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3 years ago

What is the work required to lift a 10 kg box off the ground to a height of 5.0 m?

Physics

1 answer:

FinnZ [79.3K]3 years ago

7 0

Answer:

490J

Explanation:

Given parameters:

Mass of box = 10kg

Height = 5m

Unknown:

Work done = ?

Solution:

Work done can be derived from the product of force and distance;

Work done = force x distance

Force here is applied by gravity and it is the weight

The distance is the height

Work done = weight x height

Work done = mass x acceleration due to gravity x height

Work done = 10 x 9.8 x 5 = 490J

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If i have a kinematic equation vf^2=vi^2-2*a(xf-xi), how can i solve for xi step by step

azamat

Answer:

Explanation:

$v_f^2 = v_i^2-2a(x_f-x_i)$

Subtract both sides by $v_i^2$ :

$- v_i^2+v_f^2 = -2a(x_f-x_i)$

Divide both sides by -2*a:

$\frac{v_i^2 - v_f^2}{2a} =x_f-x_i$

Add both sides by $x_i$ :

$x_i+\frac{v_i^2 - v_f^2}{2a} =x_f$

Subtract both sides by $\frac{v_i^2 - v_f^2}{2a}$ :

$x_i=x_f-\frac{v_i^2 - v_f^2}{2a}$

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3 years ago

A motorboat accelerates uniformly from a velocity of 10.5 m/s to the west to a velocity of 6.5 m/s to the west. If its accelerat

BlackZzzverrR [31]

Answer:

7.4m ............................

3 0

3 years ago

A girl pulls on a 10-kg wagon with a constant horizontal force of 30 n. if there are no other horizontal forces, what is the wag

olga2289 [7]

Force = mass * acceleration
F = ma

Given m = 10 kg, F = 30 N;

F = ma
30 = 10a

Solving for a:
a = 3 m/s^2

The acceleration is 3 meters per second squared.

8 0

3 years ago

A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m

elena-14-01-66 [18.8K]

Answer:

1,230N

Explanation:

1. Name of the variables:

$f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration$

2. Formulae:

$f=\dfrac{number\text{ }of\text{ }revolutions}{time}$

$\omega=2\pi f$

$a_c=\omega^2 r$

$F_c=m\times a_c$

3. Solution (calculations)

$f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}$

$\omega=2\pi\times0.\overline{60}\approx 3.808rad/s$

$a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2$

$F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N$

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3 years ago

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3 years ago

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