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3 years ago

Eating disorder Symptoms

Physics

1 answer:

weeeeeb [17]3 years ago

4 0

Answer:

some symptoms can be lack of energy, and dizziness.

Explanation:

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A ball is released from the top of a tower of height h meters. it takes T seconds to reach the ground . what is the position of

alekssr [168]

Answer:the

8/9 h

Explanation:

Height = 1/2 a T^2 now change to T/3

now height = 1/2 a (T/3)^2 =<u> 1/9</u> 1/2 a T^2 <===== it is 1/9 of the way down or 8/9 h

3 0

2 years ago

The gravitational acceleration is 9.81 m/s2 here on Earth at sea level. What is the gravitational acceleration at a height of 35

azamat

To solve this problem it is necessary to apply the definition of severity of Newtonian laws in which it is specified that gravity is defined by

$g= \frac{GM}{R^2}$

Where

G= Gravitational Constant

M = Mass of Earth

R= Radius from center of the planet

According to the information we need to find the gravity 350km more than the radius of Earth, then

$g_{ss} = \frac{GM}{R+h^2}$

$g_{ss} = \frac{6.67*10^{-11}*5.972*10^{24}}{(6371*10^3+350*10^3)^2}$

$g_{ss} = 8.82m/s^2$

Therefore the gravitational acceleration at 350km is $8.82m/s^2$

5 0

3 years ago

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.310 m and then accelerates through this dista

Nastasia [14]

Answer:A)u =4.295m/s , B)a = 29.746m/s² C) F=3,153N

Explanation:

Using the kinematic expression

v² = u² - 2as

where

u = initial velocity

v = final velocity

s = distance

g = acceleration due to gravity .

Given that he reaches a height of 0.940 m above the floor,

the final velocity = 0

Here, acceleration due to gravity is acting in opposite the initial direction of motion. So, a=-9.81 m/s.

v² = u² + 2as

0² - u² = 2 (- 9.81) × 0.940

- u² = 2 × - 9.81 × 0.920

- u² = -18.4428

cancelling the minus in both sides , we have that

u² = 18.4428

u = √18.4428

u =4.295m/s

(b) His acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.310 m. m/s2

Using v² = u² + 2as

where u = initial speed of basketball player before lengthening = 0 m/s,

v = final speed of basketball player after lengthening = 4.295m/s,

a = acceleration while straightening his legs

s = distance moved during lengthening = 0.310m

v² = u² + 2as

a = (v² - u²)/2s

a = (4.29m/s)² - (0 m/s)²)/(2 × 0.310m)

a = (18.4428 m²/s² - 0 m²/s²)/(0.62 m)

a = (18.4428 m²/s²/(0.62 m)

a = 29.746m/s²

c) The force (in N) he exerts on the floor to do this, given that his mass is 106 kg. N

Force= mass x acceleration.

F = 106 kg X 29.746m/s²

F = 3,153.076 rounded to 3,153N

8 0

3 years ago

What is true about energy that is added to a closed system

arsen [322]

It does work or increases thermal energy

4 0

4 years ago

Read 2 more answers

To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.7 kg and length 3.0 m

Zanzabum

Answer:

$T_{1}$ = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

τ(net) = 0

Taking moment about $T_{1}$ we have:

(M + m) g * 0.5L - $T_{2}$ (L - d) = 0

⇒ $T_{2}$ = [(M + m) g * 0.5L] ÷ (L - d)

$T_{2}$ = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

$T_{2}$ = 59.535 ÷ 2.4

$T_{2}$ = 24.80625 N ≈ 24.81 N

Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N

Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N

Using sum of equilibrium in the vertical direction, we have:

$T_{1}$ + $T_{2}$ = W + w ------- Eqn 1

Substituting T2, W & w into the Eqn 1

$T_{1}$ + 24.81 = 26.46 + 13.23

$T_{1}$ = <u>14.88</u> N

8 0

3 years ago