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Sedbober [7]
3 years ago
14

6. The slope of a speed-time linear graph is

Physics
1 answer:
Tpy6a [65]3 years ago
6 0

Answer:

The slope of SPEED- TIME graph represents the acceleration of the particular body. If the Speed-Time is a straight line then the acceleration is zero. ... More than saying it as speed time graph, we can say it as Velocity Time graph, since acceleration is a vector quantity.

Explanation:

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The position of an ant and a spider in a room are A(2,3,5) and B(6,0,8) determine the distance
marta [7]

Answer:

|d|  =  \sqrt{ {(x_{2} -x_{1})}^{2} + {(y_{2} -y_{1})}^{2} + {(z_{2} -z_{1})}^{2}}  \\  =  \sqrt{ {(6 - 2)}^{2} + {(3 - 0)}^{2} +{(5 - 8)}^{2}  }  \\  =  \sqrt{16 + 9 + 9}  \\   = \sqrt{34}

4 0
3 years ago
What creates the energy of the Sun?
Svet_ta [14]

Answer:

C

Explanation:

During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons. Hydrogen nuclei fuse to form one helium atom.

5 0
3 years ago
Why do people make this to the seashore?
Nadya [2.5K]

Answer:

Protecting the Seashore

Seashore Conservation Code

When investigating a seashore it is important to remember that the welfare of the plants and animals must come first. If seaweeds are attached to rocks do not try to pull them off, just examine them where they are growing. Handle animals with great care, returning them to the place where they were found. Anemones and limpets are normally firmly stuck to rocks and any attempt to remove them may result in their death. Replace any large stones that are removed – animals that live on the underside soon die if left exposed.

Pollution Problems

Seas and seashores are under continual threat from pollution. Here are some examples:-

oil rigOil spills, either accidental or deliberate, have a devastating effect on marine life, especially seabirds. Black tar-like oil washed up on a rocky shore would obviously completely destroy the entire community of living things. Chemicals used to disperse oil spills on a shore may effectively clean up the oil, but they can also cleanse the shore of life! The best policy for marine life is to physically remove as much oil as possible and leave the rest to degrade naturally. A rocky shore will recover much more quickly if chemicals are not used.

Sewage and litter also upsets seashore ecosystem, making them unsightly and perhaps contaminated. Some litter can be dangerous to wildlife.

Toxic chemicals and radioactive waste - a cocktail of toxic chemicals have found their way into the sea over the years, affecting both marine and human life. For many years potentially deadly radioactive waste from nuclear power stations and other sources was disposed of in the sea, and beaches have been contaminated by accidental leakages from nuclear power stations.

Explanation:

5 0
3 years ago
Read 2 more answers
Froghopper insects have a typical mass of around 11.1 mg 11.1 mg and can jump to a height of 55.3 cm. 55.3 cm. The takeoff veloc
Molodets [167]

Answer:

a = 2710m/s²

Explanation:

See attachment below.

4 0
3 years ago
Read 2 more answers
Light of wavelength 550 nm falls on a
Brums [2.3K]

Answer:

The first diffraction maximum fringe will be at approximately 2.7 meters from the central maximum.

Explanation:

We can describe single slit diffraction phenomenon with the equation:

a\sin\theta=m\lambda (1)

with θ the angular position of the minimum of order m respect the central maximum, a the slit width and λ the wavelength of the incident light. Because the distances between the first minima and the central maximum (y_{m}) are small compared to the distance between the screen and the slit (x), we can approximate \sin\theta\approx\tan\theta=\frac{y}{x}, using this on (1):

a\frac{y_{m}}{x}=m\lambda

solving for y

y_{m}= \frac{mx\lambda}{a}

Note that y_{m}is the distance between a minimum and the central maximum but we need the position of a maximum not a minimum, here we can use the fact that a maximum is approximately between two minima, so the first diffraction maximum fringe is between the minima of order 1 and 2, so we should find y_{1}, y_{2} add them and divide by two:

y_{1}= \frac{(1)(10.0m)(550\times10^{-9}\,m)}{3.00\times10^{-6}\,m}

y_{1}= 1.8 m

y_{2}= \frac{(2)(10.0m)(550\times10^{-9}\,m)}{3.00\times10^{-6}\,m}

y_{1}= 3.6 m

maximum = \frac{1.8+3.6}{2}=2.7m

7 0
3 years ago
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