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3 years ago

I need help please I will give points

Physics

1 answer:

stira [4]3 years ago

3 0

Answer:

-5 N of force

Explanation:

Hope it helps!

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A thin, metallic spherical shell of radius 0.227 m has a total charge of 6.03 × 10 − 6 C placed on it.

KATRIN_1 [288]

Answer:

Explanation:

Given

radius $r=0.227 m$

Charge on surface $Q=6.03\times 10^{-6} C$

Point Charge inside sphere $q=1.15\times 10^{-6} C$

Electric Field at $r=0.735 m$

Treating Surface charge as Point charge and applying Gauss law

$E_{total}A=\frac{q_{enclosed}}{\epsilon _0}$

where A=surface area up to distance r

$E_{total}=\frac{Q+q}{4\pi r^2}$

$E_{total}=\frac{6.03\times 10^{-6}+1.15\times 10^{-6}}{4\pi (0.735)^2\times 8.85\times 10^{-12}}$

$E_{total}=1.194\times 10^{5} N/C$

3 0

4 years ago

During the motion of the slinky in a transverse wave, what do the particles of the slinky coil do?

Mazyrski [523]

Answer:

C.) The slinky particles move up and down

Explanation:

Transverse Wave-

A wave that has a disturbance perpendicular to the wave motion

Hello! This is the correct answer! Have a blessed day! :)

If you are in K12, please review the lesson! :) It will give you some very helpful definitions! I hope this helped!

5 0

3 years ago

Im bad at work problems can any one help with this problem ?

lyudmila [28]

We will put the number of trips in the first column, the miles driven in the second column and gallons of fuel used in the third column.

8 7,680 1,010
7 9,940 1,330
12 14,640 1,790
12 13,920 2,050

4 0

4 years ago

An automobile starter motor has an equivalent resistance of 0.055 Ï and is supplied by a 12.0 v battery which has a 0.0305 Ï int

LiRa [457]

12 V is the f.e.m. $\epsilon$ of the battery. The potential difference that is applied to the motor is actually the fem minus the voltage drop on the internal resistance r:
$\epsilon - Ir$
this is equal to the voltage drop on the resistance of the motor R:
$RI$
so we can write:
$\epsilon - Ir = RI$
and using $r=0.0305~\Omega$ and $R=0.055~\Omega$ we can find the current I:
$I= \frac{\epsilon}{R+r}=140~A$

8 0

3 years ago

Multiple-Concept Example 9 reviews the concepts that are important in this problem. A drag racer, starting from rest, speeds up

Mademuasel [1]

Answer:

V = 90.51 m/s

Explanation:

From the given information:

Initial speed (u) = 0

Distance (S) = 391 m

Acceleration (a) = 18.9 m/s²

Using the relation for the equation of motion:

v² - u² = 2as

v² - 0² = 2as

v² = 2as

$v = \sqrt{2as}$

$v = \sqrt{2*18.9*391}$

v = 121.57 m/s

After the parachute opens:

The initial velocity = 121.57 m/ss

Distance S' = 332 m

Acceleration = -9.92 m/s²

How fast is the racer can be determined by using the relation:

$V= \sqrt{v^2 + 2aS'}$

$V = \sqrt{121.57^2+ 2 (-9.92)(332)}$

V = 90.51 m/s

6 0

3 years ago