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xeze [42]
3 years ago
8

A point charge of 4.0 µC is placed at a distance of 0.10 m from a hard rubber rod with an electric field of 1.0 × 103 . What is

the electric potential energy of the point charge?
× 10–4 J

What is the electric potential energy of the point charge at 1.3 m?


× 10–3 J
Physics
2 answers:
Afina-wow [57]3 years ago
4 0

The electric potential energy of the point charge is 4.

the electric potential energy of the point charge at 1.3 m is 5.2.

Juliette [100K]3 years ago
4 0

Answer:

Part 1)

U = 4 \times 10^{-4} J

Part 2)

U = 5.2 \times 10^{-3} J

Explanation:

Electric field due to rubber rod is given by

E = 1.0 \times 10^3 N/c

now we also know that

V = E. r

V = (1.0 \times 10^3)(0.10)

V = 1.0 \times 10^2

now potential energy is given as

U = qV

U = (4 \mu C)(1.0 \times 10^2)

U = 4 \times 10^{-4} J

Similarly the electric potential at distance r = 1.3 m

so we have

V = (1.0 \times 10^3)(1.3)

V = 1.3 \times 10^3 Volts

now potential energy is given as

U = (4\mu C)(1.3 \times 10^3)

U = 5.2 \times 10^{-3} J

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A mass of 6 kg with initial velocity 16 m/s travels through a wind tunnel that exerts a constant force 8 N for a distance 1.6 m.
strojnjashka [21]

Answer:

D=99.4665307m \approx 99.5m

Explanation:

From the question we are told that

Mass  m=6kg

Velocity of mass  V_m=16

Force of Tunnel  F_t=8N

Length of Tunnel L_t=1.6

Height of frictional incline H_i=2.9

Angle of inclination  \angle =16 \textdegree

Acceleration due to gravity  g=9.8m/s^2

First Frictional surface has a coefficient  \alpha_1 =0.21\ for\ d_c=1

Second Frictional surface has a coefficient \alpha _2=0.1

Generally the initial Kinetic energy is mathematically given by

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}(6)(16)^2

K.E=768

Generally the work done by the Tunnel is mathematically given as

w_t=F_t*d_t

w_t=8*1.6

w_t=12.8J

Therefore

Total energy\ E_t=Initial\ kinetic energy\ K.E*Work done\ by\ tunnel\ W_t

E_t=K.E+E_t\\E_t=768J+12.8J

E_t=780.8J

Generally the energy lost while climbing is mathematically given as

E_c=mgh

E_c=(6)(9.8)(2.9)

E_c=170.52J

Generally the energy lost to friction is mathematically given as

E_f=\alpha *m*g*cos\textdegree*d_c

E_f=0.21*6*9.8*cos16*1

E_f=11.86965942 \approx 12J

Generally the energy left in the form of mass Em is mathematically given as

E_m=E_t+E_c+E_f

E_m=(768J)-(170.52)-(12)

E_m=585.48J

Since

E_m=\alpha_2*g*m*d

Therefore

It slide along the second frictional region

D=\frac{585.46}{0.1*9.81*6}

D=99.4665307m \approx 99.5m

6 0
3 years ago
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