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Archy [21]
3 years ago
9

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

Physics
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

3 0
3 years ago
Classify the following situations as involving balanced or unbalanced forces.
Murrr4er [49]

Answer:

11212121221212121212121212121212121212121212121221212

Explanation:

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5 0
3 years ago
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:

8 0
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.


Im'ma guess cause didn't correct the answer but says Flowing away from pole was wrong though.


My guess is crashing with warm water. Might be wrong too though...

6 0
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                              

8 0
3 years ago
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