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hammer [34]
3 years ago
14

How much net force is needed to accelerate a 15 kg mass at 2.8 m/s^2

Physics
1 answer:
Lera25 [3.4K]3 years ago
4 0
Refer to the 2th Law of Newton

F = m. a

F = 15 x 2.8 = 42 N
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You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
Anarel [89]

Answer:

R = 24.3 m

Explanation:

As we know that the orbital speed is given as

v = \sqrt{\frac{GM}{R + h}}

here we know that

v = 5500 m/s

R = 4.48 \times 10^6 m

h = 630 km

now we have

5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}

M = 2.34 \times 10^24 kg

now acceleration due to gravity of planet is given as

a = \frac{GM}{R^2}

a = \frac{(6.6 \times 10^{-11})(2.34 \times 10^{24})}{(4.48\times 10^6)^2}

a = 7.7 m/s^2

now range of the projectile on the surface of planet is given as

R = \frac{v^2 sin2\theta}{g}

R = \frac{14.6^2 sin(2\times 30.8)}{7.7}

R = 24.3 m

3 0
3 years ago
Coherent light of wavelength 525 nm passes through two thin slits that are 4.15×10^(−2) mm apart and then falls on a screen 7
IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

sin \theta=\frac{n \lambda}{d}

where

n is the order of the maximum

\lambda is the wavelength

a is the distance between the slits

In this problem,

n = 5

\lambda=525 nm =5.25\cdot 10^{-7} m

a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m

So we find

\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}

And given the distance of the screen from the slits,

D=75.0 cm = 0.75 m

The distance of the 5th  bright fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}

And the distance of the 8th dark fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm

5 0
3 years ago
If two connected points objects pass through the same set of three points, the shapes created by each will be identical, regardl
NeX [460]

Answer:

This is false

Explanation:

This is why the answer to this question is false. If these types of objects (2 points connected) should pass through same sets of 3 points, irrespective of the order that each object was plotted, we will not get identical shapes created.

The shape that is going to be created will be dependent on the pattern/order that was involved in the plotting. If it was identical, then we will have identical shapes. But if the order during plotting is different then we will have different shapes created.

Thank you!

5 0
3 years ago
Why won't a very bright beam of red light impart more energy to an ejected electron than a feeble beam of violet light?
bearhunter [10]
This is related to the energy carried by photons of light the energy of each photon is proportional to the frequency of the light since red light has a lower frequency then violet light and photons of red light carry less energy than the photons of violet light as a result the red protons eject electrons that have less energy than the ejected electrons by Violet photons
3 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
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