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Leni [432]
3 years ago
14

At highway speeds a car can accelerate at 1.7 m/s2.

Physics
1 answer:
Vaselesa [24]3 years ago
6 0

Answer: 2.5 seconds

Explanation:

We know that the acceleration is:

a(t) = 1.7 m/s^2

To get the velocity function, we must integrate over time, and we will get:

v(t) = (1.7m/s^2)*t + v0

Where v0 is the initial velocity, in this case, we assume that we start at 23.6m/s, then the initial velocity is:

v0 = 23.6 m/s

Then the velocity equation is:

v(t) =  (1.7m/s^2)*t + 23.6 m/s

Now we want to find the value of t such v(t) = 27.8 m/s

Then:

v(t) =  27.8 m/s =  (1.7m/s^2)*t + 23.6 m/s

        27.8 m/s  - 23.6 m/s =  (1.7m/s^2)*t

         4.2 m/s = (1.7m/s^2)*t

         4.2m/s/(1.7m/s^2) = t = 2.5 s

Then at that acceleration, you need 2.5 seconds.

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George rides his bike to his friend’s house that is 5 kilometers from his house. If he rides his bike at an average speed of 15
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Read 2 more answers
A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi
bogdanovich [222]

Answer:

1) W_{object} = 400 N

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

\sum F_{Tot} = 0

In this case we just have y-direction forces.

\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0

Now, let's solve the equation for W(object).

W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N

2 ) To find the position of the heaviest weight we need to use the torque definition.

\sum \tau = 0

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

L = 1.50 m  ; total length of the bar.

D_{AP} = x  ; distance between Tension A and P point.

D_{BP} = L-x ; distance between Tension B and P point.

D_{W_{bar}P} = \frac{L}{2}-x ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0

Finally we just need to solve it for x.

x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}

x = 0.28 m

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

8 0
4 years ago
A 100 kg boat is floating in water. half of the boat is submerged under water. what is the weight of the boat?
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W=MG
w is weight 
m is mass
g is gravity 
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hope this helps
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3 years ago
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