{op}-2(0.7)\\V_l=12.74V$

As we said before only two diodes will work at a time, because the signal is half positive and half negative,so two of them will work only half of the cycle.

The averague voltage on a full wave rectifier is given by:

$V_{avg}=2*\frac{V_l}{\pi}\\V_{avg}=8.11V$

Using Ohm's law:

$I_{avg}=\frac{V_{avg}}{R}\\\\I_{avg}=8.11mA$

7 0

3 years ago