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2 years ago

Which of the following is the suffix that means “berry shaped?"

Physics

1 answer:

Ipatiy [6.2K]2 years ago

7 0

Answer:

Coccus

Explanation:

Berry shaped bacterium. Streptococci. Berry shaped bacteria occurring in twisted chains

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it would take 100s for the train to crossthe bridge

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2 years ago

How long would this acceleration last

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2 years ago

Beryllium has a charge of 2, and bromine has a charge of –1. which is the best name for the ionic bond that forms between them?

Otrada [13]

The best name for the ionic bond that forms between them is Beryllium Bromide.

We have been provided with data,

Beryllium charge, q = 2

Bromine charge, q = -1

As we know the valance electron of Be is +2 and the valance electron of bromine is -1. Since one is metallic and the other is non-metallic.

Now, when they combine they exchange valance electron, and bromine change into bromide so they form Beryllium Bromide.

So, the best name for the ionic bond that forms between them is Beryllium Bromide.

Learn more about ionic bonds here:

brainly.com/question/21464719

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8 0

2 years ago

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

2 years ago