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3 years ago

10

A plane passes over Point A with a velocity of 8000 m/s north. Forty seconds later it passes over Point B at a velocity of 10,00

0 m/s north. What is the planes’ acceleration from A to B?

1 answer:

Airida [17]3 years ago

7 0

Answer:

The planes’ acceleration from A to B is 500m/s^2

Explanation:

Given that the initial velocity u is 8000m/s

and also given the final velocity v=10,000 m/s

the time taken to move from A to B = 40 second

The acceleration is defined as the rate of change of velocity with time

we know that the expression for acceleration is given as

a=(v-u)/t

substituting our given data into the expression for a we have

a=(10000-8000)/40

a=2000/40

a=500m/s^2

The planes’ acceleration from A to B is 500m/s^2

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A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.

Luba_88 [7]

Answer:

The final equilibrium temperature of the system is $T = 12.48^oC$

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice $M_{i}$ = 129 g = 0.129 kg

Mass of the steam $M_s$ = 19 g = 0.019 kg

Initial temperature is $T_i$ = 0°C

Temperature of steam $T_s$ = 100°C

Following the change of state of water in the question

The energy required by ice to change to water is mathematically given as

$Q_A = M_iL_f$

Where $L_f$ is a constant known as heat of fusion and the value is $334*10^3 J/kg$

$Q_A = 0.129 *334 *10^3 = 43086 J$

The energy been released when the steam changes to water is mathematically given as

$Q_B = M_s * L_v$

Where $L_v$ is a constant known as heat of vaporization and the value is $2256*10^3J/kg$

$Q_B = 0.019 * 2256*10^3 = 42864J$

The energy released when the temperature of water decrease from 100°C to 0°C is

$Q_C = M_s *C_water ($ 100°C)

Where $C_{water}$ is the specific heat of water which has a value $4186J/kg \cdot K$

$Q_C = 0.019 *4186*100 = 7953.4$

Looking at the values we obtained we noticed that ]

$Q_B + Q_C > Q_A$

What this means is that the ice will melt

bearing in mind the conservation of energy

looking the way at which water at different temperature were mixed according to the question

Heat lossed by the vapor = heat gained by ice

$Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T$

$T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}$

$T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}$

$T = 12.48^oC$

3 0

4 years ago

A bus took 8 hours to travel 639 km. For the first 5 hours, it

vladimir1956 [14]

Answer:

93 km/h

Explanation:

Given that a bus took 8 hours to travel 639 km. For the first 5 hours, it travelled at an average speed of 72 km/h

Let the first 5 hours journey distance = F

From the formula of speed,

Speed = distance/time

Substitute speed and time

72 = F/5

F = 72 × 5 = 360 km

The remaining distance will be:

639 - 360 = 279km

The remaining time will be:

8 - 5 = 3 hours

Speed = 279/3

Speed = 93 km/h

Therefore, the average speed for the remaining time of the journey is equal to 93 km/h

8 0

3 years ago

3. A bicycle has a momentum of 25.00 kg* m/s and a velocity of 2.5 m/s . What is the bicycle's

Anna35 [415]

Answer:

10 kg

Explanation:

The question is most likely asking for the mass of the bicycle.

Momentum is the product of an object's mass and velocity. Mathematically:

p = m * v

Where p = momentum

m = mass

v = velocity

Hence, mass is:

m = p / v

From the question:

p = 25 kgm/s

v = 2.5 m/s

Mass is:

m = 25 / 2.5 = 10 kg

The mass of the bicycle is 10 kg.

In case the question requires the Kinetic energy of the bicycle, it can be gotten by using the formula

K. E = ½ * p * v

K. E. = ½ * 25 * 2.5 = 31.25 J

5 0

4 years ago

A crane raises a crate with a mass of 75 kg to a height of 10 m. Given that the

Crazy boy [7]

Answer: C. 7,350 J

Explanation:

Just did it in Apex

7 0

3 years ago

Two parallel slits are illuminated by light composed of two wavelengths. One wavelength is λA = 622nm. The other wavelength is λ

Sergio039 [100]

Answer:

$\lambda_{B}=414.67 nm$

Explanation:

In this question we have given

$\lambda_{A}=622nm$

we have to find

$\lambda_{B}=?$

We know that

optical path difference for bright fringe is given as $=n\lambda$

Here,

n is order of fringe

and optical path difference for dark fringe is given as $=(n+.5)\lambda$

since the light with wavelength $\lambda_{A}$ produces its third-order bright fringe at the same place where the light with wavelength $\lambda_{B}$ produces its fourth dark fringe

it means

optical path difference for 3rd order bright fringe= optical path difference for forth order dark fringe

Therefore,

$3\lambda_{A}=(4+.5)\lambda_{B}$ ...............(1)

Put value of $\lambda_{A}$ in equation (1)

$3 \times 622=(4+.5)\lambda_{B}$

$1866=4.5\lambda_{B}$

$\lambda_{B}=414.67 nm$

3 0

3 years ago