What distance will be traveled if you are going at an average speed of 420 km/hr for 45 minutes?

You are snowboarding down a mountain face at a

Gelneren [198K]

I think it would be 10 m/s

6 0

3 years ago

Light travels through a vacuum at a speed of 3 x 10 m/s. What is the speed of

Andrew [12]

Answer:

v = c / n (n = 1 for air)

v = c / 1.33 = 3 * 10E8 m/s / 1.33 = 2.25 * 10E8 m/s

3 0

3 years ago

One billiard ball is shot east at 2.2m/s. A second, identical billiard ball is shot west at 1.2m/s. The balls have a glancing co

Kruka [31]

Answer:

$v_{1}=1.886 \frac{m}{s}$

β= 57.99 south of east

Explanation:

$v_{1}=2.2 \frac{m}{s} \\v_{2}=1.2 \frac{m}{s} \\m_{1}=m_{2}=m\\v_{fx}=1.6 \frac{m}{s} \\v_{fy}=$ ?

Velocity in axis x the two balls come one from east and west

$m_{1}*v_{1x}+m_{2}*v_{2x}=m_{1}*v_{fx1}+m_{2}*v_{fx2}\\m*(v_{1x}+v_{2x})=m*(v_{fx1}+v_{fx2})\\v_{fx2}=0\\v_{1x}+v_{2}=v_{f1}+0\\v_{fx1}=2.2 \frac{m}{s}+(1.2\frac{m}{s})\\ v_{fx1}=1 \frac{m}{s} \\$

Velocity in axis y initial is zero so:

$v_{y1}+v_{y2}=v_{y1f}+v_{y2f}\\v_{y1}=0\\v_{y2}=0\\v_{y1f}+v_{y2f}=0\\v_{y1f}=-v_{y2f}\\v_{y2f}=1.6\frac{m}{s}$

$v=\sqrt{v_{1fx}^{2}+v_{1fy}^{2}}\\ v=\sqrt{1^{2}+1.6^{2}}\\v=1.886 \frac{m}{s}$

Angle is find using:

tan(β)= $\frac{v_{fy}}{v_{fx}}$

$\beta =tan^{-1}*\frac{1.6}{1}=57.99$

5 0

4 years ago

Which statements provides evidence that the earth revolves around the sun

Aloiza [94]

Answer:

Different star constellations are visible from Earth at different seasons of the year.

Explanation:

The reason the fact that we can see different constellations in the sky during different seasons on earth is the most compelling reason we travel around the sun is because if the sun travelled around the earth, certain constellations would only be visible in certain places. You’d have to travel to see certain ones.

However, you don’t have to do that because we travel around the sun, therefore travelling around other stars too.

4 0

4 years ago

A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2.

Bumek [7]

Answer:

the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

Explanation:

Given that;

speed of car V = 120 km/h = 33.3333 m/s

Reaction time of an alert driver = 0.8 sec

Reaction time of an alert driver = 3 sec

extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

S_d = 73.3333 m

hence, number of car of additional car length n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

8 0

3 years ago