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3 years ago

13

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1 answer:

alexgriva [62]3 years ago

6 0

Answer:

Get your answer for $1 from www.gotit-pro.com

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A lion with a mass of 190 kg is chasing a gazelle with a mass of 15 kg. The distance between the lion and the gazelle is 2 meter

Vadim26 [7]

Answer:

Explanation:

with a mass of 190 kg is chasing a gazelle with a mass of 15 kg. the distance between the lion and the gazelle is 2 meters. how much gravitational force does the lion exert on the gazelle? ... In which figure is de bc ? a. figure 1 b. figure 2 c. figure 3 d. figure 4... Mathematics 3 21.06.2019 15:00.

7 0

4 years ago

Does anyone know how to solve this??

Scrat [10]

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wtjfavyw

7 0

3 years ago

The Berlin marathon is one of the fastest marathons courses; many world records have been set there. In 2014 Dennis Kimetto ran

Mandarinka [93]

Answer:

about 2000 klm per min

Explanation:

7 0

3 years ago

claculate the pressure exerted by the water on the bottom of a deep dam of 12m from its surface .(density of water =1000kg/m sqa

Rudiy27

Explanation:

We know that,

$hydrostatic \: pressure (p) = \alpha hg$

Where,

$\alpha = density \: of \: water \: = 1000$

h = Height at which pressure is to be calculated

$p = 1000 \times 12 \times 9.8 = 117600 \:$

8 0

3 years ago

An α-particle has a charge of +2e and a mass of 6.64 × 10-27 kg. It is accelerated from rest through a potential difference that

zzz [600]

Answer with Explanation:

We are given that

Charge on alpha particle=q=2 e= $2\times 1.6\times 10^{-19} C$

1 e= $1.6\times 10^{-19} C$

Mass of alpha particle=m= $6.64\times 10^{-27} kg$

Potential difference,V= $1.97\times 10^6 V$

Magnetic field,B=3.49 T

a.Speed of alpha particle=v= $\sqrt{\frac{2 qV}{m}}$

By using the formula

$v=\sqrt{\frac{2\times 2\times 1.6\times 10^{-19}\times 1.97\times 10^6}{6.64\times 10^{-27}}$

$v=1.38\times 10^7 m/s$

b.Magnetic force,F $=qvB=2\times 1.6\times 10^{-19}\times 1.38\times 10^7\times 3.49=1.5\times 10^{-11} N$

F= $1.5\times 10^{-11} N$

c.Radius of circular path, r= $\frac{mv^2}{F}$

$r= \frac{6.64\times 10^{-27}\times (1.38\times 10^7)^2}{1.5\times 10^{-11}}$

r=0.084 m

5 0

3 years ago