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riadik2000 [5.3K]

1 year ago

15

another way, speed is a scalar value, while reloolg isa Which ONE of the following represents a toy car that moves at a higher a

verage speed? (12) Show your calculations. A: A toy car that travels 1,50 km in 3 min B: A toy car that travels 800 m in 1,50 min C: A toy car that travels 250 km in 8 hours CAR A: CAR B: CAR C:

1 answer:

erica [24]1 year ago

4 0

Car A will have highest speed is 83.3m/s .

<h3>What is speed ? </h3>

The rate of change of position of an object in any direction.

The S.I unit is m/s . Speed is a scalar quantity it defines only magnitude not direction

.

speed = distance /time

In case of Car A ,

We have given distance 150Km in 3 min ,

First we have convert the distance km to m

150×1000m

then conversion of min to sec

38×60sec

speed = 15000/180

speed = 83.3m/sec

In case of Car B

we have given 800m in 150 min

lets convert the time into second

150×60

Speed = 800/150×60

speed = 0.88m/ s

In case of Car C

We have given here distance 250 Km and time in 8 hours

convert km to m

25000

and time into sec

88×60×60

speed = 0.86m/ s

Hence ,Car A has highest speed amongst them .

To learn more about speed click here

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A 1.0-kg block of aluminum is at a temperature of 50 Celsius. How much thermal energy will it lose when it’s temperature is redu

Ipatiy [6.2K]

Answer:

The lose of thermal energy is, Q = 22500 J

Explanation:

Given data,

The mass of aluminium block, m = 1.0 kg

The initial temperature of block, T = 50° C

The final temperature of the block, T' = 25° C

The change in temperature, ΔT = 50° C - 25° C

= 25° C

The specific heat capacity of aluminium, c = 900 J/kg°C

The formula for thermal energy,

<em>Q = mcΔT</em>

= 1.0 x 900 x 25

= 22500 J

Hence, the lose of thermal energy is, Q = 22500 J

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3 years ago

A plane has an airspeed of 142 m/s. A 30.0 m/s wind is blowing southward at the same time as the plane is flying. What must be t

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Answer:

$\theta=12.19^{\circ}$

Explanation:

Given that

The speed of the airplane ,v= 142 m/s

The speed of the air ,u = 30 m/s

Lets take angle make by airplane from east direction towards north direction is θ .

Now by using diagram ,we can say that

$sin\theta =\dfrac{u}{v}$

Now by putting the values in the above equation we get

$sin\theta =\dfrac{30}{142}$

$sin\theta=0.21$

$\theta=12.19^{\circ}$

Therefore the angle will be 12.19° .

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4 years ago

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

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Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

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