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3 years ago

A 1300 kg car traveling with a speed of 3.5 m/s executes a turn with a 8.5 m radius of curvature.

Physics

1 answer:

Y_Kistochka [10]3 years ago

7 0

Answer:

1.4 m/s/s (2.s.f)

Explanation:

The formula for centripetal acceleration is:

$a=\frac{v^{2} }{r}$ , where v is velocity and r is the radius.

In the question we are given the information that the car has a mass of 1300kg, a velocity of 2.5m/s, and a turn radius of 8.5m which are all the values we need. Therefore we can simply substitute in the values to solve the question:

$a=\frac{3.5^{2} }{8.5} \\a=1.4$

Therefore the centripetal acceleration of the car is 1.4m/s/s. (2.s.f)

Hope this helped!

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Two cars, initially at rest and 5 km apart at t=0 , simultaneously move toward each other. Car A travels at a constant speed of

Anastasy [175]

Answer:

d. 268 s

Explanation:

Constant Speed Motion

An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.

Expressed in a simple equation, we have:

$\displaystyle v=\frac{d}{t}$

Where

v = Speed of the object

d = Distance traveled

t = Time taken to travel d.

From the equation above, we can solve for d:

d = v . t

And we can also solve it for t:

$\displaystyle t=\frac{d}{v}$

Two cars are initially separated by 5 km are approaching each other at relative speeds of 55 km/h and 12 km/h respectively. The total speed at which they are approaching is 55+12 = 67 km/h.

The time it will take for them to meet is:

$\displaystyle t=\frac{5}{67}$

t = 0.0746 hours

Converting to seconds: 0.0746*3600 = 268.56

The closest answer is d. 268 s

8 0

3 years ago

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kondaur [170]

Answer:

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Explanation:

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3 years ago

Halleys comet has period of 75.3 years. Using Kepler’s third law, find it’s semimajor axis expressed in astronomical units?

natta225 [31]

Answer: 17.83 AU

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. 

$T^{2}\propto a^{3}$ (1)

Talking in general, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite, comet) orbiting a greater body in space with the size $a$ of its orbit.

However, if $T$ is measured in years, and $a$ is measured in astronomical units (equivalent to the distance between the Sun and the Earth: $1AU=1.5(10)^{8}km$ ), equation (1) becomes: