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3 years ago

Which statement is true about the circuit diagram below?

Physics

1 answer:

Snezhnost [94]3 years ago

8 0

Answer:

Explanation:

Well, the electricity starts it's move from the positive terminal to the negative terminal, So it's clockwise.

and the resistances ain't parallel, so the gon be in series.

your answer gon be choice A

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If the pressure of a gas is increased, then the volume will decre

pogonyaev

Answer:

The answer is b

Explanation:

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

A projectile is fired with an initial velocity of 450 feet per second at an angle of 70° with the horizontal.

pav-90 [236]

<h2>After 26.28 seconds projectile returns 26.28 seconds.</h2>

Explanation:

Initial velocity = 450 ft/s = 137.16 m/s

Angle, θ = 70°

Consider the vertical motion of projectile,

When the projectile return to the ground we have

Displacement, s = 0 m

Acceleration, a = -9.81 m/s²

Initial velocity, u = 137.16 x sin70 = 128.89 m/s

Substituting in s = ut + 0.5 at²

s = ut + 0.5 at²

0 = 128.89 x t + 0.5 x (-9.81) x t²

t² - 26.28 t = 0

t ( t- 26.28) = 0

t = 0 s or t = 26.28 s

After 26.28 seconds projectile returns 26.28 seconds.

5 0

3 years ago

I need help fast please help

Irina18 [472]

A and C Im pretty sure :)

5 0

3 years ago

A 10 cm diameter pulley is used to lift a bucket of cement weighing 400 N. How much force must be applied to the rope to lift th

Alina [70]

hi brainly user! ૮₍ ˃ ⤙ ˂ ₎ა

⊱┈────────────────────────┈⊰

$\large \bold {ANSWER}$

Considering that the pulley is fixed, the force applied should be equal to the weight of the object - of 400N.

$\large \bold {EXPLANATION}$

Pulleys or pulleys are mechanical tools used to assist in the movement of objects and bodies. There are two types of pulleys: fixed and movable. While the fixed pulley changes the direction of force, the moving pulley helps to decrease the force needed to move the object or body in question.

As the statement only tells us a pulley, we must consider that it is fixed, <u>because generally when it is mobile, this information is highlighted in the question</u>.

In this way, a fixed pulley only changes the direction of the applied force. Thus, the force must have the same magnitude as the weight of the object to be moved. If the bucket weighs 400N, the force applied to the pulley must be 400N.

<u>Therefore, having a fixed pulley, the force applied must be equal to the weight of the object, and will be 400N.</u>

3 0

2 years ago