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3 years ago

What do the four terrestrial planets have in common?

Physics

1 answer:

stellarik [79]3 years ago

7 0

They all have identical orbits shapes

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At what speed do a bicycle and its rider, with a combined mass of 100 kg, have the same momentum as a 1400 kg car traveling at 6

Ilia_Sergeevich [38]

Momentum of car

Given: Mass m= 1,400 Kg; V = 6.0 m/s

Formula: P = mv

P = (1,400 Kg)(6.0 m/s)

P = 8,400 Kg.m/s

Velocity of the rider to have the same momentum as a car.

Mass of rider and bicycle m = 100 Kg

P = mv

V = P/m

V = 8,400 Kg.m/s/100 Kg

V = 84 m/s

7 0

2 years ago

Whose geocentric model of the solar system was accepted for 1400 years

-BARSIC- [3]

Answer:

Plato, Aristotle developed it further and used for 1400 years till Copernicus.

Explanation:

8 0

2 years ago

Question 15 of 25

vichka [17]

Answer: conduction :it transfers heat between objects that are in direct contact with eachother

7 0

2 years ago

A 50 Q resistor in a circuit has a current flowing through it of 2.0 A. What is

Alex17521 [72]

Hello!

We can use the following equation for calculating power dissipated by a resistor:
$P = i^2R$

P = Power (? W)
i = Current through resistor (2.0 A)
R = Resistance of resistor (50Ω)

Plug in the known values and solve.

$P = (2.0^2)(50) = \boxed{\text{ B. }200 W}$

7 0

1 year ago

A car of mass m goes around a banked curve of radius r with speed v. If the road is frictionless due to ice, the car can still n

Sholpan [36]

Answer:

horizontal component of normal force is equal to the centripetal force on the car

Explanation:

As the car is moving with uniform speed in circle then the force required to move in the circle is towards the center of the circle

This force is due to friction force when car is moving in circle with uniform speed

Now it is given that car is moving on the ice surface such that the friction force is zero now

so here we can say that centripetal force is due to component of the normal force which is due to banked road

Now we have

$N sin\theta = \frac{mv^2}{R}$

$N cos\theta = mg$

so we have

$v = \sqrt{Rg tan\theta}$

so this is horizontal component of normal force is equal to the centripetal force on the car

5 0

3 years ago