Science

If a 1 kg book has 46 Joules of gravitational potential energy how high is the shelf it is on?

Mashcka [7]

Answer:

4.7m

Explanation:

Given parameters:

Mass of the book = 1kg

Gravitational potential energy = 46J

Unknown:

Height of the shelf = ?

Solution:

The potential energy is due to the position of a body above the ground.

Gravitational potential energy = mgh

m is the mass,

g is the acceleration due gravity = 9.8m/s²

h is the height which is unknown

46 = 1 x 9.8 x h

h = 4.7m

4 0

2 years ago

The pulley system has a mechanical advantage of 5. Determine how much input force is required to achieve 1000 N of lifting force

olga55 [171]

Answer:

Input force of pulley system = 200 N

Explanation:

Given:

Mechanical advantage of pulley system = 5

Output force from pulley system = 1,000 N

Find;

Input force of pulley system

Computation:

Mechanical advantage = Output force / Input force

Mechanical advantage of pulley system = Output force from pulley system / Input force of pulley system

5 = 1,000 / Input force of pulley system

Input force of pulley system = 1,000 / 5

Input force of pulley system = 200 N

3 0

2 years ago

For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate

faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue= $h_1$ =0.200 m

Metal volume flow rate,Q=0.03 $m^3/min$

Q= $\frac{0.03}{60}=5\times 10^{-4}m^3/s$ because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

$d_=0.030 m$

$h_2=0$

$A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}$

$A_1=7.065\times 10^{-4} m^2$

$v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}$

$v_1=0.708 m/s$

Pressure at the top and bottom of the sprue is atmospheric

$h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}$

Substitute the values

$0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}$

$v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264$

$v_2=\sqrt{4.421264}=2.1 m/s$

$Q=A_2v_2$

$5\times 10^{-4}=A_2\times 2.1$

$A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2$

Reynolds number= $\frac{v_2D\rho}{\eta}$

$\eta=0.004 N.s/m^2$

$\rho=2700 kg/m^3$

Substitute the values then we get

Reynolds number= $\frac{2.1\times 0.03\times 2700}{0.004}$

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0

2 years ago

A boy and a girl are balanced on a massless seesaw. The boy has a mass of 60 kg and the girl's mass is 50 kg. If the boy sits 1.

skelet666 [1.2K]

Answer: 1.8m

Explanation:

Using the principle of moment to solve the question which states that the sum of clockwise moment is equal to the sum of anticlockwise moments.

The boy moves in the clockwise direction towards the pivot while the girl moves in the anticlockwise direction towards the pivot.

Since moment = force × perpendicular distance

CW moments = 60×1.5 = 90Nm

ACW moments = 50×x = 50x

Equating both moments to get 'x'

90 = 50x

x =9/5

x = 1.8m

This shows that the girl must sit 1.8m from the other side for equilibrium.:

3 0

2 years ago

A child on a swing falls a height of 1m. What speed will the child reach?

kirill [66]

Answer: $v=4.42\ m/s$

Explanation:

Given

Child falls a height of $h=1\ m$

Applying conservation of energy we can write

Loss in Potential energy=Gain in kinetic energy

$mgh=\frac{1}{2}mv^2$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 1}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

Thus child reaches a velocity of 4.42 m/s

7 0

2 years ago