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Basile [38]
3 years ago
10

What must be your average speed in order to travel 350 km in 5.15 h? *

Physics
1 answer:
Alexxandr [17]3 years ago
3 0
V= s/t
​
=
5.15/ 350
​
=68.0 km/h
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Cables and conductors installed on the side of a pole or building shall be protected up to ___feet above finished grade.
Allisa [31]

Answer:

10 feet.

Explanation:

Conductors are considered outside the building when they installed. The requirement of NEC is above 10 feet for the clearance after the final grade for service conductors and cables where they are connected with the building, above the sidewalks and the to other areas which are accessible to the pedestrians.

8 0
4 years ago
A car travels 50 km in 25 km /h and then travels 60km with a velocity 20 km/h and then travels 60km with a velocity 20 km/h in t
Butoxors [25]

Answer:

v = 21.25 km/h

The average velocity is 21.25km/h

Explanation:

Average velocity = total displacement/time taken

v = d/t

Given;

A car travels 50 km in 25 km /h

d1 = 50km

v1 = 25km/h

time taken = distance/velocity

t1 = d1/v1

t1 = 50/25 = 2 hours

and then travels 60km with a velocity 20 km/h

d2 = 60km

v2 = 20km/h

t2 = d2/v2 = 60/20

t2 = 3 hours

and then travels 60km with a velocity 20 km/h in the same direction

d3 = 60km

v3 = 20km/h

t3 = d3/v3 = 60/20

t3 = 3 hours

Average velocity = total displacement/total time taken

v = (d1+d2+d3)/(t1+t2+t3)

v = (50+60+60)/(2+3+3)

v = 170/8

v = 21.25 km/h

The average velocity is 21.25km/h

3 0
3 years ago
Question 2
yarga [219]

Answer:

24.4 m

Explanation:

first we have to calculate how long the cat will be in the air, in vertical direction this is free fall from 4m (poor cat, hope he survived) with acceleration of g=9.81 m/s^2

time = sqrt ( 2 * height / g)

time = sqrt (2 * 4 / 9.81)

time = 0,903 s

then we know his horizontal flight was 22 m long, so ve can calculate verical speed

speed = length / time

speed = 22 m / 0,903 s

speed = 24.363 m

as you are asked to give 3 significant numbers answer is 24,4 m

8 0
3 years ago
Does the zero electric field intensity in a given region imply zero potential?​
Rama09 [41]

Answer:

No, just because the electric field is zero at a particular point, it does not necessarily mean that the electric potential is zero at that point. ... At the midpoint between the charges, the electric field due to the charges is zero, but the electric potential due to the charges at that same point is non-zero.

Explanation:

7 0
3 years ago
One particle has a charge of 2.15 x 10^ -9 while another particle has a charge of 3.22 x 10^ -9 If the two particles are separat
marin [14]

Answer:

B. 2.77 x 10^{-4} N

Explanation:

The required force can be calculated by:

F = \frac{Kq_{1}q_{2}  }{d^{2} }

Where F is the force between the particles, K is the coulomb's constant (9 x 10^{9} Nm^{2}/C^{2}), q_{1} is the charge on the first particle, q_{2} is the charge on the second particle and d^{2} is the distance between the particles.

So that:

F = \frac{9*10^{9}*2.15*10^{-9} *3.22*10^{-9}  }{(0.015)^{2} }

  = \frac{6.2307*10^{-8} }{(2.25*10^{-4} }

  = 2.7692 x 10^{-4}

 The force between the particles is 2.77 x 10^{-4} N.

3 0
3 years ago
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