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4 years ago

11

Which could result from untreated diabetes check all that apply A .Asthma B. Numbness in hands and feet C.Exhaustion D.Weight lo

ss E.Blurred vision

2 answers:

soldier1979 [14.2K]4 years ago

7 0

Answer:

B. Numbness In Hands and Feet

C. Exhaustion

D. Weight Loss

E. Blurred Vision

grigory [225]4 years ago

3 0

With the exception of Asthma, all of those things, and a lot more that you don't want, could result from untreated diabetes.

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Which statement is the best example of pseudoscience?

Nady [450]

<span>pseudoscience includes beliefs theories, or practices that have been or are considered scientific, but have no basis in scientific fact.
This could mean the were disproved scientifically, can't be tested or lack evidence to support them.
Examples:Channeling- involves communicating with a spirit through a person

Astrology- beliefs that humans are affected by the position of celestial bodies
</span>

5 0

3 years ago

In order to slide a heavy desk across the floor at constant speed in a straight line, you have to exert a horizontal force of 40

san4es73 [151]

Answer:

F = f from Newton’s first law.

Explanation:

since the desk is moved in a straight line at a constant speed, newton first law tell us that the two forces must be equal.

Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. since the table has been set in motion by the 400 N force, it will remain in motion unless it is been acted upon by an external force, and this means that the 400 N must be equal to the frictional force for it to have been in motion in the first instance.

3 0

4 years ago

A gymnast of mass 63.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t

Sergio [31]

Answer:

Explanation:

A ) When gymnast is motionless , he is in equilibrium

T = mg

= 63 x 9.81

= 618.03 N

B )

When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.

T = mg

= 618.03 N

C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2

Net force on it = T - mg , acting in upward direction

T - mg = m a

T = mg + m a

= m ( g + a )

= 63 ( 9.81 + .6)

= 655.83 N

D ) If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2

Net force acting in downward direction

mg - T = ma

T = m ( g - a )

= 63 x ( 9.81 - .6 )

= 580.23 N

6 0

3 years ago

A baseball player friend of yours wants to determine his pitching speed. you have him stand on a ledge and throw the ball horizo

zhenek [66]

Answer:

The pitching speed of your friend is 33.20 m/s

Explanation:

<em>Lets explain how to solve the problem</em>

Your friend throw the ball horizontally that means the vertical initial

component of velocity is zero ( $u_{y}=0$ ).

The ball is thrown from a height 4 meters above the ground.

The height $h=u_{y}t+\frac{1}{2}gt^{2}$

<u><em>Remember:</em></u> the height is negative value because its below the point of

thrown (initial position)

h = -4 m , $u_{y}=0$ and g = -9.8 m/s²(downward)

<em>Substitute these values in the rule above</em>

⇒ $4=0-\frac{1}{2}(9.8)t^{2}$

⇒ -4 = -4.9t² (multiply both sides by -1)

⇒ 4 = 4.9t² (divide both sides by 4.9)

⇒ 0.81633 = t² (take √ for both sides)

⇒ <em>t = 0.9035</em>

Then the time of the ball to land on the ground is 0.9035 seconds

The range of the ball on the ground is 30 m

The range $R=u_{x}*t$ , where $u_{x}$ is the horizontal

component of the initial velocity

R = 30 meters and t = 0.9035

⇒ $30=u_{x}(0.9035)$ (divide both sides by 0.9035)

⇒ $u_{x}=33.20$ m/s

<em>The pitching speed of your friend is 33.20 m/s </em>

4 0

3 years ago

A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area

weqwewe [10]

Answer:

W₂= 10000 N

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

Pressure is defined as the force (F) applied per unit area (A)

P=F/A (N/m²)

P1=P2

$\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }$

$F_{2} = \frac{F_{1}*A_{2} }{A_{1}}$ Equation (1)

Data

W₁ = weight sits on the small piston

F₁ = W₁= 500 N

A₁ = 2.0 cm²

A₂ = 40 cm²

Calculation of the weight (W₂) can the large piston support

We replace data in the equation (1)

$F_{2} = \frac{(500)*(40) }{2}$

F₂ = 10000 N

W₂= F₂= 10000 N

6 0

4 years ago