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Vesna [10]
3 years ago
12

Hi please may someone help me especially on the sketch part.

Physics
1 answer:
vaieri [72.5K]3 years ago
7 0

Ignoring the air resistance it will take about 3 seconds for the object to reach the ground.We know that the acceleration due to gravity is 10m/s2.

We also know that the final velocity is 30 m/s while the initial velocity is 0 m/s

we can use the formulae for acceleration to calculate the time taken/

(final - initial velocity)/timetaken=10

(30-0)/timetaken=10

timetaken =30/10=3 seconds

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At the presentation ceremony, a championship bowler is presented a 1.64-kg trophy which he holds at arm's length, a distance of
solong [7]

To solve the problem it is necessary to take into account the concepts of the kinetic equations for the description of the torque at the rate of force and distance.

By definition the torque is given by,

\tau = F*d

where,

F= Force

d = Distance

For the problem in question the mass of the trophy is 1.64Kg and the distance of the tropeo to the board (the shoulder) is 0.655m

PART A) For part A, the torque with the given mass and the stipulated torque in the horizontal plane must be calculated as well,

\tau = F*d

For Newton's second law

\tau = mg*d

\tau = 1.64*9.81*0.655

\tau = 10.5Nm

PART B) For part B there is an angle of 26 degrees with respect to the horizontal, therefore to know the net torque it is necessary to know the horizontal component to the formed angle, that is,

\tau = F*dcos\theta

\tau = mgdcos\theta

\tau = 1.64*9.81*0.655*cos26

\tau = 9.471Nm

3 0
4 years ago
To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp
liq [111]

Answer:

Minimum time interval (t2)=0.90 SECONDS

Explanation:

  • coefficient of friction for employees footwear = 0.5
  • coefficient of friction for typical athletic shoe = 0.810
  • frictional force = coefficient of friction X acceleration due to gravity X mass of body
  • Acceleration due to gravity is a constant = 9.81 m/s
  • Let frictional force for employee footwear = FF1
  • Let frictional force for athletic footwear =FF2

                 FF1 = O.5 X 9.81 X mass of body

                         = 4.905 x mass of body

                  FF2 = 0.810 X 9.81 X mass of body

                          = 7.9461 x mass of body

The body started from rest there by making the initial velocity zero ( u = 0)

From d= ut + 1/2 a x t^{2}

  •      d = \frac{1}{2} x a x t^{2}  .....................................i  

            where d= distance and it is given as 3.25m

  •          F =ma  ...................................ii

making acceleration subject of the formula from equation ii

  •              a =\frac{F}{m}

         Making t subject of formula from equation (i)

  • t=\sqrt{\frac{2d}{(f/m} }

    where

  • \frac{FF1}{Mass of body} = 4.905
  • \frac{FF2}{Mass of body} =7.9461

  Let

  •            t1 = minimum time taken for frictional force for employee foot wear
  •                                 t1 = \sqrt{\frac{6.5}{4.905} } =1.15 seconds

  •                                  t2 = \sqrt{\frac{6.5}{7.9461} } = 0.90 seconds

 

THANK YOU

5 0
3 years ago
which of the following is not used to protect us from the possibility of receiving an electric shock​
Juli2301 [7.4K]

Answer:water

Explanation:

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3 years ago
Pleeeeeeease help asap
meriva

Answer:

Pushups will improve your biceps. Pull ups will improve your core and abdomen

5 0
3 years ago
Use Snell's Law to solve the following:
azamat

Answer:

1.171

Explanation:

if n₁sinΘ₁=n₂sinΘ₂, then n₂=n₁sinΘ₁ / sinΘ₂;

n_2=\frac{1.5*sin45}{sin65}=\frac{1.5*0.707}{0.906} =1.1705

3 0
3 years ago
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