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3 years ago

5

Relating to Newton’s Third Law, action and reaction, what is the reaction when a rocket expels gas, smoke and flames from the no

zzle end of the rocket engine?

1 answer:

Cloud [144]3 years ago

7 0

Answer:

Propels in the opposite direction

Explanation:

You might be interested in

A charge -353e is uniformly distributed along a circular arc of radius 5.30 cm, which subtends an angle of 48°. What is the line

vladimir2022 [97]

Answer:

- 1.3 x 10⁻¹⁵ C/m

Explanation:

Q = Total charge on the circular arc = - 353 e = - 353 (1.6 x 10⁻¹⁹) C = - 564.8 x 10⁻¹⁹ C

r = Radius of the arc = 5.30 cm = 0.053 m

θ = Angle subtended by the arc = 48° deg = 48 x 0.0175 rad = 0.84 rad (Since 1 deg = 0.0175 rad)

L = length of the arc

length of the arc is given as

L = r θ

L = (0.053) (0.84)

L = 0.045 m

λ = Linear charge density

Linear charge density is given as

$\lambda =\frac{Q}{L}$

Inserting the values

$\lambda =\frac{-564.8\times 10^{-19}}{0.045}$

λ = - 1.3 x 10⁻¹⁵ C/m

4 0

3 years ago

A student sects a leaf of length 7.2 cm to draw. Her drawing is 28.8 cm in length. What is the magnification of the drawing?

Alina [70]

Answer:

A) x4

Explanation:

Magnification is equal to image size divided by the actual size, or M = I/A.

The image size is the student's drawing, which is 28.8 cm, and the actual size is 7.2 cm. Divide them, and cancel out the units, and you should get:

28.8 cm/7.2 cm = 4

7 0

3 years ago

A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a

UNO [17]

Answer:

$P_{avg} = 6.283*10^{-9} \ W$

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

$\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})$

where;

$E= \frac{d \phi}{dt}$

$E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t$

$E_{rms} = \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}$

Replacing our values into above equation; we have:

$E_{rms} = \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}$

$E_{rms} = \frac {8.98909588*10^{-4} }{\sqrt{2}}$

$E_{rms} = 6.356*10^{-4} \ V$

Then the $P_{avg$ is calculated as:

$P_{avg} = \frac{E^2}{R}$

$P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}$

$P_{avg} = 6.283*10^{-9} \ W$

6 0

4 years ago

A student sits at rest on a piano stool that can rotate without friction. The moment of inertia of the student-stool system is 4

irina [24]

Here We can use principle of angular momentum conservation

Here as we know boy + projected mass system has no external torque

Since there is no torque so we can say the angular momentum is conserved

$mvL = (I + mL^2)\omega$

now we know that

m = 2 kg

v = 2.5 m/s

L = 0.35 m

I = 4.5 kg-m^2

now plug in all values in above equation

$2\times 2.5 \times 0.35 = (4.5 + (2\times 0.35^2))\omega$

$1.75 = [4.5 + 0.245]\omega$

$1.75 = 4.745\omega$

$\omega = 0.37 rad/s$

so the final angular speed will be 0.37 rad/s

4 0

3 years ago

Rahul has 2 bulbs connected across two cells in a simple circuits shown. How can he make the bulbs glow dimmer?

Ray Of Light [21]

Answer:

in the parallel connection the light bulbs shine less than in the series connection

Explanation:

In a series circuit the current through the whole circuit is the same, therefore the power (brightness) of each bulb is

P = i² R

where R is the resistance of each bulb and i the current of the circuit.

If we connect the light bulbs and the cells in parallel, the current in the circuit is the sum of the east that passes through each light bulb,

i = i₁ + i₂

if the two light bulbs are the same

i = 2 i₁

i₁ = i / 2

so the power of each bulb is is

P = i₁² R

P = R i² / 4

P = ¼ P_initial

Therefore we see that in the parallel connection the light bulbs shine less than in the series connection

3 0

3 years ago