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MaRussiya [10]
3 years ago
9

The slowest animal ever discovered was a crab found in the Red Sea. It traveled with an average speed of 5.70km/yr. how long wou

ld it take this crab to travel 100km?
Physics
2 answers:
Vaselesa [24]3 years ago
8 0
If it takes 1 year for this crab to travel 5,70km

Then, it will takes approximatly 1000/5,70 = 175 years to travel 1000 km 
svlad2 [7]3 years ago
7 0

Answer:

Time taken by the crab, t = 17.54 years.

Explanation:

It is given that,

Speed of a crab, v = 5.70 km/yr

Distance travelled by the crab, d = 100 km

We need to find the time taken by the crab to cover this much distance. Time taken is given by :

t=\dfrac{d}{v}

t=\dfrac{100\ km}{5.70\ km/yr}

t = 17.54 years

So, the time taken by the crab is 17.54 years. Hence, this is the required solution.

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Your boat departs from the bank of a river that has a swift current parallel to its banks. If you want to cross this river in th
REY [17]

Answer:

D.

Explanation:

Given that your boat departs from the bank of a river that has a swift current parallel to its banks. If you want to cross this river in the shortest amount of time, you should direct your boat: so that it drifts with the current.

If the boat moves perpendicular to the current, the current flow will be the resistance to the movement of the boat. So, it's better for the boat to drifts perpendicularly with the current.

The best answer is therefore option D.

8 0
3 years ago
A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap
viktelen [127]

Answer:

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

V_r + V_L + V_c = V_{net}

now we will have

iR + L\frac{di}{dt} + \frac{q}{C} = 12 V

now we have

1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12

So we will have

q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}

at t = 0 we have

q = 0

0 = 3\times 10^{-6} + c_1  + c_2

also we know that

at t = 0 i = 0

0 = -4000 c_1 - 1000c_2

c_2 = -4c_1

c_1 = 1 \times 10^{-6}

c_2 = -4 \times 10^{-6}

so we have

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

5 0
3 years ago
Is pulling a dog a contact force or a non contact force
horrorfan [7]
A contact force since you are making physical contact with the dog

Mark my brainliest please
6 0
3 years ago
At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt
Harrizon [31]

Answer:

It will take you 30.8 s to travel the 120 m of the ramp.

Explanation:

Hi there!

The equation for the position of an object moving in a straight line is:

x = x0 + v * t

Where:

x = position at time t

x0 = initial position

v = velocity

t = time

In this case, we will consider the start of the ramp as the origin of our reference system so that x0 = 0.

Now, let´s calculate the speed of the person walking on the ground:

x = v * t

120 m = v * 72 s

v = 120 m / 72 s

v = 1.7 m/s

If you walk on the ramp with that speed, your total speed will be your walking speed plus the speed of the ramp because both are in the same direction. Then, using the equation for the position:

x = v * t

In this case, v = speed of the ramp + walking speed

v = 2.2 m/s + 1.7 m/s = 3.9 m/s

120 m = 3.9 m/s * t

t = 120 m / 3.9 m/s = 30.8 s

It will take you 30.8 s to travel the 120 m

8 0
3 years ago
Which of the following would experience induced magnetism most easily? A. Aluminum B. Copper C. Air D. Permalloy
galina1969 [7]
Permalloy would experience induced magnetism most easily 
5 0
3 years ago
Read 2 more answers
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