A roller coaster goes from 2.00 m/s [forward] to 10.0 m/s [forward) in 4.50 s. What is its acceleration?

1 answer:

8 0

The answer would be 1.77778:

We must use the kinematic equation v = v0 + at, then fill in the elements given and solve which equals 1.77778

You might be interested in

A beam of light, initially travelling in the air, strikes water surface at an angle of 24.5° with the normal. If the speed of li

kykrilka [37]

Answer:

Explanation:

n = 3.00e8/2.22e8 = 1.35

1.00sin24.5 = 1.35sinθ

θ = 17.9°

8 0

3 years ago

Which numbers on the ph scale indicate an acid

KiRa [710]

Answer:

0 - 6.9 --> Acidic

Explanation:

4 0

3 years ago

A 1000-kg car traveling at 70 m/s takes 3 m to stop under full braking. the same car under similar road conditions, traveling at

azamat

We assume $a=const$ (acceleration is constant. We apply the equation
$v^2=v0^2+2as$ where s is the distance to stop $v=0(m/s)$ . We find the acceleration from this equation
$a=-v0^2/(2s)=-70^2/(2*3) =-816.7 (m/s^2)
$
We know the acceleration, thus we find the distance necesssary to stop when initial speed is $v=140 (m/s)$
$s=-v0^2/(2a) =140^2/(2*816.7)=12 (m)$

5 0

4 years ago

What is the acceleration of a 10kg rock falling at a gravitational fiend of 10N/kg

Archy [21]

Answer:

10N/kg...................

6 0

3 years ago

The moon Phobos orbits Mars

shepuryov [24]

$27.9816 \times 10^{3} s$ is the period of orbit.

<u>Explanation: </u>

The equation that is useful in describing satellites motion is Newton form after Kepler's Third Law. The period of the satellite (T) and the average distance to the central body (R) are related as the following equation:

$\frac{T^{2}}{R^{3}}=\frac{4 \times \pi^{2}}{G \times M_{c e n t r a l}}$