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2 years ago

Will a substance that contains an o-h group dissolve in water?

Physics

1 answer:

Stella [2.4K]2 years ago

4 0

Usually, it increases the solubility in water.

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A metre rule is used to measure a length. Which reading is shown to the nearest millimetre? A 0.7m B 0.76m C 0.761m D 0.7614m

Cloud [144]

Answer:

none

Explanation:

~both of them show to the nearest metre.

~millimeter has (mm) unit eg 0.7mm

4 0

2 years ago

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A body is moving with a uniform acceleration of 2m/s² what does it mean?

Nastasia [14]

Answer:

the body has energy due to its constant motion. it means it moves in a uniform acceleration which has zero velocity

Explanation:

Uniform or constant acceleration is a type of motion in which the velocity of an object changes by an equal amount in every equal time period.

7 0

2 years ago

A 388 Hz tuning fork is resonating in a closed tube on a warm day when the speed of sound is 346 m/s. What is the length of the

Marizza181 [45]

Answer:

Explanation:

because u are subtracting if this is from flvs that is what i did and it was right

5 0

2 years ago

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A bicycle pump contains 200 cm3 of air and is connected to a bicycle tyre. The volume of the tyre is 800 cm3. The pressure of th

Aleksandr [31]

Answer:

The total volume of the air is 1000 cubic centimeters.

Explanation:

Since the bicycle pump and the bicycle tyre have the same pressure, then the total volume of the air is the sum of the volume of each element, then we translate this into the following artihmetical expression:

$V = 200\,cm^{3}+800\,cm^{3}$

$V = 1000\,cm^{3}$

The total volume of the air is 1000 cubic centimeters.

7 0

2 years ago

A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas

Radda [10]

Answer:

change in momentum, $\Delta p=7.65 \,kg.m.s^{-1}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

Average Force, $F=144.3396\,N$

$F_x=125.0018\,N$

$F_y=72.1698\,N$

Explanation:

Given:

angle of kicking from the horizon, $\theta= 30^{\circ}$

velocity of the ball after being kicked, $v=18 m.s^{-1}$

mass of the ball, $m=0.425\, kg$

time of application of force, $t=5.3\times 10^{-2}\,s$

We know, since body is starting from the rest

$\Delta p=m.v$ .....................(1)

$\Delta p=0.425\times 18$

$\Delta p=7.65 \,kg.m.s^{-1}$

Now the components:

$\Delta p_x= 7.65\times cos 30^{\circ}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

similarly

$\Delta p_y= 7.65\times sin 30^{\circ}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

also, impulse

$I=F\times t$ .........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

$F\times t=m.v$

$F\times 5.3\times 10^{-2}= 7.65$

$F=144.3396\,N$

Now, the components

$F_x=144.3396\times cos 30^{\circ}$

$F_x=125.0018\,N$

$F_y=144.3396\times sin 30^{\circ}$

$F_y=72.1698\,N$

6 0

2 years ago