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2 years ago

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Physics

2 answers:

Crank2 years ago

8 0

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mars1129 [50]2 years ago

4 0

Answer:

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Which property describes the ability of one substance to dissolve in another substance?

oee [108]

The answer would be solubility

5 0

3 years ago

I need HELP Please !!

aliya0001 [1]

You have to figure it out

5 0

2 years ago

Two boats - Boat A and Boat B - are anchored a distance of 24 meters apart. The incoming water waves force the boats to oscillat

ozzi

Answer:

wavelength = 24 m

Period = 10 s

f = 0.1 Hz

Amplitude = 4 m

Explanation:

Wavelength:

Since the boats are at crest and trough, respectively at the same time. Hence, the horizontal distance between them is the wavelength of the wave:

wavelength = 24 m

Period:

The period is given as:

$Period = \frac{time}{no.\ of\ cycles} \\\\Period = \frac{10\ s}{1}\\\\$

Period = 10 s

Frequency:

The frequency is given as:

$f = \frac{1}{time\ period}\\\\f = \frac{1}{10\ s}\\\\$

f = 0.1 Hz

Amplitude:

Amplitude will be half the distance between extreme points, that is, crest and trough:

Amplitude = 8 m/2

Amplitude = 4 m

5 0

3 years ago

An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect

Dvinal [7]

Answer:

The distance is $d =1.66*10^{-9}m$

Explanation:

From the question we are told that

The initial speed of the electron is $v_i = 3.2 *10^5 m/s$

The mass of electron is $m = 9.11*10^{-31}kg$

Let $d$ be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

Let $KE_i$ be the initial kinetic energy of the electron \

Let $KE_d$ be the kinetic energy of the electron at the distance $d$ from the proton

Considering that energy is conserved,

The energy at the initial position of the electron = The energy at the final position of the electron

i.e

$KE_i +U_1 = KE_d + U_2$

$U_1 \ and \ U_2$ are the potential energy at the initial position of the electron and at distance d of the electron to the proton

Here $U_1 = 0$

So the equation becomes

$\frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2 + \frac{kq_1 q_2}{d}$

Here $q_1 \ and \ q_2$ are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton

$k$ is electrostatic constant with value $8.99*10^9 N \cdot m^2 /C^2$

i.e $q = 1.602 *10^{-19}C$

$v_d$ is the velocity at distance d from the proton = 2 $v_i$

So the equation becomes

$\frac{1}{2}mv_i^2 = \frac{1}{2} m (2v_i)^2 -\frac{k(q)^2}{d}$

$\frac{1}{2} mv_i^2 = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}$

$3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}$

Making d the subject of the formula

$d = \frac{2k(q)^2}{3mv_i^2}$

$= \frac{2* 8.99*10^9 *(1.602*10^{-19}^2)}{3 * 9.11*10^{-31} *(3.2*10^5)^2}$

$=1.66*10^{-9}m$

6 0

3 years ago

A dog is chasing a cat towards a tree. The cat has a 10-yard lead and runs at 6 yards per second. The dog runs at a speed of 8 y

oksano4ka [1.4K]

Answer : Cat will reach tree first.

Explanation :

It is given that,

Total distance between dog and tree is 30 yards. The cat has 10 yards lead. So, the distance between cat and tree is 20 yards.

Speed of cat is 6 Yards/ sec.

Speed of dog is 8 Yards/sec.

speed of cat is, $S_c=\dfrac{distance\ covered\ by\ cat}{time\ taken}$

$6\ Yards/sec=\dfrac{20\ Yards}{t_c}$

$t_c=3.33\ sec$

Now, speed of dog is, $S_d=\dfrac{distance\ covered\ by\ dog}{time\ taken}$

$8\ Yards/sec=\dfrac{30\ yards}{t_d \ sec}$

$t_d=3.75\ sec$

It is clear that, cat is taking less time as compared with dog. So, the cat will reach tree first.

3 0

3 years ago