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3 years ago

6

A ball is dropped from the top of an eleven-story building to a balcony on the ninth floor. In which case is the change in the p

otential energy associated with the motion of the ball the greatest

1 answer:

lozanna [386]3 years ago

5 0

Answer:

at the top of the 9 story building i think

Explanation:

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A 500N person stands 2.5m from a wall against which a horizontal beam is attached. The beam is 6m long and weighs 200N. A cable

Vinil7 [7]

PART A)

Here by force balance along Y direction

$Tsin45 + F_y = 500 + 200$

Force balance along X direction

$Tcos45 = F_x$

now by torque balance

$Tsin45 (6) = 500 (2.5) + 200 (3)$

$T(4.24) = 1250 + 600$

$T = 436.3 N$

PART B)

now from above equations

$Tsin45 + F_y = 700$

$F_y = 391.5 N$

$F_x = Tcos45 = 308.5 N$

now net reaction force of wall is given as

$F = \sqrt{F_x^2 + F_y^2}$

$F = 498.4 N$

5 0

3 years ago

Which object has more momentum?

lakkis [162]

Answer:

3. Large butterfly in flight, flying through the air?

Explanation:

Momentum is simply defined as the quantity of motion a body possess. It is mathematically given as;

Momentum = mass x velocity

The larger the mass, the larger the momentum and also the velocity

Since the large butterfly is in flight, it has the largest velocity.

A sleeping bear and resting caterpillar have no momentum because their velocity is 0

8 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 describe an electrical motor? Check all that apply

weeeeeb [17]

Answer:

changes electrical energy into mechanical energy

5 0

3 years ago

Water has a very high specific heat capacity when compared to most other common materials. In fact, ethyl alcohol has a specific

AveGali [126]

Answer:

Lead, Ethyl alcohol and water.

Explanation:

Specific heat capacity of a substance can be define as the quantity of heat that is absorbed by a substance needed to change the temperature of a unit mass of one kilogram of the substance by one kelvin

5 0

3 years ago