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3 years ago

Can two negatively charged balloons move apart without ever touching why or why not

2 answers:

Nutka1998 [239]3 years ago

8 0

Yes, because electric charges have electric field surrounding them that allow them to exert forces on other objects without touching them.

Sloan [31]3 years ago

4 0

Your answer would indeed be, yes, because electric charges have electric field surrounding them that allow them to exert forces on other objects without touching them.

just took the test and got it right! Hope this helped!

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This diagram represents a top-down view of an experiment on a table. The 250g and 100g masses are falling and are pulling the bl

nata0808 [166]

The answer should be B: 500g of mass on bottom.

7 0

3 years ago

Read 2 more answers

Keaton is asked to solve the following physics problem:

RideAnS [48]

Answer:

The answer is C "think about the problem first, systematically consider all factors, and form a hypothesis"

Explanation:

In physics there is some basic fomula that sir Isacc Newton proposed under the topic of motion. The three formulas are below;

1) v=u+at

2)v^2=u^2+2as

3)s=ut+(1/2)(at^2)

the variables are explained below;

u= initial velocity of the body

a=acceleration/Speed of the body

t= time taken by the body while travelling

s= displacement of the body.

Therefore to solve keatons problem, the factors(variables) in the formulas above need to be systematically considered. Since the ball was dropped from the top of the building, the initial velocity is 0 because the body was at rest. Also the acceleration will be acceleration due to gravity (9.8m/s^2)

5 0

3 years ago

Hank and Harry are two ice skaters whiling away time by playing 'tug of war' between practice sessions. They hold on to opposite

Alex73 [517]

Answer:

the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1

Explanation:

Given the data in the question;

Hank and Harry are two ice skaters, since both are on top of ice, we assume that friction is negligible.

We know that from Newton's Second Law;

Force = mass × Acceleration

F = ma

Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.

Now,

Mass $_{Hank$ × Acceleration $_{Hank$ = Mass $_{Henry$ × Acceleration $_{Henry$

Mass $_{Hank$ / Mass $_{Henry$ = Acceleration $_{Henry$ / Acceleration $_{Hank$

given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,

Mass $_{Hank$ / Mass $_{Henry$ = 1 / 1.26

Mass $_{Hank$ / Mass $_{Henry$ = 0.7937 or [ 0.7937 : 1 ]

Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]

8 0

3 years ago

flipper (the dolphin) is out in the open ocean hunting tuna avec. he emits his pulse at 22khz and .42 seconds later he hears it

marusya05 [52]

First we need to find the speed of the dolphin sound wave in the water. We can use the following relationship between frequency and wavelength of a wave:
$v=\lambda f$
where
v is the wave speed
$\lambda$ its wavelength
f its frequency
Using $\lambda = 2 cm = 0.02 m$ and $f=22 kHz = 22000 Hz$ , we get
$v=(0.02 m)(22000 Hz)=440 m/s$

We know that the dolphin sound wave takes t=0.42 s to travel to the tuna and back to the dolphin. If we call L the distance between the tuna and the dolphin, the sound wave covers a distance of S=2 L in a time t=0.42 s, so we can write the basic relationship between space, time and velocity for a uniform motion as:
$v= \frac{S}{t}= \frac{2L}{t}$
and since we know both v and t, we can find the distance L between the dolphin and the tuna:
$L= \frac{vt}{2}= \frac{(440 m/s)(0.42 s)}{2}=92.4 m$

5 0

3 years ago

Two long parallel wires are separated by forty centimeters and carry oppositely-directed currents of ten amperes. Find the magni

Softa [21]

Answer:

1.04μT

Explanation:

Due to both wires have opposite currents, the magnitude of the total magnetic field is given by

$B_T=\frac{\mu_o I}{2 \pi r_1}-\frac{\mu_o I}{2 \pi r_2}$

I: electric current = 10A

mu_o: magnetic permeability of vacuum = 4pi*10^{-7} N/A^2

r1: distance from wire 1 to the point in which B is measured.

r2: distance from wire 2.

The distance between wires is 40cm = 0.4m. Hence, r1=0.2m r2=0.6m

By replacing in the formula you obtain:

$B_T=\frac{(4\pi *10^{-7}N/A^2)(10A)}{2\pi}(\frac{1}{0.4m}-\frac{1}{0.6m})=1.04*10^{-6}T =1.04\mu T$

hence, the magnitude of the magnetic field is 1.04μT

4 0

3 years ago