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igomit [66]
3 years ago
15

Match the object to the type of motion it's displaying. which letter goes with which number?

Physics
1 answer:
makkiz [27]3 years ago
8 0

<em>Hello There!!</em>

<em>I think the best answer choice is:</em>

<em>A. </em>Linear motion

<em>B. </em>Rotational motion

<em>D. </em>Vibrational motion

<em>4. </em>Runners sprinting across a straight track

<em>C. </em>Curvilinear motion

<em>3. </em>Music being\\\\\\ played on a stringed instrument, such as a guitar

<em>2. </em>A car driving up a winding mountain road

<em>1. </em>The propeller of an aircraft in midair

<em>I'm not sure this is 100% right....</em>

<em>P.S </em><em>Tell me if this is wrong....</em>

<em />Hope This Helps<em> </em>Sangwoo!!<em />

<em>#</em>Be<em> </em>Bold<em />

<em># </em>Always<em> </em>Brainly<em />

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A cable is 100-m long and has a cross-sectional area of 1.0 mm2. A 1000-N force is applied to stretch the cable. Young's modulus
Blizzard [7]

Answer:

1 m

Explanation:

L = 100 m

A = 1 mm^2 = 1 x 10^-6 m^2

Y = 1 x 10^11 N/m^2

F = 1000 N

Let the cable stretch be ΔL.

By the formula of Young's modulus

Y=\frac{F\times L}{A\times\Delta L}

\Delta L=\frac{F\times L}{A\times\Y}

\Delta L=\frac{1000\times 100}{10^{-6}\times10^{11}}

ΔL = 1 m

Thus, the cable stretches by 1 m.

4 0
3 years ago
1. Boyle's law relates the pressure of a gas to its
KIM [24]

Answer:

volume is the correct answer

Explanation:

7 0
2 years ago
A concave mirror has a focal length of 13.5 cm. This mirror forms an image located 37.5 cm in front of the mirror. Find the magn
77julia77 [94]

Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}

\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}

u = -21.09 cm

The magnification of the mirror is given by :

m=\dfrac{-v}{u}

m=\dfrac{-(-37.5)}{(-21.09)}

m = -1.77

So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

7 0
3 years ago
Calculate the number of free electrons and holes (in m-3) in an intrinsic semiconductor that has electron and hole mobilities of
allochka39001 [22]

Answer:

ni = 2.04e19

Explanation:

we know that in semiconductor like intrinsic, when electron leave the band, it leave a hole in valence band so we have

n = p = ni

from intrinsic carrier concentration

\sigma = n\left | e \right | \mu_e + n\left | e \right | \mu_h

\sigma = ni\left | e \right | \mu_e  + ni\left | e \right | \mu_h

\sigma = ni \left | e \right | ( \mu_e + \mu_h)

1.7 = ni * 1.6*10^{-19} * (.35 + .17)

ni = 2.014 *10^{19} m^{-3}

ni = 2.04e19

5 0
3 years ago
If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?
HACTEHA [7]

\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time,  t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0  +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}

5 0
2 years ago
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