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Tems11 [23]
3 years ago
15

PLEASE ANSWER ASAP!

Physics
1 answer:
slega [8]3 years ago
4 0

Answer: direct quadratic relationship

Explanation:

I know it's late but I took the test

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An object has a mass of 50.0 g and a volume of 10.5 cm3. What is the object's density?
givi [52]
Divide the objects mass by its volume. 50.0g / 10.5cm^3 =4.76cm ^3
8 0
3 years ago
A gas undergoes two processes. In the first, the volume remains constant at 0.200 m3 and the pressure increases from 1.00×105 Pa
Alborosie
<h2>The work done = - 2 x 10⁴ J</h2>

Explanation:

In the first case , the volume is kept constant and pressure varies .

In isothermal process  , the work done

W₁ = V x ΔP

here V is the volume of gas and ΔP is the change in pressure

Thus W₁ = 0

Because there is no change in volume , therefore displacement is zero .

In second case pressure is constant , but volume changes

Thus W₂ = P x ΔV

here P is the pressure  and ΔV is the change in volume

Therefore W₂ = 4 x 10⁵ x 5 x 10⁻² = 2 x 10⁴ J

The total work done W = - 2 x 10⁴ J

Because the work done in compression is negative .

7 0
4 years ago
1. Explain a stretching routine for your work environment. As you consider your routine, keep in mind the specific individuals w
weeeeeb [17]

Answer:

Explanation:

A stretching routine should target most of the main muscles in the body and at least one stretching exercise per section of the body. Also in a work environment since individual's sizes, ages, and cultures vary you should design a routine that is for beginners and very easy and effective. Regardless, each individual should be able to move at their own pace and push as far as they are physically able. One such routine would be the one in the attached picture below which targets all major areas, is easy and effective. It starts at the neck muscles and works its way down the arms, back, and legs.

3 0
3 years ago
A U-tube is open to the atmosphere at both ends. Water is poured into the tube until the water column on the vertical sides of t
zmey [24]

Answer:

The value is  \Delta h  =  0.003 \  m

Explanation:

From the question we are told that

   The  height of the water is  h_1  =  10 \ cm  =  0.10 \  m

    The  density of  oil is \rho_o  =  950 \  kg/m^3

  The  height of  oil  is  h_2  =  6 \ cm  =  0.06 \  m

Given that both arms of the tube are open then the pressure on both side is the same

So  

      P_a =  P_b

=>   Here  

             P_a  =  P_z + \rho_w  *  g *  h

where  \rho_w is the density of water with value  \rho_w =  1000 \ kg/m^3

and  P_z is the atmospheric pressure

and  

        P_b  =  P_z + \rho_o  *  g *  h_2

=>   P_z + \rho_w  *  g * h =    P_z + \rho_o  *  g *  h_2

=>    \rho_w   * h  =    \rho_o  *  h_2

=>      h  =  \frac{950 * 0.06 }{1000}

=>      h  = 0.057 \ m

The  difference in height is evaluated as    

           \Delta h  =  0.06 - 0.057

          \Delta h  =  0.003 \  m

     

6 0
4 years ago
A quarterback throws a football with an angle of elevation 55° and speed 60 ft/s. Find the horizontal and vertical components o
EastWind [94]

Answer:

The horizontal component of the velocity vector is;

vh = 34.4 ft/s

The vertical component of the velocity vector is;

vy = 49.1 ft/s

Explanation:

Given;

Velocity of football v = 60 ft/s

Angle of elevation ∅ = 55°

The horizontal component of the velocity vector is;

vh = vcos∅

Substituting the values;

vh = 60cos55°

vh = 34.41458618106 ft/s

vh = 34.4 ft/s

The vertical component of the velocity vector is;

vy = vsin∅

Substituting the values;

vy = 60sin55°

vy = 49.14912265733 ft/s

vy = 49.1 ft/s

7 0
3 years ago
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