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Kay [80]
3 years ago
13

All vehicles, including animal-drawn vehicles, must have at least one white light visible from a distance of not less than _____

_____ feet to the front.
Physics
1 answer:
Y_Kistochka [10]3 years ago
5 0
1000 feet. Hope this helps.
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A blue ball is thrown upward with an initial speed of 21.8 m/s, from a height of 0.9 meters above the ground. 2.7 seconds after
worty [1.4K]
I can think of two possible and logical questions for the problem given. First, you can calculate for the maximum height reached by the blue ball. Second, you can compute the length of time for the two balls to be at the same height. If so, the solution are as follows:

When the object is thrown upwards or when the object is dropped from a height, the only force acting upon it is the gravitational force. Because of this, it simplifies equations of motion.

1. For the maximum height, the equation is
H = v₀²/2g
where
v₀ is the initial speed
g is the acceleration due to gravity equal to 9.81 m/s²

For the blue ball, v₀ = 21.8 m/s. Substituting the values:
H = (21.8 m/s)²/2(9.81m/s²)
H = 24.22 m
The maximum height reached by the blue ball is 24.22 m + 0.9 = 25.12 m.

2. For this, you equate the y values of both balls:

y for red ball = y for blue ball
v₀t + 0.5gt² = v₀t + 0.5gt²
(10.4 m/s)t + 0.5(9.81 m/s²)(t²) + 26.6 m = (21.8 m/s)t + 0.5(9.81 m/s²)(t²) + 0.9 m
Solving for t, 
t = 2.25 seconds

Thus, the two balls would be at the same height after 2.25 seconds.
3 0
3 years ago
The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou
zlopas [31]

Answer:

q = 3.6 10⁵  C

Explanation:

To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth

          r = 6 , 37 106 m

          E = k q / r²

          q = E r² / k

          q = \frac{100 \ (6.37 \ 10^6)^2}{9 \ 10^9}

          q = 4.5 10⁵ C

Now let's calculate the charge on the planet with E = 222 N / c and radius

           r = 0.6 r_ Earth

           r = 0.6 6.37 10⁶ = 3.822 10⁶ m

           E = k q / r²

            q = E r² / k

            q = \frac{222 (3.822 \ 10^6)^2}{ 9 \ 10^9}

            q = 3.6 10⁵  C

4 0
3 years ago
What distance is covered by an airplane traveling at a velocity of 660 miles per hour in 3.5 hours?
N76 [4]

As per the question, the velocity of the airplane [v] = 660 miles per hour.

The total time taken by airplane [t] = 3.5 hours.

We are asked to determine the total distance travelled by the airplane during that period.

The distance covered [ S] by a body is the product of velocity with the time.

Mathematically distance covered = velocity × total time

                                                      S = v × t

                                                        = 660 miles/hour ×3.5 hours

                                                        = 2310 miles.

Hence, the total distance travelled by the airplane in 3.5 hour is 2310 miles.

4 0
3 years ago
A man climbs on to a wall that is 3.6m high and gains 2268J of potential energy. What is the mass of the man? *
Ilia_Sergeevich [38]
The mass of the man is 63 kilograms
There is it written down properly. Its hard to type this stuff in here

3 0
3 years ago
James Joule (after whom the unit of energy is named) claimed that the water at the bottom of Niagara Falls should be warmer than
Molodets [167]

Answer:

0.12 K

Explanation:

height, h = 51 m

let the mass of water is m.

Specific heat of water, c = 4190 J/kg K

According to the transformation of energy

Potential energy of water = thermal energy of water

m x g x h = m x c x ΔT

Where, ΔT is the rise in temperature

g x h =  c x ΔT

9.8 x 51 = 4190 x ΔT

ΔT = 0.12 K

Thus, the rise in temperature is 0.12 K.

7 0
3 years ago
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