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astra-53 [7]
3 years ago
10

A child rides her bike at a rate of 12.0 km/hr down the street. A squirrel suddenly runs in front of her so she applies the brak

es and slows to 8.0 km/hr in 0.25 sec. What is the acceleration of the bike rider during this period?
Physics
1 answer:
laiz [17]3 years ago
3 0

Answer:

C: 16 km/hr/sec

Explanation:

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A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f
VashaNatasha [74]

Answer:

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34  

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 =  X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + (\frac{L-y}{2}) ) g sinФ - F2 x (\frac{L}{2}) x gsinФ = 0

plug in the  equations of F1 and F2 into the above equation and after simplification, we get:

(L^{2} - y^{2} ) . w = L^{2} . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

(L^{2} - y^{2} ) . w = L^{2} . 5/9 x w

Hence,

(L^{2} - y^{2} ) = \frac{5L^{2} }{9}

\frac{L^{2} - y^{2}  }{L^{2} } = \frac{5}{9}

Taking L^{2} common and solving for \frac{y}{L}, we will get

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34

8 0
3 years ago
For this triangle...?
Mariulka [41]
<h3><u>Answer</u>;</h3>

= 0.6

<h3><u>Explanation</u>;</h3>

Using Pythagoras theorrem

Base² + height ² = Hypotenuse²

Thus;

Base² = 15² - 12²

          = 81

Base = √81 = 9

But; cosine = adjacent/hypotenuse

Hence; cos θ = 9/15

                      <u>= 0.6 </u>

5 0
3 years ago
Read 2 more answers
Two particles with the same charge will repel each other. True or false
FinnZ [79.3K]
True. Think of a magnet and how they only connect to the opposite charges. 
7 0
3 years ago
How does the number of valence electrons affect the reactivity?
ololo11 [35]
The less valence electrons the more reactive the element is.
6 0
3 years ago
Given A=125.0=0.4 and 25.0=0.1 calculate A-B<br><br><br><br><br>​
kicyunya [14]

Answer: Find an answer to your question given A=125.0=0.4 and 25.0=0.1 calculate A-B​

Explanation:

6 0
2 years ago
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