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3 years ago

A train A travelled a distance of 150 km in 3 hours, whereas, train 'B"

Physics

1 answer:

ozzi3 years ago

8 0

Answer:

Therefore, Train A is faster with 50km/h and a 5km/h difference

Explanation:

train A

150km/3h = 50km/h

train B

180km/4h = 45km/h

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If a star is moving away from you at a constant speed, how do the wavelengths of the absorption lines change as the star gets fa

Minchanka [31]

Answer:

they stay shifted the same amount to the red

Explanation:

Redshift is given by

$z=\dfrac{\lambda_o-\lambda_e}{\lambda_e}$

Where,

$\lambda_o$ = Wavelength observed

$\lambda_e$ = Wavelength emitted

Also

Transverse redshift is given by

$1+z=\dfrac{1}{\sqrt{1-v^2/c^2}}$

v = Velocity of object

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

So, if the velocity is constant the redshift remains the same

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3 years ago

The normal boiling point of a certain liquid is , but when of urea () are dissolved in of the solution boils at instead. Use thi

Kruka [31]

Answer:

100 Degrees is boiling point.

Explanation:

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3 years ago

1. Three kids in a parking lot launch a rocket that rises

Mnenie [13.5K]

Answer:

Average speed equals distance / time

380 / 40 = 9.5 m/s.

Explanation:

3 0

3 years ago

At what speed does a 1,248 kg compact car have the same kinetic energy as a 18,777 kg truck going 26 km/h?

notka56 [123]

The speed of car is 100.8km/h

$KE = \frac{1}{2} m(v \: truck ) {}^{2}$

$= 0.5 \times 18777 \times \frac{26}{3.6} \times \frac{26}{3.6}$

$= 489708.79j$

$= \frac{1}{2} \times 1248 \times( v \: car) {}^{2}$

$= 624v {}^{2}$

$so \: v {}^{2} = \frac{489708.7}{624}$

$= 784$

$v = \sqrt{784}$

$v = 28m/s$

v car= 28×3.6

=100.8km/h

Hence, the speed of the car is 100.8km/h

learn more about speed from here:

brainly.com/question/28326855

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3 0

1 year ago

Force → F = ( − 8.0 N ) ˆ i + ( 6.0 N ) ˆ j acts on a particle with position vector → r = ( 3.0 m ) ˆ i + ( 4.0 m ) ˆ j . What a

andrey2020 [161]

Explanation:

It is given that,

Force, $F=(-8\ N)i+(6\ N)j$

Position vector, $r=(3i+4j)\ m$

(a) The torque on the particle about the origin is given by :

$\tau=F\times r\\\\\tau=(-8i+6j)\times (3i+4j)\\\\\tau=(-50k)\ N-m$

(b) To find the angle between r and F use dot product formula as :

$F{\cdot} r=|F||r|\ \cos\theta\\\\\cos\theta=\dfrac{F{\cdot} r}{|F| |r|}\\\\\cos\theta=\dfrac{(-8i+6j){\cdot} (3i+4j)}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=\dfrac{-24+24}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=0\\\\\theta=90^{\circ}$

Hence, this is the required solution.

8 0

3 years ago

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