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3 years ago

5

Which of the following actions does NO work?

2 answers:

aliya0001 [1]3 years ago

8 0

Answer:

Droping a baseball

Explanation:

becasue you are useing no sort of force to drop a baseball but you are for everything else

Mashutka [201]3 years ago

6 0

Answer:

Dropping a baseball

Explanation:

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sasho [114]

Answer:

C?

Explanation:

Please give brainliest.

5 0

2 years ago

A child on a tricycle is moving at a speed of 1.40 m/s at the start of a 2.25 m high and 12.4 m long incline. The total mass is

goblinko [34]

Answer:

The work done by the child as the tricycle travels down the incline is 416.96 J

Explanation:

Given;

initial velocity of the child, $v_i$ = 1.4 m/s

final velocity of the child, $v_f$ = 6.5 m/s

initial height of the inclined plane, h = 2.25 m

length of the inclined plane, L = 12.4 m

total mass, m = 48 kg

frictional force, $f_k$ = 41 N

The work done by the child is calculated as;

$\Delta E_{mech} = W - f_{k} \Delta L\\\\W = \Delta E_{mech} + f_{k} \Delta L\\\\W = (K.E_f - K.E_i) + (P.E_f - P.E_i) + f_{k} \Delta L\\\\W = \frac{1}{2} m(v_f^2 - v_i^2) + mg(h_f - h_i) + f_{k} \Delta L\\\\W = \frac{1}{2} \times 48(6.5^2 - 1.4^2) + 48\times 9.8(0-2.25) + (41\times 12.4)\\\\W = 966.96 \ - \ 1058.4 \ + \ 508.4\\\\W = 416.96 \ J$

Therefore, the work done by the child as the tricycle travels down the incline is 416.96 J

5 0

3 years ago

Hurricanes are especially destructive because of the abrupt decrease in air pressure that they bring with the winds. Consider a

ivann1987 [24]

Answer:

2.96 × 10^4 N

Explanation:

1 atm = 101325 N/m², pressure inside the airtight room = 1.02 atm, pressure outside due to hurricane = 0.91 atm

net pressure directed outward = P inside - P outside

net pressure = 1.02 - 0.91 = 0.11 atm

where 1 atm = 101325N/m²

0.11 atm = 0.11 × 101325 N/m² = 11145.75 N/m²

area of the square wall = l × l where l is the length of the wall in meters = 1.63 × 1.63 = 2.6569

net pressure = net force / area

make net force subject of the formula

net force = net pressure × area = 11145.75 × 2.6569 = 2.96 × 10 ^4 N

8 0

2 years ago

A gymnast dismounts off the uneven bars in a tuck position with a radius of 0.3m (assume she is a solid sphere) and an angular v

kifflom [539]

Here we will say that there is no external torque on the system so we will have

$L_i = L_f$

here we know that

$L_i = I_1\omega_1$

where we know that

$I_1 = \frac{2}{5}mr^2$

Also we know that

$I_2 = \frac{1}{12}mL^2$

initial angular speed will be

$\omega_1 = 2\pi(2rev/s) = 4\pi rad/s$

now from above equation

$\frac{2}{5}mr^2 (4\pi) = \frac{1}{12}mL^2 \omega$

$0.4(0.3)^2(4\pi) = \frac{1}{12}(1.5)^2\omega$

$0.452 = 0.1875 \omega$

now we have

$\omega = 2.41 rad/s$

so final speed will be 2.41 rad/s

6 0

2 years ago

A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are

igor_vitrenko [27]

Answer:

0.231 N

Explanation:

To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

$\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2$

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

$I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2$

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

$T = I\alpha = 0.303*0.6454 = 0.196 Nm$

So the force acting on the other end to generate this torque mush be:

$F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N$

4 0

3 years ago