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mafiozo [28]
3 years ago
8

How long will it take a truck to travel 132 kilometers if it is traveling 45 km per hour?

Physics
2 answers:
Sloan [31]3 years ago
8 0

Answer:

2 hours and 56 minutes

Explanation:

pentagon [3]3 years ago
5 0

Answer:

2 hours and 50 minutes

Explanation:

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Jill is pushing a box across the floor. Which represents the upward force perpendicular to the floor?
Georgia [21]

Answer:

When Jill is pushing a box across a floor, it represents the upward motion and it is natural force is applied.

So it is represented as FN and normal force takes place in considering the force perpendicular to the floor.

It seems to support that forced applied on an object when the object is in contact with other.

Explanation:

5 0
2 years ago
Any child is pushing a shopping cart at a speed of 1.5 m/s.how long will it take this child to push the cart down the aisle with
NARA [144]
1.5 m/s is the velocity. 9.3 m is the length of aisle, over which Distance will be covered. Time is demanded in which the child will move the cart over the aisle with 1.5 m/s. v=S/t and, t=S/v Put values, t=9.3/1.5=6.2 s
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2 years ago
For safety reasons, a worker’s eye travel time in a certain operation must be separated from the manual elements that follow. Th
iogann1982 [59]

Answer:

The answer is 12.67 TMU

Explanation:

Recall that,

worker’s eyes travel  distance must be = 20 in.

The perpendicular distance from her eyes to the line of travel is =24 in

What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?

Now,

We solve for the given problem.

Eye travel is = 15.2 * T/D

=15.2 * 20 in/24 in

so,

= 12.67 TMU

Therefore, the MTM -1 of normal time that should be allowed for the eye  travel element is = 12.67 TMU

7 0
3 years ago
The answer please? I have exam tomo
makkiz [27]

Please be determined and being hardworking person do not rely on the other people to make your problems solved

Explanation:

Ok?

5 0
2 years ago
Find the magnitude of the force at point x = 15cm if k=10500 N/m when the work done by this force is WD= 1/2k x2
Ann [662]

Answer:

sqdqk3

Explanation:

qdjqưbdkq ưqdhud j

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3 years ago
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