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jeka94
3 years ago
13

Surgical repair of a muscle

Physics
1 answer:
Misha Larkins [42]3 years ago
6 0
A myoplasty is the surgical repair of a muscle.
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A wild pig attacks a hunter with a velocity of 5 m/s. At the instant when the pig is at a distance of 100 m the hunter shoots an
castortr0y [4]

Answer:

31.55 m/s

Explanation:

Let the initial velocity of the arrow is u metre per second.

Angle of projection, θ = 40 degree

range = 100 m

Use the formula for the range.

R = \frac{u^{2}Sin2\theta }{g}

100 = u^2 Sin(2 x 40) / 9.8

100 x 9.8 = u^2 Sin 80

u^2 = 995.11

u = 31.55 m/s

4 0
3 years ago
what can you say about the motion of an object if its speed time graph is a straight line parallel to the time axis ? Also draw
MaRussiya [10]

Answer:

It says about the motion and the graph of the object is stationary, basically travelling at the same speed at any time of the graph. It will never change.

Explanation:

To draw a diagram:

1. Draw an object and represent the speed as stationary and constant at any time.

5 0
2 years ago
You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and han
sleet_krkn [62]

Answer:

2.2nC

Explanation:

Call the amount by which the spring’s unstretched length L,

the amount it stretches while hanging x1

and the amount it stretches while on the table x2.

Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,

we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,

applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,

where ke is the Coulomb constant. Combining these,

we get q = √(mgx2(L+x2)²/x1ke =2.2nC

4 0
3 years ago
3. Your friend says your body is made up of more than 99.9999% empty space. What do you think?
Nesterboy [21]

Answer:

I would agree with the statement. it's not just the body, but everything that we see is almost 99.9999% empty space

8 0
3 years ago
At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107
dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

K = - V\cdot \frac{dP}{dV}

Where:

K - Bulk module, measured in pascals.

V - Sample volume, measured in cubic meters.

P - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

-\frac{K \,dV}{V} = dP

This resultant expression is solved by definite integration and algebraic handling:

-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP

-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}

\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}

\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }

The final volume is predicted by:

V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }

If V_{o} = 1\,m^{3}, P_{o} - P_{f} = -10132500\,Pa and K = 2.3\times 10^{9}\,Pa, then:

V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }

V_{f} \approx 0.996\,m^{3}

Change in volume due to increasure on pressure is:

\Delta V = V_{o} - V_{f}

\Delta V = 1\,m^{3} - 0.996\,m^{3}

\Delta V = 0.004\,m^{3}

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0
3 years ago
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