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harkovskaia [24]
2 years ago
14

Johnny was playing baseball with his friends and they noticed a bolt of lightning. They heard thunder seven seconds later. How f

ar away is the storm?
Physics
2 answers:
Thepotemich [5.8K]2 years ago
4 0

That particular strike was very roughly 2.4 km (1.5 miles) away from them.

That's if you use 340 m/s (1120 ft/sec) for the speed of sound. 
But the air in the region for several thousand feet around a thunderstorm
is doing weird things to sounds that pass through it, so you can't use any
exact number for the speed of sound in a stormy area.

The only thing you can be absolutely sure of is that Johnny and his friends
need to round up their equipment and get in the house.  NOW ! 
murzikaleks [220]2 years ago
3 0

2,317 meters on edg!!

You might be interested in
a torch bulb is rated 2.5V and 750mA. Calculate its power,its resistance and the energy consumed if this bulb lighted for 4 hour
Hatshy [7]
Using Ohm's Law, we can derived from this the value of resistance. If I=V/R, therefore, R = V/I
Substituting the values to the given, 
P = Power = ?
R = Resistance = ?
V = Voltage = 2.5 V
I = Current = 750 mA

R = V/I = 2.5/ (750 x 10^-3)
R = 3.33 ohms

Calculating the power, we have P = IV

P = (750 x 10^-3)(2.5) 
P = 1.875 W

The power consumption is the power consumed multiply by the number of hours. In here, we have;
1.875W x 4 hours = 7.5 watt-hours
3 0
2 years ago
When the boy crashes his bumper car into the girl's bumper car, the momentum from his car is transferred to hers. What evidence
Darya [45]

Answer:

Where is the text?

Explanation:

If you refer to the short sentence you wrote as text, I believe the answer is probably the word "crashes" because it shows how the momentum was transferred.

8 0
2 years ago
Read 2 more answers
A punted football is observed to have velocity components vhorizontal = 15 m/s to the right and vvertical = 1.25 m/s directed do
enot [183]

Answer:

v₀ₓ = 15 m / s,  v_{oy} = 5.2 m / s

v = 15.87 m / s ,   θ = 19.1

Explanation:

This is a projectile launch problem. The horizontal speed that is constant throughout the entire path is worth 15 m / s, instead the vertical speed changes in value due to the acceleration of gravity, let's look for the initial vertical speed

                      Vy² =v_{oy}² - 2 g y

                      v_{oy}² = v_{y}² + 2 g y

                       v_{oy} = √ (v_{y}² + 2 gy

Let's calculate

                    v_{oy} = √ (1.25² + 2 9.8 1.3)

                    v_{oy} = √ (27.04)

                    v_{oy} = 5.2 m / s

 The initial speed can be calculated by the initial speed

                   v = √ v₀ₓ² + v_{oy}²

                   v = RA (15² + 5.2²)

                   v = 15.87 m / s

We look for the angle with trigonometry

                 tan θ = voy / vox

                 θ = tan⁻¹ I'm going / vox

                θ = tan⁻¹ 5.2 / 15

                θ = 19.1

The answer is

              v₀ₓ = 15 m / s

              v_{oy} = 5.2 m / s

5 0
2 years ago
If the angle of elevation of the cannon is decreased from 35 to 30 degrees, the vertical component of the ball's initial velocit
stealth61 [152]

Answer:

Determine the initial horizontal and vertical velocity for a dart launched with a ... an angle of 30 degrees. Vi=4m/s Vix Vicoso Mas cos 30 = = 30°. Viy=visine ... What is the resultant direction (angle) of the cannon ball (relative to a fixed frame)? ... Which of the following launch angles would result in the greatest time of flight iſ the.

Explanation:

I hope it helped

4 0
2 years ago
Official (Closed) - Non Sensitive
Pavlova-9 [17]

Answer:

The minimum running time is 319.47 s.

Explanation:

First we find the distance covered and time taken by the train to reach its maximum speed:

We have:

Initial Speed = Vi = 0 m/s    (Since, train is initially at rest)

Final Speed = Vf = 29.17 m/s

Acceleration = a = 0.25 m/s²

Distance Covered to reach maximum speed = s₁

Time taken to reach maximum speed = t₁

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi)/a

t₁ = (29.17 m/s - 0 m/s)/(0.25 m/s²)

t₁ = 116.68 s

Using 2nd equation of motion:

s₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

s₁ = (0 m/s)(116.68 s) + (0.5)(0.25 m/s²)(116.68 s)²

s₁ = 1701.78 m = 1.7 km

Now, we shall calculate the end time and distance covered by train, when it comes to rest on next station.

We have:

Final Speed = Vf = 0 m/s    (Since, train is finally stops)

Initial Speed = Vi = 29.17 m/s     (The train must maintain max. speed for min time)

Deceleration = a = - 0.7 m/s²

Distance Covered to stop = s₂

Time taken to stop = t₂

Using 1st equation of motion:

Vf = Vi + at₂

t₂ = (Vf - Vi)/a

t₂ = (0 m/s - 29.17 m/s)/(- 0.7 m/s²)

t₂ = 41.67 s

Using 2nd equation of motion:

s₂ = (Vi)(t₂) + (0.5)(a)(t₂)²

s₂ = (29.17 m/s)(41.67 s) + (0.5)(- 0.7 m/s²)(41.67 s)²

s₂ = 607.78 m = 0.6 km

Since, we know that the rest of 7 km, the train must maintain the maximum speed to get to the next station in minimum time.

The remaining distance is:

s₃ = 7 km - s₂ - s₁

s₃ = 7 km - 0.6 km - 1.7 km

s₃ = 4.7 km

Now, for uniform speed we use the relation:

s₃ = vt₃

t₃ = s₃/v

t₃ = (4700 m)/(29.17 m/s)

t₃ = 161.12 s

So, the minimum running time will be:

t = t₁ + t₂ + t₃

t = 116.68 s + 41.67 s + 161.12 s

<u>t = 319.47 s</u>

5 0
2 years ago
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