The perimeter of the regular hexagon is 624 cm. If the length of one side is 7x

the kinetic energy of an object of mass in moving with a velocity of 5 MS -1 is 25j what will be its kinetic energy when its vel

MAXImum [283]

Answer:

<em>When the speed is doubled, K = 100 J, when the speed is tripled, K = 225 J</em>

Explanation:

<u>Kinetic Energy </u>

Is the type of energy an object has due to its speed. It's proportional to the square of the speed.

The equation for the kinetic energy is:

$\displaystyle K=\frac{1}{2}mv^2$

Where:

m = mass of the object

v = speed at which the object moves

The kinetic energy is expressed in Joules (J)

The object has a kinetic energy of K=25 J when moving at v=5 m/s, thus the mass can be calculated by solving for m:

$\displaystyle m=\frac{2K}{v^2}$

$\displaystyle m=\frac{2*25}{5^2}=2$

m = 2 Kg

If the speed is doubled, v=10 m/s, the new kinetic energy is:

$\displaystyle K=\frac{1}{2}2\cdot 10^2$

K = 100 J

If the speed is tripled, v=15 m/s, the new kinetic energy is:

$\displaystyle K=\frac{1}{2}2\cdot 15^2$

K = 225 J

When the speed is doubled, K = 100 J, when the speed is tripled, K = 225 J

3 0

3 years ago

Consider the minute hand on a clock. (a) Compute the frequency of its motion in cycles per second. State your answer to three si

kicyunya [14]

Answer:

(a)0.0002778

(b) $1.16\times10^{-5}$

Explanation:

(a) The minute hand has a period of 60 minutes ( or 60 * 60 = 3600 seconds) for 1 circle. Its frequency per second would be

1 / 3600 = 0.0002778

(b) The hour hand has a period of 24 hours ( or 24*60 * 60 = 86400

seconds) for 1 circle. Its frequency per second would be

$1 / 86400 = 1.16\times10^{-5}$

5 0

3 years ago

A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho

fiasKO [112]

Answer:

$v_f=8.17\frac{m}{s}$

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

$W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m$

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

$W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$