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A 200-g hockey puck is launched at an initial speed of 16 m/s up a metal ramp that is inclined at a 30° angle. The coefficients

rusak2 [61]

Answer:

h = 8.588 m

Explanation:

Given:

Mass of hockey puck: m = 200 g = 0.2 K g

Angle of incline: θ = 30º (with respect to horizontal)

Coefficient of static and kinetic friction between the puck and ramp:

μ s = 0.4 and μk = 0.3

initial speed:

v = 3.5 m /s

Let h be the vertical height reached by the puck, above the ground. This corresponds to a distance of

d = h *sin 30º = 2 *h

along the incline. As the puck is moving, only kinetic friction comes into play as it climbs up the incline. Let N be the normal reaction exerted by the metal ramp on the ball. Then,

N = m*g *cos θ = 0.2 *9.8 1*cos 30 º = 1.6991 N

Therefore, kinetic friction acting on the puck is:

F k = μ k *N = 0.3 *1.6974 = 0.50974 N

From work energy theorem, the change in kinetic energy should equal the work done by friction and gravity. Therefore,

0.5* 0.2 *16² = 0.50974 *2h + 0.2*9.81*h

⇒ h = 8.588 m

8 0

3 years ago

Using the college registration example from section 6.7.3 as a starting point, do the following:

Arada [10]

Answer:

Explanation:

NCLUDE Irvine32.inc

TRUE = 1

FALSE = 0

.data

gradeAverage WORD ?

credits WORD ?

oKToRegister BYTE ?

str1 BYTE "Error: Credits must be between 1 and 30" , 0dh,0ah,0

main PROC

call CheckRegs

exit

main ENDP

CheckRegs PROC

push edx

mov OkToRegister,FALSE

; Check credits for valid range 1-30

cmp credits,1 ; credits < 1?

jb E1

cmp credits,30 ; credits > 30?

ja E1

jmp L1 ; credits are ok

; Display error message: credits out of range

E1:

mov edx,OFFSET str1

call WriteString

jmp L4

L1:

cmp gradeAverage,350 ; if gradeAverage > 350

jna L2

mov OkToRegister,TRUE ; OkToRegister = TRUE

jmp L4

L2:

cmp gradeAverage,250 ; elseif gradeAverage > 250

jna L3

cmp credits,16 ; && credits <= 16

jnbe L3

mov OkToRegister,TRUE ; OKToRegister = TRUE

jmp L4

L3:

cmp credits,12 ; elseif credits <= 12

ja L4

mov OkToRegister,TRUE ; OKToRegister = TRUE

L4:

pop edx ; endif

ret

CheckRegs ENDP

END main

8 0

4 years ago

All of the following are organization habits except:

atroni [7]

A. multitasking
the rest of them are

6 0

3 years ago

P36. In Section 3.5.4, we saw that TCP waits until it has received three duplicate ACKs before performing a fast retransmit. Why

KatRina [158]

Answer:

The Packets can be arrived out of an order from the Internet Protocol layer.

So, whenever the out of an order packets would be received then, it would be generated the duplicate ACK's, if we perform the re-transmission after the first duplicate ACK would be lead the senders to introduced too many redundant packets in the networks.

Explanation:

All the bytes, in the Transmission Control Protocol connections, are the numbered, from the beginning at the randomly choose the initial sequence number (ISN).

The SYN packets consumes the one sequence number, so the data will be actual and it begins at the ISN+1.

The receivers ack's sequences the number x acknowledged the receipts of all the data bytes that is less than byte the number x.

7 0

3 years ago

Write a loop that sets each array element to the sum of itself and the next element, except for the last element which stays the

pantera1 [17]

Answer:

The solution code is written in Python

numList = [10, 20, 30, 40]
for i in range(0, len(numList) - 1):
numList[i] = numList[i] + numList[i + 1]
print(numList)

Explanation:

Firstly, create a sample number list, numList (Line 1)

Create a for-loop that will traverse through the array element from 0 till the second last of the element (len(numList) - 1) (Line 3)

Set the current element, numList[i], to the sum of the current element, numList[i] and the next element, numList[i+1]

Print the modified numList (Line 6) and we can see the output as follows:

[30, 50, 70, 40]

4 0

3 years ago