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3 years ago

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

1 answer:

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 $t^{2}$