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Luda [366]
3 years ago
5

When boating in shallow areas or seagrass beds, you see a mud trail in your wake where your boat has churned up the bottom. If y

ou see this trail, you should:
Physics
1 answer:
cluponka [151]3 years ago
7 0

Answer:

when boating in the shallow area or sea grass beds and the mud trail in our wake we should stop our vessel and try to tilt the motor out through the water.

Explanation:

If we see a trail like, when boating in the shallow area or sea grass beds and the mud trail in our wake we should stop our vessel and try to tilt the motor out through the water. And with the help of a pole (or long rod) we should move our vessel out of the shallow water region. If not with pole we should try to walk out our vessel out of the region. since, the running motor of the vessel will attract more problems like clouded water and grass getting stuck in the motor.

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It took 1700 J of work to stretch a spring from its natural length of 2m to a length of 5m. Find the spring's force constant.
Alenkasestr [34]
The work to stretch a spring from its rest position is

               (1/2) (spring constant) (distance of the stretch)²

           E = 1/2 k x² .

You said it takes 1700 joules to stretch the spring 3 meters from its rest position, so we can write

                 1700 joules  =  1/2 k (3m)²

1 joule = 1 newton-meter

                 1700 N-m =  1/2 k  (3m)²

Multiply each side by 2:   3400 N-m  =  k · 9m²   

Divide each side by  9m²     k  =  3400 N-m / 9m²

                                                   =  (377 and 7/9)   newton per meter    

7 0
3 years ago
Read 2 more answers
When a falling meteoroid is at a distance above the Earth's surface of 3.40 times the Earth's radius, what is its acceleration d
Alchen [17]

Answer:

g = 0.85 ms^{-2}

Explanation:

g = \frac{GM}{h^{2} }

were; g is the acceleration due to Earth's gravity, G is Newton's gravitation constant (6.674 x 10^{-11} Nm^{2}kg^{-2}), M is the mass of the earth (5.972 x 10^{24} kg), and h is the distance of meteoroid to the earth.

h = 3.40 x R

  = 3.40 x 6371 km

h = 21661.4 km

  = 21661400 m

Thus,

g = \frac{6.674*10^{-11}*5.972*10^{24}  }{(21661400)^{2} }

  = \frac{3.9857 *10^{14} }{4.6922*10^{14} }

  = 0.84944

g = 0.85 ms^{-2}

The acceleration due to the Earth's gravitation is 0.85 ms^{-2}.

6 0
3 years ago
Why is earth the only planet on which water exists in a liquid state
Gennadij [26K]
This is due to earths location in the solar system. Earth is in the habitat zone or the Goldie locks zone, in this zone it's not too hot or not too cold for water to exist. Other planets in different star systems have liquid oceans due to them being in the habitat zone.
7 0
3 years ago
n 38 g rifle bullet traveling at 410 m/s buries itself in a 4.2 kg pendulum hanging on a 2.8 m long string, which makes the pend
Kitty [74]

Answer:

68cm

Explanation:

You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get

p_f=p_i\\mv_1+Mv_2=(m+M)v

m: mass of the bullet

M: mass of the pendulum

v1: velocity of the bullet = 410m/s

v2: velocity of the pendulum =0m/s

v: velocity of both bullet ad pendulum joint

By replacing you can find v:

(0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}

this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:

E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2

g: 9.8/s^2

h: height

By doing h the subject of the equation and replacing you obtain:

(0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m

hence, the heigth is 68cm

4 0
3 years ago
A 0.290 kg block on a vertical spring with a spring constant of 5.00 ✕ 103 N/m is pushed downward, compressing the spring 0.110
True [87]

Answer:

The height at point of release is 10.20 m

Explanation:

Given:

Spring constant : K= 5 x 10 to the 3rd power n/m

compression x = 0.10 m

Mass of block m= 0.250 kg

Here spring potential energy converted into potential energy,

mgh = 1/2 kx to the 2 power

For finding at what height it rise,

0.250 x 9.8 x h = 1/2 x 5 x 10 to the 3 power x (0.10)to the 2 power) - ( g= 9.8 m/8 to the 2 power

h= 10.20

Therefore, the height at point of release is 10.20 m

4 0
3 years ago
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