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Citrus2011 [14]

2 years ago

12

Choose the true statements about electric fields and forces. (Choose 3)

1 answer:

aleksley [76]2 years ago

3 0

Answer:

The three best answers are:

Electric fields can be positive or negative (depending on source)

Electric fields can attrace or repel objets (assuming the object is charged)

Electric forces are strongest near the charged object E ∝ 1 / R^2

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F(x)= 10x-5<br>What is the value of f-1(-4) ?

Afina-wow [57]

Answer:

$f^{-1}(-4) = \frac{1}{10}$

Explanation:

Firstly finding $f^{-1}(x)$

So,

$f(x) = 10x-5$

Substitute $y = f(x)$

$y = 10x-5$

Exchange the values of x and y

$x = 10y-5$

Solving for y

$x = 10y-5$

Adding 5 to both sides

$10y = x+5$

Dividing both sides by 10

$y = \frac{x+5}{10}$

Replace $y = f^{-1}(x)$

$f^{-1}(x) = \frac{x+5}{10}$

For x = -4

$f^{-1}(-4) = \frac{-4+5}{10}$

$f^{-1}(-4) = \frac{1}{10}$

6 0

3 years ago

Read 2 more answers

A tennis ball is tossed upwards into the air with an initial velocity of +5m/s, how much time does it take for the tennis ball t

Sever21 [200]

HOPE THIS HELPS YOU!!!!!

3 0

3 years ago

Parker (73.2 kg) is being dragged down the hall with an applied force of 123 N. If the frictional force is 27.4 N, what is the c

levacccp [35]

Answer:

The coefficient of friction in the hall is 0.038

Explanation:

Given;

mass of the Parker, m = 73.2 kg

applied force on the parker, F = 123 N

frictional force, Fs = 27.4 N

the coefficient of friction in the hall = ?

frictional force is given by;

Fs = μN

Where;

μ is the coefficient of friction

N is normal reaction = mg

Fs = μmg

μ = Fs / mg

μ = (27.4) / (73.2 x 9.8)

μ = 0.038

Therefore, the coefficient of friction in the hall is 0.038

6 0

3 years ago

A proton moves with a velocity of v with arrow = (3î − 5ĵ + k) m/s in a region in which the magnetic field is B with arrow = (î

Allisa [31]

Answer:

The magnitude of the magnetic force this particle experiences is $2.6\times10^{-9}\ N$ .

Explanation:

Given that,

Velocity v= (3i-5j+k) m/s

Magnetic field B=(i+2j-k) T

We need to calculate the value $\vec{v}\times\vec{B}$

$(\vec{v}\times\vec{B})=3i+4j+11k$

We need to calculate the magnitude of the magnetic force this particle experiences

Using formula of magnetic force

$\vec{F}=q(\vec{v}\times\vec{B})$

Put the value into the formula

$\vec{F}=1.6\times10^{-19}\times(3i+4j+11k)$

$\vec{F}=(4.8i+6.4j+1.76k)\times10^{-19}$

$|F|=2.6\times10^{-9}\ N$

Hence, The magnitude of the magnetic force this particle experiences is $2.6\times10^{-9}\ N$ .

7 0

3 years ago

A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a

stira [4]

Answer:

Explanation:

Given

acceleration is given by

$a=-g-c_Dv^2$

where $\ddot{y}=a$

$\dot{y}=v$

Also acceleration is given by

$a=v\frac{\mathrm{d} v}{\mathrm{d} s}$

$ds=\frac{v}{a}dv$

$\int ds=\int \frac{v}{-g-0.001v^2}dv$

$\Rightarrow Let -g-0.001v^2=t$

$-0.001\times 2vdv=dt$

$vdv=-\frac{dt}{0.002}$

$at\ v_0=50\ m/s,\ t=-g-0.001(50)^2$

$t=-g-2.5$

at $v=0,\ t=-g$

$\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}$

$\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}$

$s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}$

$s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})$

$s=113.608\ m$

when air drag is neglected maximum height reached is

$h=\frac{v_0^2}{2g}$

$h=\frac{50^2}{2\times 9.8}$

$h=127.55\ m$

3 0

3 years ago