If the weight of 1kg is 10 N in water what is the density of the stone

GREYUIT [131]

Answer:

h = 1.8 m

Explanation:

The initial velocity of the glove, u =- 6 m/s

We need to find the maximum height of the glove. Let it is equal to h. Using equation of kinematics. At the maximum height v = 0

$v^2-u^2=2ah$ , h is the maximum height and a = -g

$0^2-(6)^2=2\times (-10)\times h\\\\h=\dfrac{36}{20}\\\\h=1.8\ m$

Hence, it will go up to a height of 1.8 m.

4 0

2 years ago

The centers of two 15.0 kg spheres are separated by 3.00 m. The magnitude of the gravitational force between the two spheres is

kompoz [17]

we have to use newtons law of gravitation which is
F=GMm/r^2
G=6.67 x 10^-11N kg^2/m^2
M=(15kg)
m=15 kg
r=(3.0m)^2
putting values we have
=(6.67 x 10^-11N kg^2/m^2)(15kg)(15kg)/(3.0m)^2 
=1.67 x 10^-9N

7 0

3 years ago

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i

RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is $2.7 rad/s^2$

Explanation:

a)

In a uniform circular motion, the centripetal acceleration is given by

$a_c=\omega^2 r$

where:

$\omega$ is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

$\omega=1.7 rad/s$ is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity ( $g=9.8 m/s^2$ ), so:

$a_c=3.2 g = 3.2(9.8)=31.4 m/s^2$

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

$r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m$

b)

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

$a=4.4 g$