Force can be represented by

A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision

Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is $a=2,800m/s^{2}$

Explanation:

Known Data

We will asume initial speed has a negative direction, $v_{i}=-30m/s$ , final speed has a positive direction, $v_{f}=26m/s$ , $\Delta t=20ms=0.020s$ and mass $m_{b}$ .

Initial momentum

$p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s$

final momentum

$p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s$

Impulse

$I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s$

Average Force

$F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}$

Average acceleration

$F=ma$ , so $a=\frac{F}{m_{b}}$ .

Therefore, $a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}$

8 0

2 years ago

How did the astronomy of hipparchus and ptolemy violate the principles of early greek philosophers?

Mariana [72]

Answer: Hipparchus & Ptolemy believed sun and moon travel around circles with Earth in the middle (geocentric) while Plato and Aristotle believed Perfect unchanging heavens

Explanation: thus hipparchus and ptolemy violate early greek philosophy

8 0

2 years ago

Predict which molecule will show a similar relationship between its heat of fusion and its heat vaporization that water does

BaLLatris [955]

Question:

A) C6H6

B) CH3CH2CH2CH2CH2COH6

C) NaCl

D) NH3

Answer:

The correct option is;

A) C₆H₆

Explanation:

Heat of fusion = 6.02 kJ/mol

Heat of vaporization =40.8 kJ/mol

Here, we analyze each of the options as follows

A) C₆H₆

Benzene has a melting point of 5.5° C and a boiling point of

80.1 ° C similar to water

Heat of fusion = 9.92 kJ/mol

Heat of vaporization =30.8kJ/mol

B) CH₃CH₂CH₂CH₂CH₂COH₆

The above compound is more likely solid

C) NaCl solid

D) NH₃ melting point = -77.73 °C boiling point = -33.34 °C

Of the above, the compounds the one that closely resembles water is C₆H₆

8 0

2 years ago

Read 2 more answers

The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0

katen-ka-za [31]

Answer:

a) F = 2.66 10⁴ N, b) h = 1.55 m

Explanation:

For this fluid exercise we use that the pressure at the tap point is

Exterior

P₂ = P₀ = 1.01 105 Pa

inside

P₁ = P₀ + ρ g h

the liquid is water with a density of ρ=1000 km / m³

P₁ = 0.85 1.01 10⁵ + 1000 9.8 5

P₁ = 85850 + 49000

P₁ = 1.3485 10⁵ Pa

the net force is

ΔP = P₁- P₂

Δp = 1.3485 10⁵ - 1.01 10⁵

ΔP = 3.385 10⁴ Pa

Let's use the definition of pressure

P = Fe / A

F = P A

the area of a circle is

A = pi r² = [i d ^ 2/4

let's reduce the units to the SI system

d = 100 cm (1 m / 100 cm) = 1 m

F = 3.385 104 pi / 4 (1) ²

F = 2.66 10⁴ N

b) the height for which the pressures are in equilibrium is

P₁ = P₂

0.85 P₀ + ρ g h = P₀

h = $\frac{P_o ( 1-0.850)}{\rho \ g}$

h = $\frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}$

h = 1.55 m

4 0

2 years ago

A ball hits a wall. What is true about the magnitude of the force experienced by the ball compared with the force experienced by

Ivahew [28]

This is best understood with Newtons Third Law of Motion: for every action there is an equal and opposite reaction. That should allow you to see the answer.

5 0

2 years ago