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3 years ago

9

D. If a dog has a mass of 12 kg, what is its weight on Neptune? 11.7N/kg

1 answer:

steposvetlana [31]3 years ago

4 0

Answer:

133.8 N

Explanation:

Recall that the acceleration of gravity in Neptune is estimated as 11.15 m/s^2

Therefore, the weight of the dog on this planet would be:

Weight = mass x acceleration of gravity = 12 kg x 11.15 m/s^2 = 133.8 N

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A car moves in a straight line at 22.0 m/s for 10.0miles, then at 30.0 m/s for another 10.0miles. Calculate the car’s average sp

maw [93]

Answer: 25.38 m/s

Explanation:

We have a straight line where the car travels a total distance $D$ , which is divided into two segments $d=10 miles$ :

$D=d+d=2d$ (1)

Where $d=10mi \frac{1609.34 m}{1 mi}=16093.4 m$

On the other hand, we know speed is defined as:

$S=\frac{d}{t}$ (2)

Where $t$ is the time, which can be isolated from (2):

$t=\frac{d}{S}$ (3)

Now, for the first segment $d=16093.4 m$ the car has a speed $S_{1}=22m/s$ , using equation (3):

$t_{1}=\frac{d}{S_{1}}$ (4)

$t_{1}=\frac{16093.4 m}{22m/s}$ (5)

$t_{1}=731.518 s$ (6) This is the time it takes to travel the first segment

For the second segment $d=16093.4 m$ the car has a speed $S_{1}=30m/s$ , hence:

$t_{2}=\frac{d}{S_{2}}$ (7)

$t_{2}=\frac{16093.4 m}{30m/s}$ (8)

$t_{2}=536.44 s$ (9) This is the time it takes to travel the secons segment

Having these values we can calculate the car's average speed $S_{ave}$ :

$S_{ave}=\frac{d + d}{t_{1} + t_{2}}=\frac{2d}{t_{1} + t_{2}}$ (10)

$S_{ave}=\frac{2(16093.4 m)}{731.518 s +536.44 s}$ (11)

Finally:

$S_{ave}=25.38 m/s$

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3 years ago

Classify the following situations as involving balanced or unbalanced forces.

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Explanation:

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A 62.2 kg woman lifts herself 31.8 cm each time she does a chin-up. What force does she exert in lifting herself to the chin pos

RideAnS [48]

Answer:31

Explanation:

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2 years ago

What happens after the Gulf Stream travels north?

Ierofanga [76]

Incorrect didn't flow away from pole cause warmer climate I know cause I tried.

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3 years ago

A 5.0 μf and a 7.0 μf capacitor are connected in series across an 8.0-v dc source. What is the voltage across the 5.0 μf capacit

Semmy [17]

Answer:4.67 V

Explanation:

Given

$C_1=5\ \muf$

$C_2=7\ \muf$

Voltage $V=8\ V$

For capacitance in series Net capacitance is given

$C_{net}=\frac{C_1\cdot C_2}{C_1+C_2}$

$C_{net}=\frac{5\times 7}{5+7}=\frac{35}{12} \muf$

$q=C_{net}V$

$q=\frac{35}{12}\times 8=\frac{70}{3} \muf$

As capacitors are in series so charge flow will be same in both capacitors

Voltage drop in $5 \muf$ capacitor

$V=\frac{70}{3}\times \frac{1}{5}=\frac{14}{3} V\approx 4.67 V$

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3 years ago