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nata0808 [166]
2 years ago
9

An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary

, the tension in the cable was 7000 N. When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N drag force.
Physics
2 answers:
Gekata [30.6K]2 years ago
5 0
The tension in the cable when the craft was being lowered to the
seafloor = <span>7500 N - 1800 N = 5700 N</span>
Eva8 [605]2 years ago
5 0
I think you gave an incomplete question. The two questions would be:

A.What was the tension in the cable when the craft was being lowered to the seafloor? 

<span>B.What was the tension in the cable when the craft was being raised from the seafloor?
</span>
The details for answering the question are already given.

A. Tension in the cable when
 the craft was being lowered to the seafloor = 7000 - 1800 newton
                                                                      = 5200 newton

B. Tension in the cable when
 the craft was being raised from the seafloor = 7000 + 1800 newton
                                                                       = 8800 newton.
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A satellite moves in a circular orbit a distance of 1.6×10^5 m above Earth's surface. (radius of Earth is 6.38 x 10^6m and its m
Rashid [163]

Answer:

The speed of the satellite is 7809.52 m/s        

Explanation:

It is given that,

Radius of Earth, r=6.38\times 10^6\ m

Mass of earth, M=5.98\times 10^{24}\ kg

A satellite moves in a circular orbit a distance of, d=1.6\times 10^5\ m  above Earth's surface.

We need to find the speed of the satellite. It is given by :

v=\sqrt{\dfrac{GM}{R}}

R = r + d

R=(6.38\times 10^6\ m+1.6\times 10^5\ m)=6540000\ m

So, v=\sqrt{\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 5.98\times 10^{24}\ kg}{6540000\ m}}

v = 7809.52 m/s

So, the speed of the satellite is 7809.52 m/s. Hence, this is the required solution.

5 0
3 years ago
A golf ball is hit horizontally off the edge of a 30 m high cliff and lands a distance of 25 m from the edge of the cliff. What
Ratling [72]

Answer:

V₀y = 0 m/s

t = 2.47 s

V₀ₓ = 61.86 m/s

Vₓ = 61.86 m/s

Explanation:

Since, the ball is hit horizontally, there is no vertical component of velocity at initial point. So, the initial vertical velocity (V₀y) will beL

<u>V₀y = 0 m/s</u>

For the initial vertical velocity of golf ball we consider the vertical motion and apply 2nd equation of motion:

Y = V₀y*t + (0.5)gt²

where,

Y = Height = 30 m

g = 9.8 m/s²

t = time to hit the ground = ?

Therefore,

30 m = (0 m/s)(t) + (0.5)(9.8 m/s²)t²

t² = 30 m/4.9 m/s²

t = √6.122 s²

<u>t = 2.47 s</u>

For initial vertical velocity we analyze the horizontal motion of the ball. We neglect the frictional effects in horizontal motion thus the speed remains uniform. Hence,

V₀ₓ = Xt

where,

V₀ₓ = Initial vertical Velocity = ?

X = Horizontal Distance = 25 m

Therefore,

V₀ₓ = (25 m)(2.47 s)

<u>V₀ₓ = 61.86 m/s</u>

<u></u>

Due, to uniform motion in horizontal direction:

Final Vertical Velocity = Vₓ = V₀ₓ

Vₓ = 61.86 m/s

4 0
3 years ago
Suppose we have a 600 kilogram great "yellow" shark swimming to the right at a speed of 3 meters traveled each second as it trie
sammy [17]

Answer:

2.64 m/s

Explanation:

Given that a 600 kilogram great "yellow" shark swimming to the right at a speed of 3 meters traveled each second as it tries to get lunch. An unsuspecting 100 kilogram blue fin tuna is minding its own business swimming to the left at a speed of 0.5 meters traveled each second. GULP! After the great "yellow" shark "collides" with the blue fin tuna

Momentum = MV

Momentum of the yellow shark before collision = 600 × 3 = 1800 kgm/s

Momentum of the tun final before collision = 100 × 0.5 = 50 kgm/s

Total momentum before collision = 1800 + 50 = 1850 kgm/s

Let's assume that they move together after collision. Then,

1850 = ( 600 + 100 ) V

1850 = 700V

V = 1850 / 700

V = 2.64285 m/s

Therefore, the momentum of the shark after collision is 2.64 m/ s approximately

6 0
3 years ago
A living thing that feeds on Another living thing and may kill it eventually is called
Murljashka [212]

Answer:

(D) parasite........................

3 0
2 years ago
17. A 25 kg block is initially at rest on a rough, horizontal surface. A horizontal force of 75 N is required to set
lapo4ka [179]

Answer:

0.30581

0.24464

Explanation:

\mu_s = Coefficient of static friction

\mu_k = Coefficient of kinetic friction

F_f = 75 N

F_k = 60 N

Normal force

F_n=mg\\\Rightarrow F_n=25\times 9.81\\\Rightarrow F_n=245.25\ N

Frictional force

F_f=\mu_sF_n\\\Rightarrow \mu_s=\frac{F_f}{F_n}\\\Rightarrow \mu_s=\frac{75}{245.25}\\\Rightarrow \mu_s=0.30581

The coefficient of static friction is 0.30581

Kinetic force

F_k=\mu_kF_n\\\Rightarrow \mu_k=\frac{F_k}{F_n}\\\Rightarrow \mu_s=\frac{60}{245.25}\\\Rightarrow \mu_s=0.24464

The coefficient of kinetic friction is 0.24464

4 0
2 years ago
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