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rosijanka [135]
2 years ago
14

Choose 3 ways to keep our bodies healthy

Physics
2 answers:
Alina [70]2 years ago
7 0

Answer:

take daily showers . eat vegies .sanitize your hands

Lemur [1.5K]2 years ago
3 0
Sanitize your hands, eat veggies, take daily showers. Hope this helped! (No explanation needed lol)



This was fun to answer.
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In a local bar, a customer slides an empty beer mug down the counter for a refill. the height of the counter is 1.15 m. the mug
yarga [219]

<span>A.    </span>Let’s say the horizontal component of the velocity is vx and the vertical is vy. <span>
Initially at t=0 (as the mug leaves the counter) the components are v0x and v0y. 
<span>v0y = 0 since the customer slides it horizontally so applied force is in the x component only.

<span>The equations for horizontal and vertical projectile motion are:
x = x0 + v0x t 
y = y0 + v0y t - 1/2 g t^2 = y0 - 1/2 g t^2 </span></span></span>

Setting the origin to be the end corner of the counter so that x0=0 and y0=0, hence:

x = v0x t

y = - 1/2 g t^2 

Given value are: x=1.50m and y=-1.15m (y is negative since mug is going down)

<span>1.50m = v0x t    ---->  v0x= 1.50/t</span>

<span>-1.15m = -(1/2) (9.81) t^2    -----> t =0.4842 s</span>

Calculating for v0x:

v0x = 3.10 m/s

<span>B.    </span>v0x is constant since there are no other horizontal forces so, v0x=vx=3.10m/s

vy can be calculated from the formula:

<span>vy = v0y + at         where a=-g (negative since going down)</span>

vy = -gt = -9.81 (0.4842)

vy = -4.75 m/s

 

Now to get the angle below the horizontal, tan(90-Ø) = -vx/vy

tan(90-Ø )= 3.1/4.75

Ø = 56.87˚<span> below the horizontal</span>

4 0
4 years ago
Read 2 more answers
John ( body mass= 65 kg) is taking off for a long jump . Horizontal accerleration ax is 5m/s^2 and vertical acceleration ay is 0
cestrela7 [59]

Answer:

(a) The horizontal ground reaction force  F_{g,x}=325\, N

(b)  The vertical ground reaction force F_{g,y}=696\, N

(c)   The resultant ground reaction force F_g=768\, N

Explanation:  

Given

John mass , m = 65 kg

Horizontal acceleration , a_x= 5.0 \frac{m}{s^{2}}

Vertical acceleration , a_y=0.9 \frac{m}{s^{2}}

(a) Using Newton's 2nd law in horizontal direction

F_{g,x}=ma_x

=>F_{g,x}=65\times 5\, N=325\, N

Thus the horizontal ground reaction force  F_{g,x}=325\, N

(b) Using Newton's 2nd law in vertical direction

F_{g,y}-mg=ma_y

=>F_{g,y}=mg+ma_y

=>F_{g,y}=65\times (9.81+0.9)\, N=696\, N

Thus the vertical ground reaction force F_{g,y}=696\, N

(c)  Resultant ground reaction force is

F_g=(F_{g,x}^{2}+F_{g,y}^{2})^{\frac{1}{2}}

=>F_g=(325^{2}+696^{2})^{\frac{1}{2}}\, N=768\, N

=>F_g=768\, N

Thus  the resultant ground reaction force F_g=768\, N

5 0
3 years ago
A tiger travels 3m/s^2 for 4.1s,what was its initial speed if it's final speed was 55k/h?
zaharov [31]
Acceleration a=3m/s^2
time t= 4.1seconds
Final velocity V= 55km/h
initial velocity U= ?
First convert V to m/s
36km/h=10m/s
55km/h= 55*10/36=15.28m/s
Using the formula V= U+at
U= V-at
U= 15.28-3*4.1=15.28-12.3=2.98m/s
Initial velocity U= 2.98m/s or 10.73km/h (Using the conversion rate 36km/h=10m/s)
8 0
3 years ago
A ball is thrown vertically upwards with a velocity
zhuklara [117]

Answer:

Explanation:

The acceleration of gravity is 9.8m/s^2.

So to calculate the time it will take to make the ball stop(which btw means the ball now reach its greatest height), use the formula V1=V0+at. V1 is the final velocity(which is 0), V0 is the starting velocity(which is 30m/s), and the a(cceleration) is 9.8m/s^2.

(You can ignore the fact "at" is -30 instead 30, it's because the directions two velocity travel are opposite. )

We can now know the time it takes to make the ball stop just by the gravitational force is about 3 sec.

Use another formula S=1/2at^2, to find out the S(height) is 1/2*9.8*3^2=44.1, which is approximately D.45m .

6 0
3 years ago
The curved section of a speedway is a circular arc having a radius of 190 m. this curve is properly banked for racecars moving a
Anestetic [448]

The banking angle of the curved part of the speedway is determined as 32⁰.

<h3>Banking angle of the curved road</h3>

The banking angle of the curved part of the speedway is calculated as follows;

V(max) = √(rg tanθ)

where;

  • r is radius of the path
  • g is acceleration due to gravity

V² = rg tanθ

tanθ = V²/rg

tanθ = (34²)/(190 x 9.8)

tanθ = 0.62

θ = arc tan(0.62)

θ = 31.8

θ ≈ 32⁰

Learn more about banking angle here: brainly.com/question/8169892

#SPJ1

8 0
2 years ago
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