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STatiana [176]
3 years ago
12

a 2.00 kg object is moving east at 4.00 m/s when it collides with a 6 kg object that is initially at rest. after the collision t

he larger object moves east at 1 m/s. what is the final velocity of the smaller object after the collision
Physics
1 answer:
baherus [9]3 years ago
6 0

Answer:

v = 1.00 m/s east

Explanation:

Conservation of momentum

Let east be the positive direction

2.00(4.00) + 6.00(0.00) = 2.00(v) + 6.00(1.00)

v = 1.00 m/s east

The two items have stuck together.

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A hydraulic lift raises a 4000 kg automobile when a 500N force is applied to the smaller piston. If the smaller piston has an ar
MariettaO [177]

Answer:

The cross-sectional area of the larger piston is 800 cm².

Explanation:

Given;

mass of the automobile, m = 4000 kg

force applied on the small piston, F₁ = 500 N

area of the smaller piston, A₁ = 10 cm²

load lifted by the larger piston, F₂ = 4000 x 10 = 40,000 N

Pressure experienced by each piston is given as;

\frac{F_1}{A_1} = \frac{F_2}{A_2}

Where;

A₂ is the cross-sectional area of the larger piston

\frac{F_1}{A_1} = \frac{F_2}{A_2} \\\\A_2 = \frac{F_2A_1}{F_1} \\\\A_2 = \frac{40,000 \ \times\ 10}{500} \\\\A_2=800 \ cm^2

Therefore, the cross-sectional area of the larger piston is 800 cm².

8 0
3 years ago
2 characteristics of constant speed
svetoff [14.1K]

Answer:

it has no acceleration

Explanation:

8 0
3 years ago
The tension of a guitar string is increased by 40%. By what factor odes the fundamental frequency of vibration change? a. 1.13 b
bogdanovich [222]

Answer:

<h3> b. 1.18</h3>

Explanation:

The fundamental frequency in string is expressed as;

F1 = 1/2L√T/m .... 1

L is the length of the string

T is the tension

m is the mass per unit length

If the tension is increased by 40%, the new tension will be;

T2 = T + 40%T

T2 = T + 0.4T

T2 = 1.4T

The new fundamental frequency will be;

F2 = 1/2L√1.4T/m ..... 2

Divide 1 by 2;

F2/F = (1/2L√1.4T/m)/1/2L√T/m)+

F2/F = √1.4T/m ÷ √T/m

F2/F = √1.4T/√m ×√m/√T

F2/F = √1.4T/√T

F2/F = 1.18√T/√T

F2/F = 1.18

F2 = 1.18F

Hence the fundamental frequency of vibration changes by a factor of 1.18

8 0
3 years ago
I need help with this please
kondaur [170]

Answer:

1. The respiratory system functions when our involuntary nervous system sends impulses to the muscles in the diaphragm; thereby, causing the lungs to expand and contract.

2. The respiratory system oxygenates the blood which is vital for bodily function as oxygenated blood is carried from your lungs to the left side of your heart, to be circulated throughout the body. Furthermore deoxygenated blood is carried back to the right side of your heart to get oxygenated once more.

3. The other body systems that are crucial for the lungs to function are the nervous system and the muscular system.

4. without the raspatory system the body wouldn't receive any oxygen and the brain would slowly die. therefore, without the brain the heart would stop functioning and atrophy etc.

Explanation:

6 0
3 years ago
A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 m/s^{2}

           L = length of the rod = 0.2 m

Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

              x = spring extension

The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

             = 0.1236 m

       k = 25 N/m

So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0
3 years ago
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