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Svetradugi [14.3K]
3 years ago
6

Vì sao các bộ đốt nóng của bếp điện , nồi cơm điện ... thường làm từ những hợp kim nikêlin hoặc constantan mà không làm bằng dây

đồng hoặc dây nhôm
Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
6 0

Answer:

hbsbdd

Explanation:

hsbbs jsbbs jhsb jzb. sjsb vsjzzbs vsjbs hhsvd xbvd d jjxbdjdb kxbrbdjrb jxbd

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Bobby is riding his bike down the road at a speed of 10 miles per hour. Ahead of him, he sees another rider, moving in the same
vekshin1
They will be travelling slower than 10mph.
if they were travelling at the same speed then they would stay an equal distance apart.
if they were travelling fatser then they would be getting further away more quickly than Bobby is catching up.
maybe they are travelling at 5mph but I'd say it's a safer option to chose under 10mph
4 0
3 years ago
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Add these two velocity vectors to find the magnitude of their resultant vector.
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The  magnitude of their resultant vector is 4.6 meters/seconds

Since we are to add the  velocity vectors in order to  find the magnitude of their resultant vector.

Hence:

Resultant vector magnitude=5.8 meters/seconds + (1.2 meters/seconds)

Resultant vector magnitude=5.8 meters/seconds-1.2 meters/seconds

Resultant vector magnitude 4.6 meters/seconds

Inconclusion The  magnitude of their resultant vector is 4.6 meters/seconds

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3 years ago
Balanced or Unbalanced?
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3 years ago
Which technological advance allows scientists to handle these objects enough to feel their properties while still protecting the
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5 0
4 years ago
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A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the
andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and  how far away from the overpass is the roadrunner when the coyote drops the net.

d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

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3 years ago
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