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3 years ago

2. What is the magnitude of the force a 1.5 C charge exerts on a 3.2 C charge

Physics

1 answer:

lesya [120]3 years ago

4 0

The magnitude of the force is F=1.68×10 ^20 N

Explanation:

Given data

q1 =1.5 10 ^6 q2=3.2 10^6 r=1.5

We have the formula

By the coulomb's law

F= K. q1 ×q2 / r²

The K value is given by

8.99 10^9 Nm²/ c²

substitute the values we get,

F= ( 8.99× 10^9 Nm²/ c²) ×(1.5 ×10 ^6)×(3.2 ×10 ^6)/ (1.6 m² )

F=1.68×10 ^20 N

The magnitude of the force is F=1.68×10 ^20 N

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When the switch is open, the model predicts that bulb ___ will have the same brightness as bulb ___. When the switch is closed,

Arte-miy333 [17]

please show me the question paper .( Pic)

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2 years ago

Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire

Over [174]

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

$F = ma \rightarrow a = \frac{F}{m}$

Replacing,

$a =\frac{42N}{83000kg}$

$a =5.06*10^{-4}m/s^2$

The total speed change

$\Delta v = v_f -v_0 \rightarrow v_f =\text{Final velocity and } v_0 = \text{Initial velocity }$ we have that the value is 0.71m/s

If we know that acceleration is the change of speed in a fraction of time,

$a= \frac{\Delta v}{t} \rightarrow t = \frac{\Delta v}{a}$

We have that,

$t= \frac{0.71m/s}{5.06*10^{-4}m/s^2 }$

$t = 1403.16s$

Therefore the Rocket should be fired around to 1403.16s

7 0

2 years ago

A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 35.0 min

Lubov Fominskaja [6]

Explanation:

Solution:

Let the time be

t1=35min = 0.58min

t2=10min=0.166min

t3=45min= 0.75min

t4=35min= 0.58min

let the velocities be

v1=100km/h

v2=55km/h

v3=35km/h

a. Determine the average speed for the trip. km/h

first we have to solve for the distance

S=s1+s2+s3

S= v1t1+v2t2+v3t3

S= 100*0.58+55*0.166+35*0.75

S=58+9.13+26.25

S=93.38km

V=S/t1+t2+t3+t4

V=93.38/0.58+0.166+0.75+0.58

V=93.38/2.076

V=44.98km/h

b. the distance is 93.38km

6 0

3 years ago

3 Below, someone is trying to balance a plank with

belka [17]

Answer:

a. The moment of the 4 N force is 16 N·m clockwise

b. The moment of the 6 N force is 12 N·m anticlockwise

Explanation:

In the figure, we have;

The distance from the point 'O', to the 6 N force = 2 m

The position of the 6 N force relative to the point 'O' = To the left of 'O'

The distance from the point 'O', to the 4 N force = 4 m

The position of the 4 N force relative to the point 'O' = To the right of 'O'

a. The moment of a force about a point, M = The force, F × The perpendicular distance of the force from the point

a. The moment of the 4 N force = 4 N × 4 m = 16 N·m clockwise

b. The moment of the 6 N force = 6 N × 2 m = 12 N·m anticlockwise.

8 0

2 years ago

If f(x) = -3x +4 and g(x) = 2, solve for the value of x for which (fx) = g(x) is true.

maksim [4K]

Answer:

$x=-\frac{2}{3}$

Explanation:

Let:

$f(x)=-3x+4\\\\And\\\\g(x)=2$

We need to know for which value of $x$ the function $f(x)$ is equal to $g(x)$ :

$f(x)=g(x)$

Therefore, we need to solve for the previous equation for $x$ :

Replacing the values of $f(x)$ and $g(x)$ :

$-3x+4=2$

Subtract 4 from both sides:

$-3x+4-4=2-4\\\\-3x=-2$

Multiply both sides by -1

$-3x(-1)=-2(-1)\\\\3x=2$

Divide both sides by 3:

$\frac{3x}{3} =\frac{2}{3} \\\\x=\frac{2}{3}$

Therefore the value of $x$ for which $f(x)=g(x)$ is $x=\frac{2}{3}$ .

Verify the result:

$-3(\frac{2}{3} )+4=2\\\\-2+4=2\\\\2=2$

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3 years ago