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4 years ago

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A toy floats in a swimming pool. The buoyant force exerted on the toy depends on the volume ofa. none of these choices.b. all th

e water in the pool.c. the toy under water.d. the pool.

1 answer:

Citrus2011 [14]4 years ago

6 0

Answer:c

Explanation:

The buoyancy force on toy depends upon the volume of toy under pool water.

According to Archimedes principle buoyant force on a floating body depends upon the weight of displaced liquid by object.

Buoyancy force is given by

$F_b=\rho _f\times V\times g$

where $\rho _f$ =density of fluid

V=volume of object under water

g=acceleration due to gravity

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A plate of uniform areal density is bounded by the four curves: where and are in meters. Point has coordinates and . What is the

Natali5045456 [20]

The question is incomplete. The complete question is :

A plate of uniform areal density $\rho = 2 \ kg/m^2$ is bounded by the four curves:

$y = -x^2+4x-5m$

$y = x^2+4x+6m$

$x=1 \ m$

$x=2 \ m$

where x and y are in meters. Point $P$ has coordinates $P_x=1 \ m$ and $P_y=-2 \ m$ . What is the moment of inertia $I_P$ of the plate about the point $P$ ?

Solution :

Given :

$y = -x^2+4x-5$

$y = x^2+4x+6$

$x=1$

$x=2$

and $\rho = 2 \ kg/m^2$ , $P_x=1 \$ , $P_y=-2 \$ .

So,

$dI = dmr^2$

$dI = \rho \ dA \ r^2$ , $r=\sqrt{(x-1)^2+(y+2)^2}$

$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \ dy \ dx$

$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$

$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$

$I=\frac{32027}{21} \times 2$

$= 3050.19 \ kg \ m^2$

So the moment of inertia is $3050.19 \ kg \ m^2$ .

4 0

3 years ago

A 4-79 permalloy solenoid coil needs to produce a minimum inductance of 1.1 . If the maximum allowed current is 4 , how many tur

daser333 [38]

The related concept to solve this exercise is given in the expressions that the magnetic field has both as a function of the number of loops, current and length, as well as inductance and permeability. The first expression could be given as,

The magnetic field H is given as,

$H = \frac{nI}{l}$

Here,

n = Number of turns of the coil

I = Current that flows in the coil

l = Length of the coil

From the above equation, the number of turns of the coil is,

$n = \frac{Hl}{I}$

The magnetic field is again given by,

$H = \frac{B}{\mu_t}$

Where the minimum inductance produced by the solenoid coil is B.

We have to obtain n, that

$n = \dfrac{\frac{B}{\mu_t}l}{I}$

Replacing with our values we have that,

$n = \dfrac{\frac{1.1Wb/m^2 }{200000}(2m)}{4mA}$

$n = \dfrac{(\frac{1.1Wb/m^2 }{200000})(\frac{10^4 guass}{1Wb/m^2})(2m)}{4mA(\frac{10^{-3}A}{1mA})}$

$n = 27.5 \approx 28$

Therefore the number of turn required is 28Truns

4 0

4 years ago

A 1000 kg car moves with a constant speed 20.0 m/s

skad [1K]

Answer:

13.33 m/s^2

Explanation:

Velocity^2 then divide that by the radius

6 0

3 years ago

Two cars, A and B , travel in a straight line. The distance of A from the starting point is given as a function of time by xA(t)

OLga [1]

A) Car A is initially ahead

B) The two cars are at the same point at the times: t = 0, t = 2.27 s and

t = 5.73 s

C) The distance between the two cars is not changing at t = 1.00 s and t = 4.33 s

D) The two cars have same acceleration at t = 2.67 s

Explanation:

A)

The position of the two cars at time t is given by the following functions:

$x_A(t) = \alpha t + \beta t^2$

with

$\alpha = 2.60 m/s\\\beta = 1.20 m/s^2$

Substituting,

$x_A(t)=2.60t+1.20 t^2$

And

$x_B(t)=\gamma t^2 - \delta t^3$

with

$\gamma=2.80 m/s^2\\\delta = 0.20 m/s^3$

Substituting,

$x_B(t)=2.80t^2-0.20t^3$

Here we want to find which car is ahead just after they leave the starting point. To find that, we just need to calculate the position of the two cars after a very short amount of time, let's say at t = 0.1 s. Substituting this value into the two equations, we get:

$x_A(0.1)=2.60(0.1)+1.20(0.1)^2=0.27 m$

$x_B(0.1)=2.80(0.1)^2-0.20(0.1)^3=0.03 m$

So, car A is initially ahead.

B)

The two cars are at the same point when their position is the same. Therefore, when

$x_A(t)=x_B(t)$

which means when

$2.60t+1.20t^2 = 2.80t^2-0.20t^3$

Re-arranging the equation, we find

$0.20t^3-1.6t^2+2.60t=0\\t(0.20t^2-1.6t+2.60)=0$

One solution of this equation is t = 0 (initial point), while we have two more solutions given by the equation

$0.20t^2-1.6t+2.60=0$

which has two solutions:

t = 2.27 s

t = 5.73 s

So, these are the times at which the cars are at the same point.

C)

The distance between the two cars A and B is not changing when the velocities of the two cars is the same.

The velocity of car A is given by the derivative of the position of car A:

$v_A(t) = x_A'(t)=(2.60t+1.20t^2)'=2.60+2.40t$

The velocity of car B is given by the derivative of the position of car B:

$v_B(t)=x_B'(t)=(2.80t^2-0.20t^3)'=5.60t-0.60t^2$

Therefore, the distance between the two cars is not changing when the two velocities are equal:

$v_A(t)=v_B(t)\\2.60+2.40t=5.60t-0.60t^2\\0.60t^2-3.20t+2.60=0$

This is another second-order equation, which has two solutions:

t = 1.00 s

t = 4.33 s

D)

The acceleration of each car is given by the derivative of the velocity of the car A.

The acceleration of car A is:

$a_A(t)=v_A'(t)=(2.60+2.40t)'=2.40$

While the acceleration of car B is:

$a_B(t)=v_B'(t)=(5.60t-0.60t^2)'=5.60-1.20t$

So, the two cars have same acceleration when

$a_A(t)=a_B(t)$

And solving the equation, we find:

$2.40=5.60-1.20t\\1.20t=3.20\\t=2.67 s$

So, the two cars have same acceleration at t = 2.67 s.

Learn more about accelerated motion:

brainly.com/question/9527152

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brainly.com/question/2562700

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3 0

3 years ago

PLEASE HELP!!!!!!! MY HOMEWORK IS DUE TODAY !!!!!!!!

bogdanovich [222]

Answer:

Distance = 25000000 miles

Time = 50 hours

Explanation:

Venus is the closest planet to Earth. It is about 25 million miles away from Earth. Its precise distance depends on where both Venus and Earth are in their respective orbits

Given that

Speed V = 500000 mph

Distance d = 25 000,000 miles

Speed = distance/ time

Time = distance/speed

Time = 25000000/500000

Time = 50 hours

It will therefore take 50 hours to get to venus at that speed.

5 0

3 years ago