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anyanavicka [17]
2 years ago
13

What is the car's acceleration from 0 to 1 second?

Physics
1 answer:
dusya [7]2 years ago
3 0
10 mph/s because there is 60 seconds in a minute then divide by 6 which is 10.
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The centripetal force acting on the space shuttle
tamaranim1 [39]

Answer:

(4) weight

Explanation:

The centripetal force acting on the space shuttle in orbit is given by:

F=m\frac{v^2}{r}

where

m is the mass of the shuttle

v is the tangential speed of the shuttle

r is the radius of its circular orbit

When the shuttle orbits the Earth, the centripetal force that keeps the shuttle in circular motion is given by the gravitational attraction between the shuttle and the Earth, which corresponds to the weight of the shuttle, and it is given by:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M is the Earth's mass

And this force, therefore, corresponds to the centripetal force.

7 0
3 years ago
Read 2 more answers
The charges of two particles are as follows: Q1=2 x 10 -8 C and Q2 = 3 x 10 -7 C. Find the magnitude of the force between these
Fantom [35]

Answer:

F = 3.86 x 10⁻⁶ N

Explanation:

First, we will find the distance between the two particles:

r = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2+(z_{2}-z_{1})^2}\\

where,

r = distance between the particles = ?

(x₁, y₁, z₁) = (2, 5, 1)

(x₂, y₂, z₂) = (3, 2, 3)

Therefore,

r = \sqrt{(3-2)^2+(2-5)^2+(3-1)^2}\\r = 3.741\ m\\

Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:

F = \frac{kq_{1}q_{2}}{r^2}\\

where,

F = magnitude of force = ?

k = Coulomb's Constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of first charge = 2 x 10⁻⁸ C

q₂ = magnitude of second charge = 3 x 10⁻⁷ C

r = distance between the charges = 3.741 m

Therefore,

F = \frac{(9\ x\ 10^9\ Nm^2/C^2)(2\ x\ 10^{-8}\ C)(3\ x\ 10^{-7}\ C)}{(3.741\ m)^2}\\

<u>F = 3.86 x 10⁻⁶ N</u>

5 0
3 years ago
What is runoff water?
vitfil [10]

<u>Answer:</u> runoff water is water from rain, snow, or other sources, that flows through the land, and is a major component of the water cycle.

6 0
3 years ago
Read 2 more answers
On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const
IceJOKER [234]

Answer:

a)

Explanation:

  • Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

        x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}   (1)

  • Since the car starts from rest, v₀ =0.
  • We know the value of t = 5 sec., but we need to find the value of a.
  • Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

       a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2  (2)

  • Replacing a and t in (1):

       x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}  = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m.  (3)

  • Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
  • Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

       a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2  (4)

  • Replacing v₀, at and t in (1), we have:

       x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m   (5)

  • Therefore, as the truck travels twice as far as the car, the right answer is a).
7 0
2 years ago
C, N, Ne, Ar which contains a metal, nonmetal, noble gas , metalloid
postnew [5]

Answer:

c,carbono nonmetal

n, nitrogen nonmetal

ne, neón noble gas

ar,argon noble gas

4 0
2 years ago
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