Find the magnitude of the force at point x = 15cm if k=10500 N/m when the work done by this force is WD= 1/2k x2

A high-jump athlete leaves the ground, lifting her center of mass 1.8 m and crossing the bar with a horizontal velocity of 1.4 m

Romashka [77]

Answer:

The minimum speed when she leave the ground is 6.10 m/s.

Explanation:

Given that,

Horizontal velocity = 1.4 m/s

Height = 1.8 m

We need to calculate the minimum speed must she leave the ground

Using conservation of energy

$K.E+P.E=P.E+K.E$

$\dfrac{1}{2}mv_{1}^2+0=mgh+\dfrac{1}{2}mv_{2}^2$

$\dfrac{v_{1}^2}{2}=gh+\dfrac{v_{2}^2}{2}$

Put the value into the formula

$\dfrac{v_{1}^2}{2}=9.8\times1.8+\dfrac{(1.4)^2}{2}$

$\dfrac{v_{1}^2}{2}=18.62$

$v_{1}=\sqrt{2\times18.62}$

$v_{1}=6.10\ m/s$

Hence, The minimum speed when she leave the ground is 6.10 m/s.

6 0

3 years ago

A train moving at a constant speed on a surface inclined upward at 10.0° with the horizontal travels a distance of 400 meters in

Amiraneli [1.4K]

If the velocity of the train is v=s/t, where s is the distance and t is time, then v=400/5=80m/s. To get the vertical component of the velocity we need to multiply the velocity v with a sin(α): Vv=v*sin(α), where Vv is the vertical component of the velocity and α is the angle with the horizontal. So:

Vv=80*sin(10)=80*0.1736=13.888 m/s.

So the vertical component of the velocity of the train is Vv=13.888 m/s.

7 0

3 years ago

How long does it take (in minutes) for light to reach venus from the sun, a distance of 1.152 × 108 km?

7nadin3 [17]

Using the precise speed of light in a vacuum ( $299,792,458 \ \frac{m}{s}$ ), and your given distance of $1.152 * 10^{8} km$ , we can convert and cancel units to find the answer. The distance in m, using $\frac{1000 \ m}{1 \ km}$ , is $1.152 * 10^{11} m$ . Next, for the speed of light, we convert from s to min, using $\frac{1 \ min}{60 \ s}$ , so we divide the speed of light by 60. Finally, dividing the distance between the Sun and Venus by the speed of light in km per min, we find that it is 6.405 min.