We cab calculate work because we know the force and distance.what else would we need to know to calculate power

A jet takes off from Florida and flies 100 miles north and then 400 miles west. Determine the total distance travelled by the je

Inessa [10]

Answer:

90m

Explanation:

Distance is not a vector but a scalar quantity. depicted by a quantity only.

hence:

Distance is equal to 30+40+20 = 90 m

5 0

3 years ago

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A 66.0 kg diver is 3.10 m above the water, falling at speed of 6.10 m/s. Calculate her kinetic energy as she hits the water. (Ne

castortr0y [4]

Kinetic energy as she hits the water is 3300 joule.

To find the answer, we need to know about the Newton's equation of motion.

<h3>What's the Newton's equation of motion to determine the final velocity?</h3>

The final velocity is determined as

V²=U²+2aS

V= final velocity, U= initial velocity, a= acceleration and S= distance

<h3>What's the final velocity of the driver falling from 3.10m with initial velocity of 6.10m/s?</h3>

Here, a= 9.8m/s², U= 6.10m/s and S= 3.10m
So, V²= 6.1²+2×9.8×3.10= 98
V= √98= 10m/s

<h3>What's the kinetic energy of the driver when touches the water?</h3>

Kinetic energy= 1/2×mass×velocity²

= 1/2 × 66 × 10²

= 3300J

Thus, we can conclude that the kinetic energy of the driver is 3300 Joule.

Learn more about the kinetic energy here:

brainly.com/question/25959744

#SPJ4

3 0

2 years ago

EXPERTS/ACE/GENIUS (and people that wanna actually help)

scZoUnD [109]

Hello! Force = mass * acceleration. We can go ahead and eliminate A and B, because those answers are too small. We can multiply 4.5 (mass) * 9 (acceleration) to find the amount of force needed. 4.5 * 9 is 40.5.The amount of force needed is 40.5 N. The answer is C: 40.5 N.

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3 years ago

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One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

7 0

3 years ago

Which one of the following has exactly 3 significant figures?

lapo4ka [179]

A
Because leading zeros don’t count

7 0

3 years ago