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IgorLugansk [536]

3 years ago

15

Aliens Who stay in the country longer than their legally allowed limit are called

1 answer:

IRINA_888 [86]3 years ago

8 0

Illegal immigrants. Yep

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Sveta_85 [38]

The answer is b. All you have to do is find the intercepts.

6 0

3 years ago

A helicopter, starting from rest, accelerates straight up from the roof of a hospital. The lifting force does work in raising th

erik [133]

Answer:

$P = 36068.7 Watt$

Explanation:

As we know that average power is the ratio of total work done and total time interval

so we will have

$Power = \frac{Total \: Work}{time}$

now we know that

$Work = \frac{1}{2}mv^2 + mgh$

$W = \frac{1}{2}(700)(11^2) + 700(9.8)(9.6)$

so we will have

$W = 42350 J + 68856 J$

$W = 108206 J$

now power is given as

$P = \frac{108206}{3}$

$P = 36068.7 Watt$

6 0

3 years ago

The area of the atom surrounding the nucleus is mostly what

alexdok [17]

Empty space; an atom consists of a dense, positively charged nucleus surrounded by a "cloud" of negatively charged electrons

5 0

3 years ago

Characteristics we use to tell the difference between kinds of matter are called___ properties

d1i1m1o1n [39]

Answer:

I believe it's called qualitative

Explanation:

8 0

3 years ago

A 1.1 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an

elixir [45]

Answer with Explanation:

Mass of block=1.1 kg

Th force applied on block is given by

F(x)= $(2.4-x^2)\hat{i}N$

Initial position of the block=x=0

Initial velocity of block= $v_i=0$

a.We have to find the kinetic energy of the block when it passes through x=2.0 m.

Initial kinetic energy= $K_i=\frac{1}{2}mv^2_i=\frac{1}{2}(1.1)(0)=0$

Work energy theorem:

$K_f-K_i=W$

Where $K_f=$ Final kinetic energy

$K_i$ =Initial kinetic energy

$W=Total work done$

Substitute the values then we get

$K_f-0=\int_{0}^{2}F(x)dx$

Because work done= $Force\times displacement$

$K_f=\int_{0}^{2}(2.4-x^2)dx$

$K_f=[2.4x-\frac{x^3}{3}]^{2}_{0}$

$K_f=2.4(2)-\frac{8}{3}=2.13 J$

Hence, the kinetic energy of the block as it passes thorough x=2 m=2.13 J

b.Kinetic energy = $K=2.4x-\frac{x^3}{3}$

When the kinetic energy is maximum then $\frac{dK}{dx}=0$

$\frac{d(2.4x-\frac{x^3}{3})}{dx}=0$

$2.4-x^2=0$

$x^2=2.4$

$x=\pm\sqrt{2.4}$

$\frac{d^2K}{dx^2}=-2x$

Substitute x= $\sqrt{2.4}$

$\frac{d^2K}{dx^2}=-2\sqrt{2.4}$

Substitute x= $-\sqrt{2.4}$

$\frac{d^2K}{dx^2}=2\sqrt{2.4}>0$

Hence, the kinetic energy is maximum at x= $\sqrt{2.4}$

Again by work energy theorem , the maximum kinetic energy of the block between x=0 and x=2.0 m is given by

$K_f-0=\int_{0}^{\sqrt{2.4}}(2.4-x^2)dx$

$k_f=[2.4x-\frac{x^3}{3}]^{\sqrt{2.4}}_{0}$

$K_f=2.4(\sqrt{2.4})-\frac{(\sqrt{2.4})^3}{3}=2.48 J$

Hence, the maximum energy of the block between x=0 and x=2 m=2.48 J

3 0

4 years ago