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ohaa [14]
3 years ago
5

An enormous thunderstorm covers Dallas-Ft. Worth. Your best friend Clark is a storm chaser and heads to the center of the storm

to take some readings while you stay dry at home. While Clark is at the center of the storm, he sees and hears lightning strike a tree that is 150 m from where he is standing. You are 127 km from the tree. How long does it take for the sound to reach Clark
Physics
1 answer:
Rus_ich [418]3 years ago
6 0

Answer:

t = 0.437 s

Explanation:

The speed of sound is a constant that is worth v = 343 m / s

           v = d / t

            t = d / v

the time it takes for the sound to reach Clark at d = 150 m is

           t = 150/343

           t = 0.437 s

This same sound takes much longer to reach you

          t₂ = 127 10³/343

          t₂ = 370 s

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Can someone please answer this, ill give you brainliest Would be very appreciated.
Whitepunk [10]

Answer:

frost wedging

Explanation:

When glaciers melt and the rock beneath is weathered and broken down, this is called frost wedging.

6 0
2 years ago
On the interstate, the speed limit is 60 mi/h (about 100 km/h) a skilled driver can safely decelerate at about 6.1m/s^2. How lon
velikii [3]

Answer:

4.56 seconds

63.25 m

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 60 km/h = 100 km/h = 100000/3600 = 27.78 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -6.1 m/s²

Time taken by the car to stop

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-27.78}{-6.1}\\\Rightarrow t=4.56\ s

Time taken by the car to stop is 4.56 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow s=27.78\times 4.56+\frac{1}{2}\times -6.1\times 4.56^2\\\Rightarrow s=63.25\ m

The total stopping distance would be 63.25 m

4 0
3 years ago
PLEASE HELP NEED TO KNOW IF CORRECT!! BRAINLIEST ANSWER WILL BE REWARDED.
melisa1 [442]
The correct answer is transmission<<<<<<
8 0
3 years ago
Explain why friction makes it harder to push a box filled with groceries than an empty box
kupik [55]
It is harder to push a box with groceries in it because the mass is weighing the box down. Which makes it harder to push
5 0
2 years ago
A 0.500-kg object connected to a light spring with a spring constant of 20.0 N/m oscillates on a frictionless horizontal surface
givi [52]

Answer:

a = 0.009 J

b = 0.19 m/s

c = 0.005 J and 0.004 J

Explanation:

Given that

Mass of the object, m = 0.5 kg

Spring constant of the spring, k = 20 N/m

Amplitude of the motion, A = 3 cm = 0.03 m

Displacement of the system, x = 2 cm = 0.02 m

a

Total energy of the system, E =

E = 1/2 * k * A²

E = 1/2 * 20 * 0.03²

E = 10 * 0.0009

E = 0.009 J

b

E = 1/2 * k * A² = 1/2 * m * v(max)²

1/2 * m * v(max)² = 0.009

1/2 * 0.5 * v(max)² = 0.009

v(max)² = 0.009 * 2/0.5

v(max)² = 0.018 / 0.5

v(max)² = 0.036

v(max) = √0.036

v(max) = 0.19 m/s

c

V = ±√[(k/m) * (A² - x²)]

V = ±√[(20/0.5) * (0.03² - 0.02²)]

V = ±√(40 * 0.0005)

V = ±√0.02

V = ±0.141 m/s

Kinetic Energy, K = 1/2 * m * v²

K = 1/2 * 0.5 * 0.141²

K = 1/4 * 0.02

K = 0.005 J

Potential Energy, P = 1/2 * k * x²

P = 1/2 * 20 * 0.02²

P = 10 * 0.0004

P = 0.004 J

4 0
2 years ago
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