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3 years ago

PLEASE HURRY!

Physics

2 answers:

Schach [20]3 years ago

6 0

Answer:

B) Weak electrolytes dissolve and partially dissociate in water providing charged ions to conduct electricity.

Explanation:

olga nikolaevna [1]3 years ago

5 0

The answer is B. Weak electrolytes are weak because they only partially dissociate into ions in water.

Hope this helps!

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Astaten whats 2+2 answer

alexandr402 [8]

Answer:

2+2=4

Explanation:

6 0

3 years ago

Convert 60 km/ hr into m/s

elixir [45]

Answer:

1 KM per minute is the real speed in minutes, turn that into 1000 meters per minute and divided by 60, you get a good number of 16.6666666667 which means you could go 50 meters per 3 seconds

Explanation:

so it would be 16.6666666667 meters per second

4 0

3 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago

What is the magnetic force on a proton that is moving at 2.5 107 m/s to the left through a magnetic field that is 3.4 T and poin

evablogger [386]

Answer: C. 1.4 10-11 N up

Explanation:

The magnetic force, F on a charge q moving with velocity v in a magnetic field B at an angle θ is given by:

F = q v B sin θ

Charge of proton, q = 1.6 × 10⁻¹⁹ C

Strength of magnetic field, B = 3.4 T pointing outwards

velocity of the proton, v = 2.5 × 10⁷ m/s towards left

Magnetic force is given by:

F = 1.6 × 10⁻¹⁹ C× 2.5 × 10⁷ m/s ×3.4 T× sin 90 = 13.6 × 10⁻¹² N = 1.4 × 10⁻¹¹ N up

The direction of the force is given by Lorentz Right hand rule. The fingers point magnetic field, the thumb points towards velocity, then the force on the proton is given by the direction perpendicular to the palm.

The magnetic field acts outwards with velocity of the proton towards left. The force would act perpendicular to the two -upwards.

6 0

3 years ago

A particle P with speed 140 m s–1begins to decelerate uniformly at a certain instant while another particle Q starts from rest 6

Natasha2012 [34]

Answer:

i) The motion of both particles are shown on the same speed-time curve included

ii) Approximately 19.5 seconds

Explanation:

We are given that;

Initial velocity of particle, P = 140 m/s

Start time of particle P = 6 s before start time of particle Q

Position of particle Q when velocity is 25 m/s = 125 m

Therefore, from the equation of motion, we have for particle Q;

v² = u² + 2·a·s

Where:

v = Final velocity = 25 m/s

u = Initial velocity = 0 m/s

a = Acceleration

s = Distance covered = 125 m

Therefore;

25² = 0² + 2×a×125

Which gives a = 25²/(2×125) = 2.5 m/s²

The time taken for particle Q to reach 125 m is found from the relation;

s = u·t + 1/2·a·t²

Where:

t = Time of journey

Therefore;

125 = 0×t + 1/2×2.5×t²

Which gives 125 = 1.25 × t²

Hence, t² = 125/1.25 = 100

t = √(100) = 10 s

The equation for particle Q is v = 0 + 2.5×t

Hence, since particle P starts deceleration 6 seconds before the commencement of motion of particle Q, the amount of seconds after the commencement of deceleration of the first particle P that it takes for particle P to come to rest is found as follows;

Hence, at t = 6 + 10 = 16 seconds particle P speed = 25 m/s

From the equation of motion, for particle P (decelerating) we have

v = u - a·t

Where:

v = 25 m/s

u = 140 m/s

t = 16 s

Hence, 25 = 140 - a×16

∴ 16·a = 140 - 25 = 115

a = 115/16 = 7.1875 m/s²

Therefore, the time it takes before particle P comes to rest is found from the same equation of motion, where v = 0 as follows;

v = u - a·t

0 = 140 - 7.1875 × t

∴7.1875·t = 140

t = 140/7.1875 = 19.48 s ≈ 19.5 seconds.

4 0

3 years ago