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frez [133]
3 years ago
12

A truck travels down the highway at a speed of 110 km/hr. How long does the trip last if the truck covered 2200km?

Physics
1 answer:
vivado [14]3 years ago
8 0
It'll last 20 hours. If it travels 110 miles in one hours, 110 times 20 equals 2,200.
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A force of 500 N is used to slide a box across a smooth surface; the box moved 5 m in 1.2 seconds. What power is used?
Katen [24]
F = 500N
S = 5m
T = 1.2s
----------------------------------
P = ?

We can use formula P = F * V, just because it was a smooth slide we can assume that average speed was V = S/T = 5 / 1.2 = 50/12

So the final answer would be:

P = F * V = 500 * 50/12

Are you sure those are correct numbers? The answer don't look nice :D



8 0
3 years ago
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The volume of a rectangular prism is given by the formula V = lwh, where l is the length of the prism, w is the width, and h is
Verizon [17]
<span> For this case the volume of the box is given by:
</span> V = lwh&#10;
<span> Substituting values we have:
</span> V = (2a + 11) (5a - 12) (a + 6)&#10;
<span> Rewriting we have:
</span> V = (10a ^ 2-24a + 55a-132) (a + 6)&#10;&#10;V = 10a ^ 3-24a ^ 2 + 55a ^ 2-132a + 60a ^ 2-144a + 330a-792
<span> Grouping terms of equal degree we have:
</span> V = 10a ^ 3 + 60a ^ 2 -24a ^ 2 + 55a ^ 2 -132a -144a + 330a-792
<span> Adding terms of equal degree we have:
</span> V = 10 ^ 3 + 91a ^ 2 + 54a-792&#10;
<span> Answer:
 the volume of the box is:
 All the expressions given.</span>
4 0
3 years ago
A person approaches a flat mirror at a speed of 0.25 m / s. How fast is he approaching his image?
Bezzdna [24]
  • Flat mirror is given here
  • We know that in a flat mirror the distance of object from the mirror is equal to distance of image from the mirror i.e v=u
  • Only the side of image changes ime left is seemed right.

So the speed remains same 0.25m/s.

3 0
3 years ago
A box is being dragged with a horizontal force of 65 N for 12 meters. If there is a force of friction acting on it
asambeis [7]

Answer:

A. 780 J

B. 120 J

C. 660 J

Explanation:

From the question given above the following data were obtained:

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

A. Determination of the work done by the dragging force.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Workdone (Wd) by dragging force =?

Wd = Fₔ × s

Wd = 65 × 12

Wd = 780 J

Therefore, the work done by the dragging force is 780 J

B. Determination of the work done by friction.

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Workdone (Wd) by friction =?

Wd = Fբ × s

Wd = 10 × 12

Wd = 120 J

Therefore, the work done by friction is 120 J

C. Determination of the net work done on the box.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Net work done (Wd) =?

Next, we shall determine the net force acting on the box. This can be obtained as follow:

Dragging force (Fₔ) = 65 N

Force of friction (Fբ) = 10 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fբ

Fₙ = 65 – 10

Fₙ = 55 N

Thus, the net force acting on the box is 55 N

Finally, we shall determine the net work done on the box as follow:

Distance (s) = 12 m

Net force (Fₙ) = 55 N

Net work done (Wd) =?

Wd = Fₙ × s

Wd = 55 × 12

Wd = 660 J

Therefore, the net work done on the box is 660 J

4 0
3 years ago
Two positive charged particles will ?
Vika [28.1K]

Answer:

make a negative

Explanation:

yep

6 0
3 years ago
Read 2 more answers
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