Name two elements would be incorrectly placed if the periodic table was ordered by atomic mass rather than atomic number

What does a fern ultimately depend on as an energy source?

Anna35 [415]

It depends on Sunlight.

6 0

3 years ago

John says that the value of the function cos[ω(t + T) + ϕ], obtained one period T after time t, is greater than cos(ωt + ϕ) by 2

Svetllana [295]

Answer:

No one is right

Explanation:

John Case:

The function $cos(\omega t +\phi)$ is defined between -1 and 1, So it is not possible obtain a value $2\pi$ greater.

In addition, if you move the function cosine a T Value, and T is the Period, the function take the same value due to the cosine is a periodic function.

Larry case:

Is you have $f=1+cos(\omega t +\phi)$ , the domain of this is [0,2].

it is equivalent to adding 1 to the domain of the $f=1+cos(\omega t +\phi)$ , and its mean that the function $f=cos(\omega t +\phi)$ , in general, is not greater than $cos(\omega t +\phi)$ .

3 0

3 years ago

HELP ASAP!!!! Which is used to name an ionic compound?

Lina20 [59]

<span>B)if the positive ion ends in -ide </span>

7 0

3 years ago

Read 2 more answers

A rock is thrown off a cliff at an angle of 53° with respect to the horizontal. The cliff is 100 m high. The initial speed of th

Nadusha1986 [10]

(a) 129.3 m

The motion of the rock is a projectile motion, consisting of two indipendent motions along the x- direction and the y-direction. In particular, the motion along the x- (horizontal) direction is a uniform motion with constant speed, while the motion along the y- (vertical) direction is an accelerated motion with constant acceleration $g=-9.8 m/s^2$ downward.

The maximum height of the rock is reached when the vertical component of the velocity becomes zero. The vertical velocity at time t is given by

$v(t) = v_0 sin \theta +gt$

where

$v_0 = 30 m/s$ is the initial velocity of the rock

$\theta=53^{\circ}$ is the angle

t is the time

Requiring $v(t)=0$ , we find the time at which the heigth is maximum:

$0=v_0 sin \theta + gt\\t=\frac{-v_0 sin \theta}{g}=-\frac{(30)(sin 53^{\circ})}{(-9.8)}=2.44 s$

The heigth of the rock at time t is given by

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

Where h=100 m is the initial heigth. Substituting t = 2.44 s, we find the maximum height of the rock:

$y=100+(30)(sin 53^{\circ})(2.44)+\frac{1}{2}(-9.8)(2.44)^2=129.3 m$

(b) 44.1 m

For this part of the problem, we just need to consider the horizontal motion of the rock. The horizontal displacement of the rock at time t is given by

$x(t) = (v_0 cos \theta) t$