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Katarina [22]
2 years ago
10

Two forces F1 = -7.90i + 5.40j and F2 = 9.00i + 7.60j are acting on an object with a mass of m = 6.30 kg. The forces are measure

d in newtons, i and j are the unit vectors. What is the magnitude of the object's acceleration?
Physics
1 answer:
IceJOKER [234]2 years ago
7 0

Answer:

Answer:

Propotionality is important

Explanation:

Explanation:

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an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le
storchak [24]
<span>On what:

f (is the focal length of the lens) = ? 
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

</span><span>Using the Gaussian equation, to know where the object is situated (distance from the point).
</span>
\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}
\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}
\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}
\frac{1}{f} =  \frac{10}{33.18}
Product of extremes equals product of means:
10*f = 1*33.18
10f = 33.18
f =  \frac{33.18}{10}
\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark

6 0
3 years ago
Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du
Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time t is

\theta=\dfrac\alpha2t^2

It rotates 44.5 rad in this time, so we have

44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}

b. Since acceleration is constant, the average angular velocity is

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2

where \omega_f is the angular velocity achieved after 6.00 s. The velocity of the disk at time t is

\omega=\alpha t

so we have

\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}

making the average velocity

\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}

Another way to find the average velocity is to compute it directly via

\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}

c. We already found this using the first method in part (b),

\omega=14.8\dfrac{\rm rad}{\rm s}

d. We already know

\theta=\dfrac\alpha2t^2

so this is just a matter of plugging in t=12.0\,\mathrm s. We get

\theta=179\,\mathrm{rad}

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2

Then for t=6.00\,\rm s we would get the same \theta=179\,\rm rad.

7 0
3 years ago
Each plate of a parallel‑plate capacitor is a square of side 3.63 cm, and the plates are separated by 0.473 mm. The capacitor is
NARA [144]

Answer:

E= 55.53 x 10³ V/m

Explanation:

Given that

a=  3.63 cm

Area ,A= a²

distance ,d= 0.473 mm

Stored energy ,U = 8.49 nJ

Value of capacitor given as

C=\dfrac{\varepsilon _oA}{d}

By putting the values

C=\dfrac{8.85\times 10^{-12}\times 3.63^2\times 10^{-4}}{0.473\times 10^{-3}}

C=2.46 x 10⁻¹¹ F

U=\dfrac{1}{2}CV^2

V=Voltage difference

V=\sqrt{\dfrac{2U}{C}}

V=\sqrt{\dfrac{2\times 8.49\times 10^{-9}}{2.46\times 10^{-11}}}

V=26.27 V

V= E d

E=Electric filed

26.27 = E x 0.473 x 10⁻³

E= 55.53 x 10³ V/m

7 0
3 years ago
Making quick glances to the roadway in front of your vehicle is called?
umka2103 [35]

Answer:

safety check.

Explanation:

5 0
3 years ago
Read 2 more answers
Select all that apply.
Nana76 [90]

Answer: It is both B and D

Select all that apply.

At night, thermal energy moves _____.

from space to the atmosphere

from the land to the atmosphere

from the atmosphere to the land

from the atmosphere to space

3 0
3 years ago
Read 2 more answers
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