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elena-14-01-66 [18.8K]

3 years ago

11

which of the following is true? a)speed is velocity w a vector b)speed is a scalar w direction c)speed is velocity w direction d

)velocity is speed w a direction

1 answer:

Aleksandr [31]3 years ago

8 0

The answer is D velocity is speed w a direction

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two cars start at the same point and drive in a straight line for 5km. At the end of the drive their distances are the same but

Anna11 [10]

A 'displacement' always consists of a magnitude and a direction. The two cars you just described have displacements with the same magnitude ... 5 km. But if they didn't both drive in the same direction, then their displacements are different.

Remember:

-- 10 m/s² up and 10 m/s² down are different accelerations

-- 30 mph East and 30 mph West are the same speed but different velocity.

-- 5 km North and 5 km South are the same distance but different displacement.

7 0

3 years ago

What is kinematics..???

swat32

Answer:

Kinematics is the study of motion, without any reference to the forces that cause the motion. It basically means studying how things are moving, not why they're moving. It includes concepts such as distance or displacement, speed or velocity, and acceleration, and it looks at how those values vary over time.

7 0

2 years ago

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

4 years ago

A helicopter, starting from rest, accelerates straight up from the roof of a hospital. The lifting force does work in raising th

LuckyWell [14K]

Answer:

25032.47 W

Explanation:

Power is the time rate of doing work, hence,

P = Work done(non conservative) / time

Work done (non conservative) is given as:

W = total K. E. + total P. E.

Total K. E. = 0.5mv²- 0.5mu²

Where v (final velocity) = 7.0m/s, u (initial velocity) = 0m/s

Total P. E. = mgh(f) - mgh(i)

Where h(f) (final height) = 7.2m, h(i) (initial height) = 0 m

=> W = 0.5mv² - mgh(f)

P = [0.5mv² - mgh(f)] / t

P = [(0.5*790*7²) - (790*9.8*7.2)] / 3

P = (19355 + 55742.4) / 3 = 75097.4/3

P = 25032.47 W

6 0

3 years ago

Which statement is true regarding a chemical reaction?

liberstina [14]

Answer:

The total number of atoms does not change, so mass is conserved in the reaction.

Explanation:

If you like my answer than please mark me brainliest thanks

7 0

2 years ago