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olga55 [171]
3 years ago
8

HURRY ! THE BEST I WILL GIVE THE BRAINLIEST​

Physics
1 answer:
erik [133]3 years ago
6 0

The speed of both cars is the same ... 80 km per hour.

But their velocities are different, because DIRECTION is part of velocity, and their directions are different.

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A car is driving at 30 m/s it has a mass of 1800 kg what is the cars momentum?
Ostrovityanka [42]

Answer:

Explanation:

Momentum is a concept and is defined as,

Momentum = mass × velocity

So to calculate the momentum of the car

momentum of the car = mass of the car × velocity of the car

So we get,

momentum of the car = 1800 × 30

= 54000 Ns

5 0
3 years ago
Read 2 more answers
an ice skater applies a horizontal force to a 20.-kilogram block on frictionless, level ice, causing the block to accelerate uni
stiv31 [10]
The only vertical forces are weight and normal force, and they balance since the surface is horizontal. The horizontal forces are the applied force (uppercase F) in the direction the block slides and the frictional force (lowercase f) in the opposite direction.

Apply Newton's 2nd Law in the horizontal direction:
ΣF = ma
F - f = ma
where f = µmg

F - µmg = ma
F = m(a +µg)
F = (20 kg)(1.4 m/s² + 0.28(9.8 m/s²)

F = 83 N
3 0
2 years ago
What are some INDIRECT applications of centripetal force in real life?
Eddi Din [679]

Answer:

The circular turning of roads

Explanation:

please mark brainliest

8 0
2 years ago
Read 2 more answers
The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

We know that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v remain same at both side. so we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2 ..............1

let final pressure is P and temp  T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3} ..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3} .............3

divide 2 equation by 3rd equation

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, temperature on left side is 1.48 times the temperature on right

6 0
2 years ago
already did the work, i just need someone to see if i did the tangent line right? the line has to touch the point 0.6! thank you
Dmitrij [34]

The tangent looks good.

The curve is a bit crooked, at the 0.9 and 1.

But overall, cool graph.

5 0
2 years ago
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