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3 years ago

When picking up a load the correct fork spacing should be?

2 answers:

8 0

Answer:

When picking up a load the correct fork spacing should be<span> 6 inches and also </span>the forks must be spaced a<span>s wide apart as necessary for the maximum support of the load</span><span>.

Explanation:
</span>
When picking up a load the forks should be positioned so that the load is evenly distributed in which the center stringer of the pallet and the balance of the load should be spaced evenly in which makes the picking up the load to be correct and well-balanced. Other types of load, for example, a metal stillage, may have the forks set wider as there is enough strength in the bottom to prevent damage.

Licemer1 [7]3 years ago

3 0

<span>When picking up a load, the correct fork spacing must be spaced in an evenly manner in which the centre stringer of the pallet and the balance of the load should be spaced evenly in which makes the picking up the load to be correct and well-balanced.</span>

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The maximum value of magnetic in an electric field 3.2 *10^4

Svetradugi [14.3K]

Answer:

the answer is 12 because if your magnetic value and Electric field is 3.2 the answer will be 12

6 0

2 years ago

On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s

dsp73

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

= 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2

= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

4 0

3 years ago

Considering the various theories, the energy used in forming organic molecules in the primitive atmosphere could have come from

OLEGan [10]

Answer:

<h2>e. sound. </h2>

Explanation:

Such type of atmosphere in which oxygen was not present or was present in little amount is known as the primitive atmosphere and such type of atmosphere was present in the initial stage of the earth formation.
During this period, water vapor, nitrogen, hydrogen and carbon dioxide gases were present.
These gases interact with the help of energy that comes from many sources such as lightning, ultraviolet radiation, electric spark and some other.
When these inorganic molecules react in the then the formation of organic compounds takes place that becomes the basis of the organization of the life on the earth and called an organic evolution of life.

8 0

3 years ago

Can you help and explain these for me?

ollegr [7]

Yes I can do you want me to

8 0

3 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago