A 24-volt DC battery is connected

A ball is dropped from rest from the top of a cliff that is 30 m high. From ground

trasher [3.6K]

The distance below the top of the cliff that the two balls cross paths is 7.53 meters.

<u>Given the following data:</u>

Initial velocity = 0 m/s (since the ball is dropped from rest).

Height = 30 meters.

<u>Scientific data:</u>

Acceleration due to gravity (a) = 9.8 $m/s^2$ .

To determine how far (distance) below the top of the cliff that the two balls cross paths, we would apply the third equation of motion.

<h3>How to calculate the velocity.</h3>

Mathematically, the third equation of motion is given by this formula:

$V^2 = U^2 +2aS$

<u>Where:</u>

V is the final velocity.

U is the initial velocity.

a is the acceleration.

S is the distance covered.

Substituting the parameters into the formula, we have;

$V^2 = 0^2 +2(9.8) \times 30\\\\V^2 = 588\\\\V=\sqrt{588}$

V = 24.25 m/s.

<u>Note:</u> The final velocity of the first ball becomes the initial velocity of the second ball.

The time at which the two balls meet is calculated as:

$Time = \frac{S}{U} \\\\Time = \frac{30}{24.25}$

Time = 1.24 seconds.

The position of the ball when it is dropped from the cliff is calculated as:

$y_1 = h-\frac{1}{2} at^2\\\\y_1 = 30-\frac{1}{2} \times 9.8 \times 1.24^2\\\\y_1 = 30-7.53\\\\y_1=22.47\;meters$

Lastly, the distance below the top of the cliff is calculated as:

$Distance = 30-22.47$

Distance = 7.53 meters.

Read more on distance here: brainly.com/question/10545161

4 0

3 years ago

As the junior engineer at the Mesabi Range Hydraulic Engineering Company located in Ely, Minnesota, you have been tasked with de

katen-ka-za [31]

yes it will

Explanation:

5 0

4 years ago

If the maximum allowable shear stress is 70 MPa, find the shaft diameter needed to transmit 40 kW when the shaft speed is 250 rp

victus00 [196]

Answer:

The diameter is 50mm

Explanation:

The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.

T=(P×60)/(2×pi×N)

T is the Torque

P is the the power to be transmitted by the shaft; 40kW or 40×10³W

pi=3.142

N is the speed of the shaft; 250rpm

T=(40×10³×60)/(2×3.142×250)

T=1527.689Nm

Diameter of a shaft can be obtained from the formula

T=(pi × SS ×d³)/16

Where

SS is the allowable shear stress; 70MPa or 70×10⁶Pa

d is the diameter of the shaft

Making d the subject of the formula

d= cubroot[(T×16)/(pi×SS)]

d=cubroot[(1527.689×16)/(3.142×70×10⁶)]

d=0.04808m or 48.1mm approx 50mm

7 0

4 years ago

Which is/are not a mechanism commonly associated with tool wear (mark all that apply)?a. Adhesion b. Attrition c. Abrasion d. Co

butalik [34]

Answer: d)Coercion

Explanation:Tool wear is defined as the situation when the cutting tool is subjected to the regular process of cutting metal then they tend to wear because of the continuous action of cutting and facing stresses and pressure . The mechanism that does not happen during this process are coercion that means the process of exerting forces on any material forcefully against the will or need. Therefore, adhesion,attrition and abrasion are the process of tool wear .So the correct option is (d)

5 0

3 years ago

In a TDM communication example, 15 voice signals are badlimited to 5kHz and transmitted simultaneously using PAM. What is a prel

MA_775_DIABLO [31]

Answer:

Option D

160 kHz

Explanation:

Since we must use at least one synchronization bit, total message signal is 15+1=16

The minimum sampling frequency, fs=2fm=2(5)=10 kHz

Bandwith, BW required is given by

BW=Nfs=16(10)=160 kHz

5 0

3 years ago