Answer: 0.03 N/C
Explanation:
We use the current density formula to solve this question.
I/A = σ * E
Where,
I = current flowing in the circuit = 0.3 A
A = cross sectional area of the wire = 1 mm²
σ = resistivity of the wire = 1*10^7 Ω^-1·m^-1
E = strength of the electric field required
I/A = σ *E
E = I/(A * σ)
First we convert area from mm to m, so that, 1*10^-3 mm = 1*10^-6 m
E = 0.3 A / (1*10^-6 m * 1*10^7 Ω^-1·m^-1)
E = 0.3 A / 10 Ω^-1
E = 0.03 N/C
Answer:
1. Speed and velocity both involve a numeric rate describing the distance traveled by a body in a unit of time. However, speed describes the rate of a body traveling in any direction in a unit of time, while velocity describes the rate of a body traveling in a particular direction in a unit of time.
2. Answers may vary, but should resemble the following:
Average velocity explains the velocity the body traveled overall, not taking into consideration each spot in the trip. If a car moves at 65 km/h on average, it may have slowed down for some parts and sped up for others. Overall though, it would have made a certain distance of travel within a specified unit of time that totals the average velocity of 65 km/h.
Instantaneous velocity explains the velocity of a body at a particular instant of the trip. The instantaneous velocity of a car stopped at a stop sign would be 0 m/s even if it was moving before and will continue to move after this stop. The velocity at that particular instant is the instantaneous velocity.
Uniform velocity is when the distance being covered is changing uniformly with time. For example, if a car moves 20 km every 30 minutes and continues to do so in the same direction, it's traveling with a uniform velocity.
3. a=v2−v1t
a=20 m/s−60 m/s6 s
a=−406
a = –6.7 m/s2
4. v2 = v1 + at
v2 = 14 m/s + (3 m/s2 × 6 s)
v2 = 14 + 18
v2 = 32 m/s
5. v=st
v=375 km5 h
v = 75 km/h
6. First, convert the minutes to seconds. Since there are 60 seconds in one minute, multiply:
60 × 15 (minutes) = 900 seconds
s = v × t
s = 6 m/s × 900 s
s = 5,400 m
7. t=sv
t=80 km35 km/hr
t = 2.29 hr
8. a=v2−v1t
a=50 m/s−15 m/s4 s
a=35 m/s4 s
a = 8.75 m/s2
9. vav=v1+v22
vav=15 m/s+50 m/s2
vav=65 m/s2
vav = 32.5 m/s
10. a=v2−v1t
a=0 m/s−11.5 m/s3.5 s
a = –3.29 m/s2
Explanation:
We can first obtain time of flight from vertical fall
Initial velocity U=0, d = 6 m, a = 9.8 m/s²
d = ut + 1/2 at²
6.0 = 0 + (1/2 × 9.80 t²)
t = √(12/9.8)
= 1.106 sec
Horizontal velocity = Vh = Dh/t
= 24.0 /1.106 s
= 21.69 m/s
The ball was thrown at a speed of 21.69 m/s
Answer:
(i) 
, (ii) 
, (iii)
Explanation:
(i) 
and 
represent the points where particle has a velocity of zero and spring reach maximum deformation, Given the absence of non-conservative force and by the Principle of Energy Conservation, the position where particle is at maximum speed is average of both extreme positions:
(ii) Maximum accelerations is reached at and .
(iii) Greatest net forces exerted on the particle are reached at and .
