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Romashka [77]
2 years ago
5

Which of the following calls one might hear a refree make?

Physics
2 answers:
NemiM [27]2 years ago
8 0
The answer is d all of the above
mestny [16]2 years ago
7 0

Answer: D) All of the above

Explanation:

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A student switches the torch on and sees that it gives out a bright light.
garik1379 [7]

Answer:

no more fire duhh

Explanation:

3 0
2 years ago
Read 2 more answers
Please help with this and explain it,if you can.
vladimir2022 [97]

Answer:

displacement at 45 s  =  30

                           65 s  = 50

So the average speed over the interval from 45 s to 65 s is

(50 - 30) cm / 20 s = 1 cm / sec

As a check an average speed of 1 cm / sec for 20 sec will produce a

displacement of 1 cm / sec * 20 sec = 20 cm  or from 30 to 50 cm

4 0
3 years ago
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
What makes these image a represent of what research <br>​
ycow [4]
The economic down fall
3 0
3 years ago
As shown in the picture above, a sled of mass 67 kilograms is pulled
Scorpion4ik [409]

Explanation:

F net of sled = Tension force by rope - Kinetic friction between ground.

F normal of sled = mg = (67kg)(9.81kg/m^2) = 657.27N.

Kinetic friction = 0.18 (I cannot see the value) * Normal force of sled = 0.18 * 657.27N = 118.31N

So F net of sled = 800N - 118.31N = 681.69N.

(I cannot see what the question is asking for, please check on your own!)

3 0
3 years ago
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