Does all mass take up space

1. A runner has an initial speed of 2 [m/s] and slowly speeds up with a constant acceleration of

zvonat [6]

Answer:

Final Velocity = 4.9 m/s

Explanation:

We are given;. Initial velocity; u = 2 m/s

Constant Acceleration; a = 0.1 m/s²

Distance; s = 100 m

To find the final velocity(v), we will use one of Newton's equations of motion;

v² = u² + 2as

Plugging in the relevant values to give;

v² = 2² + 2(0.1 × 100)

v² = 4 + 20

v² = 24

v = √24

v = 4.9 m/s

5 0

2 years ago

PLZ HELP

lianna [129]

The air movements toward the equator are called trade winds, which are warm, steady breezes that blowalmost continuously. The Coriolis Effect makes the trade winds appear to be curving to the west, whether they are traveling to the equator from the south or north. Answer trade wind

7 0

2 years ago

A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle

liq [111]

Answer:

0.0031792338 rad/s

Explanation:

$\theta$ = Angle of elevation

y = Height of balloon

Using trigonometry

$tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta$

Differentiating with respect to t we get

$\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta$

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

$h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft$

Now y = 2500 ft

$cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974$

$\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s$

The angle is changing at 0.0031792338 rad/s

6 0

3 years ago

When landing after a spectacular somersault, a 35.0 kg gymnast decelerates by pushing straight down on the mat. calculate the fo

Sliva [168]

The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.

Now from Newton`s first law, the net force on gymnast,

$F_{net} =F-W=ma$

Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast ( $9\times g$ acceleration due to gravity)

Therefore,

$F= ma+W$ OR $F=ma+mg=m(g+a)$

Given $m = 30 kg$ and $a=9\times g=9\times 9.8 m/s^{2} =88.2 m/s^{2}$

Substituting these values in above formula and calculate the force exerted by the gymnast,

$F=(40 kg) (88.2 m/s^{2} +9.8 m/s^{2} )$

$F=3.537\times10^{3}N$

6 0

3 years ago

(1.08×10-3)(9.3×10-4)

ira [324]

First you do the first parenthesis, (1.08 x 10 - 3) and you do it in the order of operations! (parenthesis, exponents, multiplication/division, add/subtract) to get 7.8. Then you take the second parenthesis (9.3 x 10 - 4) and do the same thing to get 89! You then times 7.8 by 89 to get 694.2! If it needs more elaboration just ask ^.^

4 0

3 years ago