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3 years ago

When we dissolve a substance in water, a -------- is formed

Physics

2 answers:

damaskus [11]3 years ago

5 0

Answer:

When a substance dissolves in water, you can’t see it anymore, it’s still there, but has mixed with the water to make a transparent liquid called a solution. The bonds in salt compounds are called ionic because they both have an electrical charge—the chloride ion is negatively charged and the sodium ion is positively charged.

Explanation:

Maru [420]3 years ago

3 0

Answer:

solution

Explanation:

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A baseball pitcher throws a baseball horizontally at a linear speed of 49.4 m/s. Before being caught, the baseball travels a hor

RideAnS [48]

Answer:

$v = 3.61 m/s$

Explanation:

As we know that ball travels horizontal distance of 24.7 m with uniform speed 49.4 m/s

so we will have

$t = \frac{x}{v_x}$

$t = \frac{24.7}{49.4}$

$t = 0.5 s$

now in the same time ball is turned by angle

$\theta = 52.7 rad$

now we know that

$\theta = \omega t$

$52.7 = \omega (0.5)$

$\omega = 105.4 rad/s$

now the tangential speed of a point at equator is given as

$v = r\omega$

$v = 0.0343(105.4)$

$v = 3.61 m/s$

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3 years ago

Scientists were studying three different embryos to determine their relationship to humans. The notes about each embryo are show

Sholpan [36]

Answer:

The awnser is B

Explanation:

I took the test.

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3 years ago

Read 2 more answers

Suppose a car manufacturer tested its cars for front-en4 collisions by hauling them up on a crane and dropping then; from a cert

Brrunno [24]

Answer:

Generally from third equation of motion we have that

$v^2 = u^2 + 2a[s_i - s_f]$

Here v is the final speed of the car

u is the initial speed of the car which is zero

$s_i$ is the initial position of the car which is certain height H

$s_i$ is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

$v^2 = 0 + 2g[H - 0]$

=> $v = \sqrt{ 2 g H}$

$H = 9.86 \ m$

Explanation:

Generally from third equation of motion we have that

$v^2 = u^2 + 2a[s_i - s_f]$

Here v is the final speed of the car

u is the initial speed of the car which is zero

$s_i$ is the initial position of the car which is certain height H

$s_i$ is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

$v^2 = 0 + 2g[H - 0]$

=> $v = \sqrt{ 2 g H}$

When $v = 50 \ km/h = \frac{50 *1000}{3600} = 13.9 \ m/s$ we have that

$13.9 = \sqrt{ 2 g H}$

=> $H = \frac{13.9^2}{2 * 9.8}$

=> $H = 9.86 \ m$

6 0

3 years ago

Two metra trains approach each other on separate but parallel tracks. one has a speed of 90 km/hr, the other 80 km/hr. initially

Gennadij [26K]

The trains take 57.4 s to pass each other.

Two trains A and B move towards each other. Let A move along the positive x axis and B along the negative x axis.

therefore,

$v_A=90 km/h\\ v_B=-80 km/h$

The relative velocity of the train A with respect to B is given by,

$v_A_B=v_A-v_B\\ =(90km/h)-(-80km/h)\\ =170km/h$

If the train B is assumed to be at rest, the train A would appear to move towards it with a speed of 170 km/h.

The trains are a distance d = 2.71 km apart.

Since speed is the distance traveled per unit time, the time taken by the trains to cross each other is given by,

$t= \frac{d}{v_A_B}$

Substitute 2.71 km for d and 170 km/h for $v_A_B$

$t= \frac{d}{v_A_B}\\ =\frac{2.71 km}{170 km/h} \\ =0.01594 h$

Express the time in seconds.

$t=(0.01594h)(3600s/h)=57.39s$

Thus, the trains cross each other in 57.4 s.

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2 years ago

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