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photoshop1234 [79]
2 years ago
5

Your science teacher has provided the following materials to investigate the factors that affect the strength of electric and ma

gnetic forces.
• large iron nail – 5 mm diameter

• 100 cm of thinly wrapped copper wire

• D battery

• electrical tape

• wire strip

• thumb tacks -assorted sizes

• paperclips – assorted sizes

Design an investigation you could carry out to determine the factors that affect the strength of electrical and magnetic forces.
Physics
1 answer:
kodGreya [7K]2 years ago
3 0

The strength of the electric field can be estimated as force per electric charge, whereas the strength of a magnetic field depends on the force of electrons and velocity.

<h3>Whta is a magnetic field?</h3>

A magnetic field can be defined as magnetic work exerted on electric charges that are in motion (movement).

Conversely, an electric field is an area where an electric current is applied and electron force can be estimated per Volts per Meter.

In conclusion, the strength of the electric field can be estimated as force per electric charge, whereas the strength of a magnetic field depends on the force of electrons and velocity.

Learn more about magnetic fields here:

brainly.com/question/14411049

#SPJ1

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A ball of mass 2.0 kg falls vertically and hits the ground with speed 7.0 ms-1 as shown below.
Sauron [17]

Answer:

8.0 Ns

Explanation:

Change in momentum is given as:

Final momentum - Initial momentum

= m*v - m*u

Where m = mass of ball

v = final velocity

u = initial velocity

Change in momentum = (2.0 * 3.0) - (2.0 * 7.0)

= 6.0 - 14.0 = -8.0 Ns

The magnitude will be |-8.0| = 8.0 Ns

7 0
4 years ago
Read 2 more answers
395,000 meters in 9000
marta [7]

Answer:

43.88 meters per second

Explanation:

The computation of the speed is shown below:

As we know that

Speed = \frac{Distance}{time}

where,

Distance is 395,000 meters

Time is 9,000 seconds

Now placing these values to the formula

So, the speed is

= \frac{395,000}{9,000}

= 43.88 meters per second

As speed shows the relation between the distance and time and the same is to be considered i.e by applying the formula

6 0
4 years ago
Two large parallel conducting plates are 8.0 cm apart and carry equal but opposite charges on their facing surfaces. The magnitu
Pepsi [2]

Answer:

18.1 V

Explanation:

The electric field between two parallel plates is given by the equation:

E=\frac{\sigma}{\epsilon_0}

where

\sigma is the charge surface density

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the vacuum permittivity

For the plates in this problem,

\sigma = 2.0 nC/m^2 = 2.0\cdot 10^{-9} C/m^2

So, the magnitude of the electric field is

E=\frac{2.0\cdot 10^{-9}}{8.85\cdot 10^{-12}}=226.0 V/m

Now we can find the potential difference between the plates, which is given by

\Delta V = E d

where

d = 8.0 cm = 0.08 m is the separation between the plates

Substituting,

\Delta V=(226.0)(0.08)=18.1 V

8 0
3 years ago
An athlete performing a long jump leaves the ground at a 32.9 ∘ angle and lands 7.78 m away.part a )what was the takeoff speed ?
MatroZZZ [7]

Answer:

Explanation:

Given

Inclination \theta =32.9^{\circ}

Distance of landing point R=7.78\ m

Considering athlete to be an Projectile

range of projectile is given by

R=\frac{u^2\sin 2\theta }{g}

where u=launch velocity

7.78=\frac{u^2\sin (2\times 32.9)}{9.8}

u^2=83.58

u=\sqrt{83.58}

u=9.142\ m/s

(b)If u is increased by 8% then

new velocity is u'=9.87\ m/s

R'=\frac{u'^2\sin 2\theta }{g}

R'=9.07\ m

 

3 0
4 years ago
Read 2 more answers
Let's apply Snell's law to the refraction of light across a water–air interface. Suppose you kneel beside the fishpond in your b
Nataliya [291]

Answer:

The angle of refraction is 41.68°.

Explanation:

The refractive index for water is n_2 = 1.333, and for air n_1 = 1.00: the angle of light with the normal is 90^o-60^o = 30^o; therefore Snell's law gives

n_1sin(\theta_1)= n_2sin(\theta_2)

1.00*sin(\theta_1) = 1.33 sin(30^o)

sin (\theta_1) = \dfrac{1.33sin(30^o)}{1.00}

sin (\theta_1) = 0.665

\theta _1 = sin^{-1}(0.665)

\boxed{\theta_1 = 41.68^o}

4 0
3 years ago
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