Please help me:( I need help with this question

EARTH SCIENCE HELP FAST PLEASE

xxTIMURxx [149]

I think its C not sure tho

8 0

2 years ago

If it takes a ball dropped from rest 2.069 s to fall to the ground, from what height h was it released?

Ganezh [65]

I dont know the answer im sorry

6 0

3 years ago

If the jet is moving at a speed of 1140 km/h at the lowest point of the loop, determine the minimum radius of the circle so that

AnnZ [28]

To solve this problem we will apply the concepts related to the centripetal Force and the Force given by weight and formulated in Newton's second law. Through the two expressions we can find the radius of curve made in the hand. To calculate the normal force, we will include the concepts of sum of forces to obtain the net force on the body at the top and bottom of the maneuver. The expression for centripetal force acting on the jet is

$F_c = \frac{mv^2}{r}$

According to Newton's second law, the net force acting on the jet is

F = ma

Here,

m = mass

a = acceleration

v = Velocity

r = Radius

PART A ) Equating the above two expression the equation for radius is

$\frac{mv^2}{r} = ma$

$r = \frac{v^2}{a}$

Replacing with our values we have that

$r = \frac{(1140km/hr[\frac{1000m}{1km}\frac{1hour}{3600s}])^2}{7(9.8m/s^2)}$

$r = 1.462*10^3m$

PART B )

- The expression for effective weight of the pilot at the bottom of the circle is

$N = mg +\frac{mv^2}{r}$

$N = (69kg)(9.8m/s^2)+\frac{(69)(1140km/hr[\frac{1000m}{1km}\frac{1hour}{3600s}])^2}{1.462*10^3m}$

$N = 5408.87N$

Note that the normal reaction N is directed upwards and gravitational force mg is directed downwards. At the bottom of the circle, the centripetal force is directed upwards. So the centripetal force is obtained from the gravitational force and the normal reaction.

- The expression for effective weight of the pilot at the top of the circle is

$N = mg -\frac{mv^2}{r}$

$N = (69kg)(9.8m/s^2)-\frac{(69)(1140km/hr[\frac{1000m}{1km}\frac{1hour}{3600s}])^2}{1.462*10^3m}$

$N = 4056.47N$

Note that at the top of the circle the centripetal force is directed downwards. So the centripetal force is obtained from normal reaction and the gravitational force.

4 0

3 years ago

The wheel of a stationary exercise bicycle at your gym makes one rotation in 0.670 s. Consider two points on this wheel: Point P

lesya [120]

Answer:

a) For P: $v=0.938\frac{m}{s}$

For Q: $v = 1.876\frac{m}{s}$

b) For P:

$a_{rad}=8.80\frac{m}{s^{2}}$

for Q:

$a_{rad}=17.60\frac{m}{s^{2}}$

c) As the distance from the axis increases then speed increases too.

Explanation:

a) Assuming constant angular acceleration we can find the angular speed of the wheel dividing the angular displacement θ between time of rotation:

$\omega =\frac{\theta}{t}$

One rotation is 360 degrees or 2π radians, so θ=2π

$\omega =\frac{2\pi}{0.670} =9.38\frac{rad}{s}$

Angular acceleration is at every point on the wheel, but speed (tangential speed) is different and depends on the position (R) respect the rotation axis, the equation that relates angular speed and speed is:

$v = \omega R$