Which of the following is equal to 1W? A.1 J B.1 J/s C.1 J·s D.1 N·m

What is the direction of the force of an object moving in a circle?

Ivanshal [37]

Answer:

The answer is 45 degree angle

7 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

Which of the following is the best example of kinetic energy? A. A bowling ball rolling toward the pins B. A horse standing in a

ryzh [129]

The answer is A. Only an object in motion has kinetic energy.

8 0

3 years ago

find the period of a simple pendulum of 1m length placed on earth and on moon g on moon =1.67m/s² g on earth=10m/s²

Ierofanga [76]

Answer:

$T_{m }$ = 4.86 s

$T_{e}$ = 1.98 s

Explanation:

Given:

Length = l = 1 m

Acceleration due to gravity of moon = $g_{m}$ = 1.67 m/s²

Acceleration due to gravity of Earth = $g_{e}$ = 10 m/s²

Required:

Time period = T = ?

Formula:

T = 2π $\sqrt{\frac{l}{g} }$

Solution:

For moon

Putting the givens,

T = 2(3.14) $\sqrt{\frac{1}{1.67} }$

T = 6.3 $\sqrt{0.6}$

T = 6.3 × 0.77

T = 4.86 sec

For Earth,

Putting the givens

T = 2π $\sqrt{\frac{1}{10} }$

T = 2(3.14) $\sqrt{0.1}$

T = 6.3 × 0.32

T = 1.98 sec

3 0

3 years ago

The force of gravity on Mercury is less than the Earth because:

Vinvika [58]

Mercury has less mass than earth. So the answer is B

7 0

3 years ago

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Which of the following is equal to 1W?A.1 JB.1 J/sC.1 J·sD.1 N·m