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pickupchik [31]

2 years ago

6

What is trapezord mean

1 answer:

igor_vitrenko [27]2 years ago

5 0

A quadrilateral with only one pair of parallel sides.
<span>a small carpal bone in the base of the hand, articulating with the metacarpal of the index finger.
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What relationship do you see between a star colour and temperature

Andrei [34K]

Answer:

Stars emit colors of many different wavelengths, but the wavelength of light where a star's emission is concentrated is related to the star's temperature - the hotter the star, the more blue it is; the cooler the star, the more red it is

5 0

2 years ago

6. An aircraft is is travelling along a runway at a Velocity of 25m/s in It accelerates at a rate of 4m/s² for a distance of 750

zysi [14]

Answer:

Take-off velocity = v = 81.39[m/s]

Explanation:

We can calculate the takeoff speed easily, using the following kinematic equation.

$v_{f}^{2}=v_{i}^{2} +2*a*x$

where:

a = acceleration = 4[m/s^2]

x = distance = 750[m]

vi = initial velocity = 25 [m/s]

vf = final velocity

$v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]$

7 0

2 years ago

According to Kepler's Second Law the radius vector drawn from the Sun to a planet Multiple Choice is the same for all planets. s

Mademuasel [1]

Answer:

sweeps out equal areas in equal times.

Explanation:

As we know that there is no torque due to Sun on the planets revolving about the sun

so we will have

$\tau_{net} = 0$

now we have

$\frac{dL}{dt}= 0$

now we also know that

$Area = \frac{1}{2}r^2d\theta$

so rate of change in area is given as

$\frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt}$

so we will have

$\frac{dA}{dt} = \frac{1}{2}r^2\omega$

$\frac{dA}{dt} = \frac{L}{2m}$

since angular momentum and mass is constant here so

all planets sweeps out equal areas in equal times.

4 0

3 years ago

A wire of length 5mm and Diameter 2m extends by 0.25 when a force of 50N was use. calculate the

bazaltina [42]

Answer and Explanation:

Data provided in the question

Force = 50N

Length = 5mm

diameter = 2.0m = $2\times 10^{-3}$

Extended by = 0.25mm = $0.25\times 10^{-3}$

Based on the above information, the calculation is as follows

a. The Stress of the wire is

$= \frac{force\ applied}{area\ of \ circle}$

here area of circle = perpendicular to the are i.e cross-sectional i.e

= $\frac{\pi d^{2}}{4}$

= $\frac{\pi(2\times 10^{-3})^2}{4}$

Now place these above values to the above formula

$= \frac{4\times 50}{\pi\times 4 \times 10^{-6}} \\\\ = \frac{50}{\pi}$

= 15.92 MPa

As 1Pa = 1 by N m^2

So,

MPa = 10^6 N m^2

b. Now the strain of the wire is

$= \frac{Change\ in\ length}{initial\ length} \\\\ = \frac{0.25\times 10^{-3}}{5}$

= $5 \times 10^{-5}$

3 0

3 years ago

What condition is necessary for the sustained flow of water in a pipe?

lianna [129]

Pressure difference (voltage)
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5 0

3 years ago