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OleMash [197]
3 years ago
10

The instrument used to measure the diameter of a thin wire is​

Physics
2 answers:
BARSIC [14]3 years ago
4 0
Micrometer

Micrometer is used to measure diameter of a thin wire.
Pie3 years ago
3 0

micrometer is used to measure the diameter of a thin wire

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I NEED HELP RIGHT NOW PLEASE I will  30pts and A brainly to anyone who helps me with questions 2-13!!!!!!!
neonofarm [45]
True True False True False False True I hope I helped on the first few
3 0
3 years ago
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A 2.0-kg laptop sits on the horizontal surface of the seat of a car moving at 8.0 m/s. The driver starts slowing down to stop. F
ivanzaharov [21]

Answer: 32.65\ m

Explanation:

Given

mass of laptop m=2 kg

The velocity of car u=8 m/s

The coefficient of static friction is \mu_s=0.4

The coefficient of kinetic friction is \mu_k=0.2

As the car is moving, so the coefficient of kinetic friction comes into play

deceleration offered by friction \mu_kg=0.2\times 9.8\ m/s^2

Using the equation of motion v^2-u^2=2as\\

insert the values

0^2-8^2=2(-0.2\times 9.8)s\\\\s=\dfrac{64}{1.96}\\\\s=32.65\ m

4 0
3 years ago
You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
1 year ago
Read 2 more answers
Can someone help me with this please?
Slav-nsk [51]

Answer:

C2, C1, C4, C5 and C6 are in parallel. Therefore, we use the formula Cp = C1 + C2 + ....

Cp = C2 + C1 + C4 + C5 + C6 = ( 7 * 10 ^-3) + (18 * 10^-6) + (0.8F) + (200 * 10^-3 F) + (750 * 10^-6) = 1.008F

Now, Cp will become one capacitor and it will be aligned with C3, therefore it will now become a circuit in series.

We use the formula: 1/Cs = 1/C1 + 1/C2 + .... + ....1/Cn

Thus,

1/Cs = 1/C3 + 1/Cp

1/Cs = 1/(14 * 10^-3 F) + 1/(1.008F)

Cs = 1.4 * 10 ^-2 or if we do not round too much it will give exactly 0.0138 F

So the answer should be a)

8 0
3 years ago
As an airplane accelerates in level flight, what parameter must the pilot adjust to keep level and balance Lift = Weight?
Y_Kistochka [10]

Answer:

C. Angle of Attack.

Explanation:

The pilot must adjust the angle of attack parameter. The angle of attack of this plane to get to the desired lift coefficient.

And thus, we have

Lift = Weight

3 0
2 years ago
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