How many states have 21 year old minimum drinking age laws

Helen is riding down a freeway on her motorcycle at 95kilometersperhour. At this speed, how far will Helen ride in 2hours?

Serga [27]

The answer is 190kikometersperhour because of you multiply 95 times 2 it equals 190

3 0

2 years ago

A ball is kicked off of a roof at 23 m/s [R 25° U]. What is the height of

Ymorist [56]

Answer:

Explanation:

Considering the fact that we ave been given an angle of inclination here, we best use it! That means that the velocity of 23 m/s is actually NOT the velocity we need; I tell my students that it is a "blanket" velocity but is not accurate in either the x or the y dimension of parabolic motion. In order to find the actual velocity in the dimension in which we are working, which is the y-dimension, we use the formula:

$v_{0y}=v_0sin\theta$ and filling in:

$v_{0y}=23sin(25)$ which gives us an upwards velocity of 9.7 m/s. So here's what we have to work with in its entirety:

$v_{0y}=9.7m/s$

a = -9.8 m/s/s

t = 2.8 seconds

Δx = ?? m

The one-dimensional motion equation that utilizes all of these variables is

Δx = $v_0t+\frac{1}{2}at^2$ and filling in:

Δx = $9.7(2.8)+\frac{1}{2}(-9.8)(2.8)^2$ I am going to do the math according to the correct rules of significant digits, so to the left of the + sign and to 2 sig fig, we have

Δx = 27 + $\frac{1}{2}(-9.8)(2.8)^2$ and then to the right of the + sign and to 2 significant digits we have

Δx = 27 - 38 so

Δx = -11 meters. Now, we all know that distance is not a negative value, but what this negative number tells us is that the ball fell 11 meters BELOW the point from which it was kicked, which is the same thing as being kicked from a building that is 11 meters high.

6 0

3 years ago

A 2.5kg object oscillates at the end of a vertically hanging light spring once every 0.65s .

sesenic [268]

Answer:

Explanation:

By the general expresion for this problem, we have:

y(t) = A*cos(w*t+∅)

since: w = 2π/T = 2π/0,65

For the initial conditions:

y(0) = (0.17cm)*cos(w*0+∅) = + 0.17 m ---> cos(∅) = 1 ---> ∅ = 0°

Then:

A) y(t) = (0.17 m)*cos((2π/0,65)*t)

B part

This means, find the first solution for:

y(t) = (0.17 m)*cos((2π/0,65)*t) = 0 > (equilibrium position)

then: cos((2π/0,65)*t) = 0 ---> (2π/0,65)*t = π/2 ---> t = 0.1625 sec

C part

By definition: (Velocity) v = dy/dt

Then, deriving: v = dy/dt = - (0.17 m)*(2π/0,65)*sin((2π/0,65)*t)

The maximum velocity ocurrs when sin((2π/0,65)*t) = ±1, then (in absolute value): Vmax = 1,64 m/s

D part

By definition: (Aceleration) A = dv/dt

Then, deriving: v = dv/dt = - (0.17 m)*(2π/0,65)²*cos((2π/0,65)*t)

The maximum aceleration ocurrs when cos((2π/0,65)*t) = ±1, then (in absolute value): Amax = 15,88 m/s²

E part

For the acceleration, ocurres by all the solution when:

cos((2π/0,65)*t) = cos(phase) = ±1, this means: phase = {π, 2π, 3π, ...}, all π multiples.

Then, for the position:

y(t) = (0.17 m)*±1= {+0.17 m ; -0.17 m}

For the velocity, ocurres by all the solution when:

sin((2π/0,65)*t) = sin(phase) = ±1, this means: phase = {π/2, π, 3π/2, ...}, all π/2 multiples.

Then, for the position:

y(t) = (0.17 m)*cos(phase)= 0

6 0

3 years ago

Does the sign of the charge (positive or negative) affect how that charge is dissipated? Explain.

labwork [276]

This is an add proceeded by

5 0

3 years ago

PHYSICS 50 POINTS PLEASE HELP

tangare [24]

Answer:

One way to look at Newton’s three laws of motion is this:

The third law states what forces are. That is, all forces are interactions between two different objects. If one object is interacting with another, then equal and opposite forces act on each object. So no force acts alone. When you exert a force on something, it is exerting the identical force back on you.

The first and second laws deal with the consequences of the forces that act on an object. The first law says that in the absence of a net force on an object, it simply continues doing whatever it was already doing. If it is at rest, it will remain at rest. If it is in motion, it will continue with that same motion - at constant speed and in the direction it was already traveling.

The second law says what happens if there is a net force on the object. In that case, the object accelerates - either by changing its speed, its direction, or both - in proportion and in the direction of the net force that acts on it. The amount of acceleration depends the object’s mass. That is, the larger the mass the smaller the acceleration for a given net force. The first and second laws can be summarized in the mathematical expression

F = ma

where F is the vector sum of all the forces that act on the object at any given moment (i.e., the net force), m is the mass of the object, and a is the acceleration of the object due to the net force at that moment - and is always in the same direction of the net force.

And notice that in a way, the first law is then “contained” within the second. That is, if the net force is zero on an object, then so is the acceleration. That is, either the object is (still) at rest or, if already in motion, the velocity didn’t change, in either case, the acceleration was zero.

Explanation:

4 0

3 years ago

Read 2 more answers