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3 years ago

Because of friction, it required 6818.2 N to move a trailer 542 meters. How much work was accomplished?

Physics

1 answer:

nikklg [1K]3 years ago

5 0

Answer:

The answer to your question is W = 3695464.4 J

Explanation:

Mechanical work is the force applied to an object multiply by the distance this object move as a consequence of the force.

Formula

Work = W (J)

Force = F (N)

distance = d (m)

W = F x d

Substitution

F = 6818.2 N

d = 542 m

W = 6818.2 x 542

Simplification and result

W = 3695464.4 J

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mamaluj [8]

Answer:

22m/s

Explanation:

To find the velocity we employ the equation of free fall: v²=u²+2gh

where u is initial velocity, g is acceleration due to gravity h is the height, v is the velocity the moment it hits the ground, taking the direction towards gravity as positive.

Substituting for the values in the question we get:

v²=2×9.8m/s²×25m

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v=22.14m/s which can be approximated to 22m/s

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A student pulls a cart across the floor with a force of 45 N. If the cart travels 10 m, how much work does the cart do on the st

bixtya [17]

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If the current through a 20-ω resistor is 8.0 a , how much energy is dissipated by the resistor in 1.0 h ?

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3 years ago

A bicycle wheel has a diameter of 63.0 cm and a mass of 1.75 kg. Assume that the wheel is a hoop with all of the mass concentrat

Masteriza [31]

Answer:

F2 = 834 N

Explanation:

We are given the following for the bicycle;

Diameter; d1 = 63 cm = 0.63 m

Mass; m = 1.75 kg

Resistive force; F1 = 121 N

For the sprocket, we are given;

Diameter; d2 = 8.96 cm = 0.0896 m

Radius; r2 = 0.0896/2 = 0.0448 m

Radial acceleration; α = 4.4 rad/s²

Now moment of inertia of the wheel which is assumed to be a hoop is given by; I = m(r1)²

Where r1 = (d1)/2 = 0.63/2

r1 = 0.315 m

Thus, I = 1.75 × 0.315²

I = 0.1736 Kg.m²

The torque is given by the relation;

I•α = F1•r1 - F2•r2

Where F2 is the force that must be applied by the chain to give the wheel an acceleration of 4.40 rad/s².

Thus;

0.1736 × 4.4 = (121 × 0.315) - (0.0448F2)

>> 0.76384 = 38.115 - (0.0448F2)

>> 0.0448F2 = 38.115 - 0.76384

>> F2 = (38.115 - 0.76384)/0.0448

>> F2 = 833.73 N

Approximately; F2 = 834 N

7 0

3 years ago

A firecracker breaks up into two pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.0 m/

larisa [96]

Answer:

Total momentum, p = 21.24 kg-m/s

Explanation:

Given that,

Mass of first piece, $m_1=200\ g= 0.2\ kg$

Mass of the second piece, $m_2=300\ g= 0.3\ kg$

Speed of the first piece, $v_1=82\ m/s$ (along x axis)

Speed of the second piece, $v_2=45\ m/s$ (along y axis)

To find,

The total momentum of the two pieces.

Solve,

The total momentum of two pieces is equal to the sum of momentum along x axis and along y axis.

$p_x=m_1v_1$

$p_x=0.2\ kg\times 82\ m/s$

$p_x=16.4\ kg-m/s$

$p_y=m_2v_2$

$p_y=0.3\ kg\times 45\ m/s$

$p_y=13.5\ kg-m/s$

The net momentum is given by :

$p=\sqrt{p_x^2+p_y^2}$

$p=\sqrt{16.4^2+13.5^2}$

p = 21.24 kg-m/s

Therefore, the total momentum of the two pieces is 21.24 kg-m/s.

4 0

3 years ago