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2 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]2 years ago

7 0

Answer:

Propels in the opposite direction

Explanation:

You might be interested in

Determine the speed of sound on a rainy day with the temperature of 18 degrees celsius..

Troyanec [42]

Answer:

Answer:

= 338.2 m/s

Explanation:

vs= 331 + T (0.6m/s)

= 331 + 12 °C (0.6m/s)

= 331 + 7.2m/s

= 338.2 m/s

Explanation:

6 0

2 years ago

Consider 3.5 kg of austenite containing 0.95 wt% c and cooled to below 727°c (1341°f). (a) what is the proeutectoid phase? (b) h

vladimir2022 [97]

A. The proeutectoid phase is Fe₃c because 0.95 wt/c is greater than the eutectoid composition which is 0.76 wt/c

b. We determine how much total territe and cementite form, we apply the lever rule expressions yields.
Wx = (fe₃c-co/cfe₃ c-cx = 6.70- 0.95/6.70- 0.022 = 0.86
The total cementite
Wfe₃C = 10-Cx/ Cfe₃c -Cx = 0.95 - 0.022/6.70 - 0.022 = 0.14
The total cementite which is formed is
(0.14) × (3.5kg) = 0.49kg

c. We calculate the pearule and the procutectoid phase which cementite form the equation
Ci = 0.95 wt/c
Wp = 6.70 -ci/6.70 - 0.76 = 6.70 -0.95/6.70 - 0.76 = 0.97
0.97 corresponds to mass.
W fe₃ C¹ = Ci - 0.76/5.94 = 0.03
∴ It is equivalent to
(0.03) × (3.5) = 0.11kg of total of 3.5kg mass.

4 0

2 years ago

The driver of a car going 86.0 km/h suddenly sees the lights of a barrier 44.0 m ahead. It takes the driver 0.75 s to apply the

Lynna [10]

Part a

Answer: NO

We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.

Using the second equation of motion:

$s=ut+0.5at^2$

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.

It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, $a=-10.0 m/s^2$

$\Rightarrow s=23.9 m/s \times 0.75s+0.5\times (-10.0 m/s^2)\times (0.75 s)^2=15.11 m$

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.

Part b

Answer: 29.6 m/s

The maximum distance that car can travel is $s=44.0 m$

The acceleration is same, $a=-10.0 m/s^2$

The final velocity, v=0

Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

$v^2-u^2=2as$

$0-u^2=-2 \time 10.0m/s^2 \times 44.0 m\Rightarrow u=\sqrt{880 m^2/s^2}=29.6 m/s$

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.

7 0

3 years ago

A proton moves through a magnetic field at 26.7 % 26.7% of the speed of light. At a location where the field has a magnitude of

iogann1982 [59]

Answer:

$8.64283\times 10^{-14}\ N$

Explanation:

q = Charge of proton = $1.6\times 10^{-19}\ C$

v = Velocity of proton = $0.267\times c$

c = Speed of light = $3\times 10^8\ m/s$

B = Magnetic field = 0.00687 T

$\theta$ = Angle = $101^{\circ}$

Magnetic force is given by

$F=qvBsin\theta\\\Rightarrow F=1.6\times 10^{-19}\times (0.267\times 3\times 10^8)\times 0.00687\times sin101\\\Rightarrow F=8.64283\times 10^{-14}\ N$

The magnetic force acting on the proton is $8.64283\times 10^{-14}\ N$

4 0

3 years ago

In general, if a sound has intensity of β dB at 1 m from the source, at what distance from the source would the decibel level de

Tpy6a [65]

Answer: The required distance is given by

$r_2=1\text{ m}\cdot 10^\frac{\beta}{20}.$

Explanation: The sound intensity in dB is given by the formula

$\beta \text{ dB}=10\log\frac{I}{I_0}$

where $I_0$ is the hearing threshold in absolute units and $I$ is the absolute intensity of the sound which depends on the distance. In general, for two distances $r_1$ and $r_2$ we have that

$\frac{I(r_1)}{I(r_2)}=\frac{r_2^2}{r_1^2}=\left(\frac{r_2}{r_1}\right)^2.$

Now let us take $r_1=1\text{ m}$ and let $r_2$ be the required distance. We have

$\beta \text{ dB}=10\log\frac{I(r_1)}{I_0},\quad 0\text{ db}=10\log\frac{I(r_2)}{I_0}.$

Exponentiating these equations we obtain

$10^{\frac{\beta}{10}}=\frac{I(r_1)}{I_0},\quad 10^0=1=\frac{I(r_2)}{I_0}.$

Dividing them

$\frac{I(r_1)}{I(r_2)}=10^\frac{\beta}{10}.$

Using the previously stated identity

$\frac{r_2^2}{r_1^2}=10^\frac{\beta}{10}\Rightarrow r_2=r_110^\frac{\beta}{20}=1\text{ m}\cdot 10^\frac{\beta}{10}$

Now if we use the given example where $\beta=11$ we have

$r_2=1\text{ m}\cdot 10^{\frac{11}{20}}=3.55\text{ m}.$

7 0

2 years ago