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Snezhnost [94]
3 years ago
5

A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th

e time the clam was dropped?
Physics
1 answer:
ankoles [38]3 years ago
4 0
<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = 3 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 3 + 0.5 x 9.81 x 3²

                      s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

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A box is held at rest by two ropes that form 30° angles with the vertical. The tension T in either rope is 42 N. What is the wei
iragen [17]
<h3><u>Answer;</u></h3>

= 73 N

<h3><u>Explanation</u>;</h3>

Using the formula

2 T cos(30°) = w

Where; T is the tension on each string, while w is the weight of the box given by mg

Therefore;

W = 2Tcos 30°

    = 2 × 42 cos 30°

    = 84 cos 30°

    = 72.74

<u>   ≈ 73 N</u>

7 0
3 years ago
A 2.4 kg block is dropped onto a spring and platform of negligible mass. The block is released
statuscvo [17]

The speed of the block when the compression is 15 cm is 9.85 m/s.

The given parameters;

  • <em>mass of the block, m = 2.4 kg</em>
  • <em>height of the block, h =  5 m</em>
  • <em>compression of the spring, x = 25 cm = 0.25 m</em>

The spring constant is calculated as follows;

F = kx\\\\mg = kx\\\\k = \frac{mg}{x} \\\\k = \frac{2.4 \times 9.8}{0.25} \\\\k = 94.08 \ N/m

The speed of the block when the compression is 15 cm can be determined by applying the principle of conservation of energy;

\Delta K.E = \Delta P.E\\\\\frac{1}{2} m(v^2  - v_{0 }^2 ) = mgh - \frac{1}{2} kx^2\\\\\frac{1}{2} mv^2  = mgh -   \frac{1}{2} kx^2\\\\mv^2   = 2mgh - kx^2\\\\v^2 = \frac{2mgh - kx^2}{m} \\\\v = \sqrt{\frac{2mgh - kx^2}{m}} \\\\v = \sqrt{\frac{(2 \times 2.4 \times 9.8 \times 5) - (94.08 \times 0.15^2)}{2.4}} \\\\v = 9.85 \ m/s

Thus, the speed of the block when the compression is 15 cm is 9.85 m/s.

Learn more here:brainly.com/question/14289286

8 0
3 years ago
A three branch parallel circuit has resistors of 27 W, 56 W, and 15 W. What is the total resistance?
Mekhanik [1.2K]

Answer:8.2 Ω

Explanation:

I ASSUME you mean Ω (ohms) and not W(atts)

Re = 1/(1 / 27 + 1/56 + 1/15) = 8.2263329... ≈ 8.2 Ω

6 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
kiruha [24]
I believe the answer is up and down :)
7 0
3 years ago
A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following case
Murljashka [212]

Answer:

The answer is below

Explanation:

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the

cable for the following cases:

a. The load moves downward at a constant velocity

b. The load accelerates downward at a rate 0.4 m/s??

C. The load accelerates upward at a rate 0.4 m/s??

Solution:

Acceleration due to gravity (g) = 10 m/s²

a) Given that the mass of the crane (m) is 140 kg. If the load moves downward, the tension (T) is given by:

mg - T = ma

Since the load has a constant velocity, hence acceleration (a) = 0. Therefore:

mg - T = m(0)

mg - T = 0

T = mg

T = 140(10) = 1400 N

T = 1400 N

b)  If the load moves downward, the tension (T) is given by:

mg - T = ma

T = mg - ma = m(g - a)

T = 140(10 - 0.4) = 140(9.96) = 134.4

T = 134.4 N

c)  If the load moves upward, the tension (T) is given by:

T - mg = ma

T = ma + mg = m(a + g)

T = 140(0.4 + 10) = 140(10.4)

T = 145.6 N

2) To find the distance (s) if the load move from rest (u= 0) and accelerates for 20 seconds (t = 20). We use:

s = ut + (1/2)gt²

s = 0(20) + (1/2)(10)(20)²

s = 2000 m

7 0
3 years ago
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