What differentiates baseline actives from health enhancing activities?

What does it mean when there is a curved line going upwards on a graph?<br><br>science 8th grade :)

yulyashka [42]

Answer: it is the asymptote

Explanation: a line that continually approaches a given curve but does not meet it at any finite distance.

8 0

3 years ago

A 12,500 N alien UFO is hovering about the surface of Earth. At time , its position can be given as () = ((0.24 m/s^3)^3 + 25 m)

stepladder [879]

a) $F=(3675i-4543k)N$

b) 5843 N

Explanation:

a)

The position of the UFO at time t is given by the vector:

$r(t)=(0.24t^3+25)i+(4.2t)j+(-0.43t^3+0.8t^2)k$

Therefore it has 3 components:

$r_x=0.24t^3+25\\r_y=4.2t\\r_z=-0.43t^3+0.8t^2$

We start by finding the velocity of the UFO, which is given by the derivative of the position:

$v_x=r'_x=\frac{d}{dt}(0.24t^3+25)=3\cdot 0.24t^2=0.72t^2\\v_y=r'_y=\frac{d}{dt}(4.2t)=4.2\\v_x=r'_z=\frac{d}{dt}(-0.43t^3+0.8t^2)=-1.29t^2+1.6t$

And then, by differentiating again, we find the acceleration:

$a_x=v'_x=\frac{d}{dt}(0.72t^2)=1.44t\\a_y=v'_y=\frac{d}{dt}(4.2)=0\\a_z=v'_z=\frac{d}{dt}(-1.29t^2+1.6t)=-2.58t+1.6$

The weight of the UFO is W = 12,500 N, so its mass is:

$m=\frac{W}{g}=\frac{12500}{9.8}=1276 kg$

Therefore, the components of the force on the UFO are given by Newton's second law:

$F=ma$

So, Substituting t = 2 s, we find:

$F_x=ma_x=(1276)(1.44t)=(1276)(1.44)(2)=3675 N\\F_y=ma_y=0\\F_z=ma_z=(1276)(-2.58t+1.6)=(1276)(-2.58(2)+1.6)=-4543 N$

So the net force on the UFO at t = 2 s is

$F=(3675i-4543k)N$

b)

The magnitude of a 3-dimensional vector is given by

$|v|=\sqrt{v_x^2+v_y^2+v_z^2}$

where

$v_x,v_y,v_z$ are the three components of the vector

In this problem, the three components of the net force are:

$F_x=3675 N\\F_y=0\\F_z=-4543 N$

Therefore, substituting into the equation, we find the magnitude of the net force:

$|F|=\sqrt{3675^2+0^2+(-4543)^2}=5843 N$

7 0

3 years ago

A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At

Umnica [9.8K]

Answer:

Explanation:

Given

Charge Q is uniformly spread over large non-conducting Elastic sheet

Electric field due to non-conducting Elastic sheet

$E=\frac{\sigma }{2\epsilon }$

where $\sigma =$ surface charge density $=\frac{q}{d^2}$

$E=\frac{\frac{q}{d^2}}{2\epsilon }$

for side 2d Electric Field is given by

$E'=\frac{\frac{q}{2d^2}}{2\epsilon }$

$E'=\frac{1}{4}\times \frac{\frac{q}{d^2}}{2\epsilon }$

$E'=\frac{E}{4}$

8 0

3 years ago

You throw a ball downward from a window at a speed of 2.5 m/s. how fast will it be moving when it hits the sidewalk 2.1 m below?

Otrada [13]

List out all the variables that you do know;

acceleration=-9.8 ms⁻¹ (this remains constant on Earth)
Final velocity=?
Displacement (s)= -2.1 m
Initial Velocity(u)=2.5 ms⁻¹

v²=u²+2as
v²=(2.5)²+2(-9.8)(-2.1)
v²=47.41
v=√47.41
v=6.88549 ≈ 6.9 ms⁻¹

Hope I helped :)

4 0

3 years ago

What is the maximum weight a boat can hold if a boat can displace 60.5ml?

Valentin [98]

Answer:

a. 60.5 kg

Explanation:

Given data,

The maximum water a boat can displace is, 60.5 ml

According to the principle of buoyancy, the weight of the floating body is equal to the weight of the liquid displaced.

Under standard temperature and pressure, a unit mass of water equals one liter.

If a boat can displace a maximum of 60.5 ml of water, then it can hold a mass of a maximum of 60.5 kg of mass.

6 0

3 years ago