A major advantage of case studies is _______

A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential

castortr0y [4]

Explanation:

Given that,

Electric field = 5750 N/C

Charge $q=+10.5\times10^{-6}\ C$

Distance = 5.50 cm

(a). When the charge is moved in the positive x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot d$

$\Delta U=-q(E\cdot d)$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}$

$\Delta U=-3.32\times10^{-3}\ J$

The change in electric potential energy is $-3.32\times10^{-3}\ J$

(b). When the charge is moved in the negative x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot (-d)$

$\Delta U=-q(E\cdot (-d))$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})$

$\Delta U=3.32\times10^{-3}\ J$

The change in electric potential energy is $3.32\times10^{-3}\ J$

Hence, This is the required solution.

3 0

4 years ago

Why does the current is reduced as electrons move through a conductor

hammer [34]

Because the electrons collide with the particles inside the conductor so are therefore slowed down seen as current is the rate of flow of electrons

3 0

3 years ago

The intensity of light depends on the amount of *blank blank*

babymother [125]

Answer:

energy? or power? I'm not really sure but using the quizzes app really help with science problems

7 0

3 years ago

A tin can has a volume of 1100 cm³ and a mass of 80 g. Approximately how many grams of lead shot can it carry without sinking in

Kruka [31]

Answer:

1020g

Explanation:

Volume of can= $1100cm^3=1100\times 10^{-6}m^3$

$1cm^3=10^{-6}m^3$

Mass of can=80g= $\frac{80}{1000}=0.08kg$

1Kg=1000g

Density of lead= $11.4g/cm^3=11.4\times 10^{3}=11400kg/m^3$

By using $1g/cm^3=10^3kg/m^3$

We have to find the mass of lead which shot can it carry without sinking in water.

Before sinking the can and lead inside it they are floating in the water.

Buoyancy force = $F_b=Weight of can+weight of lead$

$\rho_wV_cg=m_cg+m_lg$

Where $\rho_w=10^3kg/m^3=$ Density of water

$m_c=$ Mass of can

$m_l=$ Mass of lead

$V_c=$ Volume of can

Substitute the values then we get

$1000\times 1100\times 10^{-6}=0.08+m_l$

$1.1-0.08=m_l$

$m_l=1.02 kg=1.02\times 1000=1020g$

$1 kg=1000g$

Hence, 1020 grams of lead shot can it carry without sinking water.

4 0

4 years ago

Two people push on the same door from opposite sides as shown.

tangare [24]

The correct answer is
<span>c. one person exerts more force than the other so that the forces are unbalanced.

In fact, the door is initially at rest. In order to move the door, a net force different from zero should be applied, according to Newton's second law:
</span> $\sum F = ma$
<span>where the term on the left is the resultant of the forces acting on the door, m is the door mass and a its acceleration.

In order to move the door, the acceleration must be different from zero. But this means that the resultant of the forces acting on it must be different from zero: this is possible only if the forces applied by the two persons are unbalanced, i.e. one person exerts more force than the other.</span>

3 0

3 years ago

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A major advantage of case studies is ________.