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goblinko [34]
2 years ago
10

A certain 48 V electric forklift can lift up to 7000 lb at a

Physics
1 answer:
Gwar [14]2 years ago
7 0

Answer:

256.25 A

Explanation:

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If a fast-moving car making a loud noise approaches and moves past the person, what will happen as the distance between the two
lesantik [10]
The pitch of the sound of the car will appear to be <span>lower than the pitch of the car next to you</span>
8 0
3 years ago
What velocity must a 1210 kg car have in order to have the same momentum as a pickup truck
Mariulka [41]

Answer: 46.5 m/s

Explanation:

Momentum = mass(m) × velocity (v)

momentum of car = momentum of truck

1210 × v = 2250 × 25

v = 46.5 m/s

6 0
3 years ago
Julianne went to a restaurant to have a taste of her favorite fried chicken and spaghetti. She drove 2 km, east and then 8.5 km,
GREYUIT [131]

Julianne’s displacement from her origin is equal to 10.015 kilometers.

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

  • Distance A = 2 km, East.
  • Distance B = 8.5 km, Northeast.

To calculate Julianne’s displacement from her origin:

<h3>How to calculate displacement.</h3>

We would denote the two (2) unit vectors along the East and Northeast directions by i and j respectively.

<u>Note:</u> Northeast is at angle of 45° with the East.

In terms of vectors, the distances becomes:

Distance A = 2i

Distance\;B=8.5 [(cos 45i + sin 45j)]\\\\Distance\;B=(\frac{8.5}{\sqrt{2} } i \;+\;\frac{8.5}{\sqrt{2} } j)

<u>For the </u><u>resultant displacement</u><u>:</u>

D^2 = [(2+\frac{8.5}{\sqrt{2} } )^2+ (\frac{8.5}{\sqrt{2} } )^2\\\\D =\sqrt{[(2+\frac{8.5}{\sqrt{2} } )^2+ (\frac{8.5}{\sqrt{2} } )^2} \\\\D=2+\frac{8.5}{\sqrt{2} } + \frac{8.5}{\sqrt{2} }

D = 10.015 kilometers.

Read more on displacement here: brainly.com/question/13416288

5 0
2 years ago
Where does a phase change occur
il63 [147K]
A phase change is occuring; the liquid water is changing to gaseous water, or steam. On a molecular level, the intermolecular forces between the water molecules are decreasing. The heat is providing enough energy for the water molecules to overcome these attractive forces.

3 0
3 years ago
Read 2 more answers
What is the gpe of a 200 kg hot air ballon 21,000 m above the ground?
stiv31 [10]

gravitational Potential energy is given by

GPE = mgh

here we have

m = 200 kg

h = 21000 m

now we will have

GPE = 200(9.8)(21000)

GPE = 4.116 \times 10^7 J

so GPE of balloon will be 41160000 J above the given height from ground

8 0
3 years ago
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