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IrinaK [193]
2 years ago
3

A storm is moving east towards your house at an average speed of 35 km/hr. If the storm is currently 80 kilometers from your hou

se in how much time do you expect it to arrive
Physics
2 answers:
gayaneshka [121]2 years ago
5 0
The storm is moving toward you at 35km/hr. How long until it travels to 80 kilometers to your house?
Distance = velocity * time
80km = 35 km/hour * x hours  Divide both sides by 35
2.29 hours = x

It will take 2.29 hours to reach your house. 
xxMikexx [17]2 years ago
5 0
The problem given above will be solved by using the speed formula.
Speed is defined as distance over time, that is,
Speed [S] = Distance [D] / Time [T].
In the question, the value of speed that we are given is 35 and the value of D is 80, we have to find the value of time.
S = 35
D = 80
T = ?
S = D /T = 35 = 80 / T
35T = 80 
T = 80 / 35 = 2.29.
Therefore, T = 2.3 hour.


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2 years ago
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An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest ini
arsen [322]
Kinetic energy is calculated through the equation,

   KE = 0.5mv²

At initial conditions,

  m₁:  KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J

  m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J

Due to the momentum balance,

   m₁v₁ + m₂v₂ = (m₁ + m₂)(V)

Substituting the known values,

   (0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)

   V = 0.2977 m/s

The kinetic energy is,
   KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
   KE = 0.03146 J

The difference between the kinetic energies is 0.0473 J. 
7 0
2 years ago
Ingrid is moving a box from the ground into the back of a truck. She uses 20 N of force to move the box 5 meters. If she uses an
Oduvanchick [21]

Answer:

C

Explanation:

4 0
2 years ago
An object has a mass of 8.00kg. What is the gravitational force on the object by the earth
Firdavs [7]
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2 years ago
A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the cen
saw5 [17]

Answer:

The electric field strength is 4.5\times 10^{4} N/C

Solution:

As per the question:

Area of the electrode, A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}

Charge, q = 50 nC = 50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m

Now,

To calculate the electric field strength, we first calculate the surface charge density which is given by:

\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}

Now, the electric field strength of the electrode is:

\vec{E} = \frac{\sigma}{2\epsilon_{o}}

where

\epsilon_{o} = 8.85\times 10^{- 12} F/m

\vec{E} = \frac{8\times 10^{- 7}}{2\times 8.85\times 10^{- 12}}

\vec{E} = 4.5\times 10^{4} N/C

7 0
2 years ago
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