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avanturin [10]
3 years ago
12

What happens to the density of gas when it compressed

Physics
1 answer:
Stella [2.4K]3 years ago
3 0
When a gas is compressed, its particles are pushed closer together so it is more dense.
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As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh
marshall27 [118]

Answer:

The resultant velocity is  v_t=10 knots

Explanation:

Apply the law of conservation of momentum

     M_L *v_L + M_f * V_f = (M_L + M_f) v_t

Where M_L is the mass of the Luxury Liner = 40,000 ton

            v_L is the velocity of Luxury Liner = 20 knots due west

            M_f mass of freighter = 60,000

           v_f is the velocity of freighter = 10 knots due north

Apply the law of conservation of momentum toward the the west direction

         v_f = 0 \ knots

So the equation would be

              M_L *v_L = (M_L + M_f) v_t

Substituting values

            40000*20 = (40000+ 60000)v_t_w

Where v_t_w the final velocity due west

Making v_t_w the subject

          v_t_w = \frac{40,000* 20}{(40000 + 60000)}

                = 8 \ knots

Apply the law of conservation of momentum toward the the north direction          

          v_L = 0 \ knots

So the equation would be

           M_f *v_f = (M_L + M_f) v_t_n

Where v_t_n the final velocity due north

     Making v_t_n the subject

          v_t_n = \frac{60,000* 10}{(40000 + 60000)}

                = 6 \ knots

The resultant velocity is

       v_t = \sqrt{v_t_w^2 + v_t_n^2}

            = \sqrt{8^2 +6^2}

           v_t=10 knots

8 0
3 years ago
A rod of length L is hinged at one end. The moment of inertia as the rod rotates around that hinge is ML2 /3. Suppose a 2.00-m r
Stels [109]

To solve this problem we will apply the concepts related to the moment of inertia and Torque, the latter both its translational and rotational expression.

According to the information given the moment of inertia of the body would be

I = \frac{1}{3} mL^2

Replacing we have

I = \frac{1}{3} (3kg)(2m)^2

I = 4 kg * m^2

Now the translational torque would be the product between the force applied (Its own Weight) and the distance (Its center of mass at the middle)

\tau = F*r

\tau = mg (\frac{L}{2})

\tau = (3)(9.8)(\frac{2}{2})

\tau = 29.4N\cdot m

Now the rotational torque is defined as the product between the moment of inertia and the angular acceleration, then,

\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}

Replacing,

\alpha = \frac{29.4}{4}

\alpha =7.35 rad / s^2

Therefore the angular acceleration is 7.35rad/s^2

4 0
3 years ago
True or false: objects fall toward earth at a rate of 9.8 m/s because of centripetal force.
Citrus2011 [14]
Sadly, no. The statement kind of has some appropriate words in it, but it's badly corrupted. Objects don't fall to Earth at a rate of 9.8 m/s, and the force that accelerates them downward is not a centripetal one.
4 0
3 years ago
An emf is induced by rotating a 1060 turn, 20.0 cm diameter coil in the Earth's 5.25 ✕ 10−5 T magnetic field. What average emf (
lara31 [8.8K]

Answer:

The average emf induced in the coil is 175 mV

Explanation:

Given;

number of turns of the coil, N = 1060 turns

diameter of the coil, d = 20.0 cm = 0.2 m

magnitude of the magnetic field,  B = 5.25 x 10⁻⁵ T

duration of change in field, t = 10 ms = 10 x 10⁻³ s

The average emf induced in the coil is given by;

E = N\frac{\delta \phi}{dt} \\\\E = N\frac{\delta B}{\delta t}A

where;

A is the area of the coil

A = πr²

r is the radius of the coil = 0.2 /2 = 0.1 m

A = π(0.1)² = 0.03142 m²

E = \frac{NBA}{t} \\\\E = \frac{1060*5.25*10^{-5}*0.03142}{10*10^{-3}} \\\\E = 0.175 \ V\\\\E = 175 \ mV

Therefore, the average emf induced in the coil is 175 mV

3 0
3 years ago
(True or False Statement)
sertanlavr [38]
False is correct......
4 0
3 years ago
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