8.

The Electric Field 9.0m from a +25μC charge is 

salantis [7]

The electric field intensity generated by a single point charge is given by

$E(r)=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, the charge is $q=25 \mu C=25 \cdot 10^{-6} C$ and we are asked to calculate the field at distance $r=9.0 m$ , so the electric field is

$E(9.0 m)=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{25 \cdot 10^{-6} C}{(9 m)^2}=2775 V/m$

5 0

3 years ago

A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a dire

svp [43]

Answer:

(A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

Explanation:

Given that,

Distance = 20.0 m

Frictional force = 35.0 N

Angle = 25.0°

(A). We need to calculate the work done on the cart by the friction

Using formula of work done

$W_{fr} = -F\cdot d$

Where, F = force

d = distance

Put the value into the formula

$W_{fr}=-35.0\times20$

$W_{fr}=−700\ J$

(B). The work done by the gravity is perpendicular to the direction of the motion

We need to calculate the work done on the cart by the gravitational force

Using formula of work done

$W=fd\cos\theta$

Put the value into the formula

$W=35.0\times20\cos90$

$W=0\ J$

(C). We need to calculate the work done on the cart by the shopper

Using formula of work done

$W_{sh}=W_{net}-W_{fr}$

Put the value into the formula

$W_{sh}=0-(-700)$

$W_{sh}=700\ J$

(D). We need to calculate the force the shopper exerts

Using formula of force

$F_{sh}=\dfrac{W_{fr}}{d\cos\theta}$

Put the value into the formula

$F_{sh}=\dfrac{700}{20\cos25}$

$F_{sh}=38.61\ N$

(E). We need to calculate the total work done on the cart

Using formula of work done

$W_{cart}=W_{fr}+W_{sh}$

Put the value into the formula

$W_{cart}=700-(-700)$

$W_{cart}=0\ J$

Hence, (A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

6 0

3 years ago

Compare the magnitudes of the electric and magnetic forces on an electron that has attained a velocity of 107 m/s. Assume an ele

Verizon [17]

Answer:

F = 1.608 x 10⁻¹⁴ N

Explanation:

Given,

The velocity of the electron, v = 10⁷ m/s

The electric field intensity, E = 10⁵ V/m

The magnetic flux density, B = 0.5 gauss

= 5 x 10⁻⁵ tesla

The magnitudes of the electric and magnetic forces on an electron is given by the Lorentz force,

F = q (E + v x B)

Substituting the given values,

F = 1.6 x 10⁻¹⁹ (10⁵ + 10⁷ x 5 x 10⁻⁵)

F = 1.608 x 10⁻¹⁴ N

Hence, the magnitudes of the electric and magnetic forces on an electron is, F = 1.608 x 10⁻¹⁴ N

3 0

3 years ago

As you lift a book into the air, what kind of energy are you increasing in the book?

frez [133]

Gravitational potential energy.

6 0

3 years ago

Read 2 more answers

At this radius, what is the magnitude of the net force that maintains circular motion exerted on the pilot by the seat belts, th

Ainat [17]

Answer:

Fc=5253 N

Explanation:

Answer:

Fc=5253 N

Explanation:

sequel to the question given, this question would have taken precedence:

"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."

so we derive centripetal acceleration first

ac (centripetal acceleration) = v^2/r

make r the subject of the equation

r= v^2/ac

ac is 6.23*g which is 9.81

v is 101m/s

substituing the parameters into the equation, to get the radius

(101^2)/(6.23*9.81) = 167m

Now for part

( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.

he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.

Fc (Centripetal Force) = m*v^2/r

So (86kg* 101^2)/(167) =

Fc=5253 N

4 0

4 years ago