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Law Incorporation [45]
4 years ago
14

What is the momentum of A 200 kg shark swimming at 15 m/s? bruh please help im failing and dont understand this:)

Physics
1 answer:
igomit [66]4 years ago
8 0

Answer:

p = 3000 kg-m/s.

Explanation:

Given that,

Mass of shark, m = 200 kg

Speed of the shark, v = 15 m/s

We need to find the momentum of the shark. It can be calcullated as follows :

<h2>p = mv</h2>

p=200\ kg\times 15\ m/s\\\\p=3000\ kg-m/s

So, the momentum of the shrak is 3000 kg-m/s.

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Are Quantum Physics, Quantum mechanics,Quantum Engagement same?
jek_recluse [69]

Answer:

<h3>The topic of quantum entanglement is at the heart of the disparity between classical and quantum physics: entanglement is a primary feature of quantum mechanics lacking in classical mechanics. ... In the case of entangled particles, such a measurement will affect the entangled system as a whole</h3>

Explanation:

3 0
4 years ago
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astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating o
devlian [24]

Answer:

The right answer is:

(a) 63.83 kg

(b) 0.725 m/s

Explanation:

The given query seems to be incomplete. Below is the attachment of the full question is attached.

The given values are:

T = 3 sec

k = 280 N/m

(a)

The mass of the string will be:

⇒ T=2 \pi\sqrt{\frac{m}{k} }

or,

⇒ m=\frac{k T^2}{4 \pi^2}

On substituting the values, we get

⇒     =\frac{280\times (3)^2}{4 \pi^2}

⇒     =\frac{280\times 9}{4\times (3.14)^2}

⇒     =68.83 \ kg

(b)

The speed of the string will be:

⇒  \frac{1}{2}k(0.4)^2=\frac{1}{2}k(0.2)^2+\frac{1}{2}mv^2

then,

⇒             v=\sqrt{\frac{k((0.4)^2-(0.2)^2)}{m} }

On substituting the values, we get

⇒                =\sqrt{\frac{280\times ((0.4)^2-(0.2)^2)}{63.83} }

⇒                =\sqrt{\frac{280(0.16-0.04)}{63.83} }

⇒                =\sqrt{\frac{280\times 0.12}{63.83} }

⇒                =0.725 \ m/s

4 0
3 years ago
Simple Pendulum: A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light
SIZIF [17.4K]

Answer:

The correct answer is "4.443 sec".

Explanation:

Given:

Mass of child,

= 34 kg

Mass of swing,

= 18 kg

Length,

= 4.9 m

The time period of pendulum will be:

T = 2 \pi \sqrt{4g}

  = 2 \pi \sqrt{\frac{4.9}{9.8} }

  = 4.443 \ sec  

5 0
3 years ago
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A cylindrical blood vessel is partially blocked by the buildup of plaque. At one point, the plaque decreases the diameter of the
densk [106]

Answer:10.4 times of initial velocity

Explanation:

Given

Diameter reduced by 69 %

it approaches with velocity v_0

suppose its velocity is v during blocked passage

suppose d is the initial diameter and d_2 diameter is

d_2=d(1-0.69)

d_2=0.31 d

A_2=\frac{\pi d_2^2}{4}

As flow is constant

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d^2v_0=d_2^2v

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6 0
3 years ago
A low C (f=65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and
WITCHER [35]

Answer:

T = 676 N

Explanation:

Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g/m^{2} = 0.005 kg

A stationary wave that is set up in the string has a frequency of;

f = \frac{1}{2L}\sqrt{\frac{T}{M} }

⇒      T = 4L^{2}f^{2}M

Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.

But M = L × ρ = (2 × 0.005) = 0.01 kg/m

T = 4 × 2^{2} ×65^{2} × 0.01

   = 4 × 4 ×4225 × 0.01

   = 676 N

Tension of the wire is 676 N.

4 0
3 years ago
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