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3 years ago

A gazelle is running at 17.46 m/s. He sees a lion and accelerates at -1.49 m/s/s,

Physics

2 answers:

natka813 [3]3 years ago

7 0

Answer=21.0211m/s

Explanation v=17.46/

Vsevolod [243]3 years ago

5 0

Answer:

V=21.0211m/s

Explanation:

Use V=vi+at

So, V=17.46m/s+(1.49m/ $s^{2}$ )(2.39s)= 21.0211m/s

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Find the momentum (in kg.m/s) of a helium nucleus having a mass of 6.68 x 10^-27 kg that is moving at 0.386c.

igomit [66]

Answer:

$7.73544\times 10^{-21}kgm/sec$

Explanation:

We have given mass of helium nucleus $m=6.68\times 10^{-27}kg$

Velocity of helium nucleus $v=0.386c=0.386\times 3\times 10^{8}m/sec=1.158\times 10^{8}m/sec$

Momentum of the helium nucleus is given by $P=mv$ where m is mass and v is velocity

So $P=mv=6.68\times 10^{-27}\times 1.158\times 10^{8}=7.73544\times 10^{-21}kgm/sec$

6 0

3 years ago

A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant

djverab [1.8K]

Answer:

0.005 V

Explanation:

We are given that

Initial circumference of circular loop=C=165 cm

Rate of circumference, $\frac{dC}{dt}=12 cm/s$

Magnetic field,B=0.5 T

We have to find the induced emf at time t=9 s

We know that induced amf,E= $\frac{Bd(A)}{dt}$

Area of circular coil,A= $\pi r^2$

$E=B\frac{d(\pi r^2)}{dt}=B(2\pi r)\frac{dr}{dt}$

Circumference of circular coil,C= $2\pi r$

$165=2\pi r$

$r=\frac{165}{2\pi}$

$\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}\times (12)=\frac{6}{\pi} cm/s=\frac{6\times 10^{-2}}{\pi} m/s$

Radius of coil at time t=9 s

$r=\frac{165}{2\pi}-(\frac{6}{\pi}\times 9)=9.08 cm=9.08\times 10^{-2} m$

$1 m=100 cm$

E= $-0.5(2\pi\times 9.08\times 10^{-2})\times \frac{6\times 10^{-2}}{\pi}=-0.005 V$

Magnitude of induced emf=0.005 V

4 0

3 years ago

Read 2 more answers

Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th

Mariulka [41]

Answer:

a) 0.0288 grams

b) $2.6*10^{-10} J/kg$

Explanation:

Given that:

A typical human body contains about 3.0 grams of Potassium per kilogram of body mass

The abundance for the three isotopes are:

Potassium-39, Potassium-40, and Potassium-41 with abundances are 93.26%, 0.012% and 6.728% respectively.

Thus; a person with a mass of 80 kg will posses = 80 × 3 = 240 grams of potassium.

However, the amount of potassium that is present in such person is :

0.012% × 240 grams

= 0.012/100 × 240 grams

= 0.0288 grams

the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is calculate as follows:

First the Dose in (Gy) = $\frac{energy \ absorbed }{mass \ of \ the \ body}$

= $\frac{1.10*10^6*1.6*10^{-14}}{80}$

= $2.2*10^{-10} \ J/kg$

Effective dose (Sv) = RBE × Dose in Gy

Effective dose (Sv) = $1.2 *2.2*10^{-10} \ J/kg$

Effective dose (Sv) = $2.6*10^{-10} J/kg$

5 0

4 years ago

The combined focal length of two thin lens is 24 cm and the focal length of one converging lens is 8

qwelly [4]

Answer: f = -12 cm

Explanation: <u>Combined</u> <u>lenses</u> is an array of simple lenses with a common axis. The combination is useful for correction of optical aberrations which cannot be corrected by simple lenses.

When two lenses are in contact and are thin, focal lengths are related as:

$\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}}$

If there is a distance between the lenses, the focal length will be:

$\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}} -\frac{d}{f_{1}f_{2}}$

Since the lenses in the question above are thin and in contact, the focal length of one of them will be:

$\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}}$

$\frac{1}{f_{2}} =\frac{1}{f_{1}} -\frac{1}{F}$

$\frac{1}{f_{2}} =\frac{1}{8} -\frac{1}{24}$

$\frac{1}{f_{2}} =\frac{-2}{24}$

$f_{2}=$ -12

The focal length of the other lens is -12 cm, with the negative sign meaning it's a converging lens.

7 0

3 years ago

Which statement correctly describes variable stars?

Kryger [21]

Answer:

Explanation:

The Period-Luminosity relationship tells us that luminosity increases with the period, and of course the more luminosity a star has the more far away they can be seen, so from this we know that:

A) False since lower luminosities can be observed when they are close.

B) False since longer periods means higher luminosities

C) False since lower luminosities can be observed when they are close.

D) True: Variable stars with shorter periods have lower luminosities, so they can only be observed when they are close.

5 0

3 years ago