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4 years ago

What is the definition of a point object?

1 answer:

4 0

Point object is an expression that is used in Kinematics. It refers the object whose dimensions are ignored or neglected and is treated as a dot object to simplify the calculations. When the object is treated as a point object it means the whole mass is summed in one point, you can imagine it as a sphere which has the mass and the radius is 0...

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Please answer ASAP for brainliest

OlgaM077 [116]

Answer:

Question #1- Scientists agree to a standard way of reporting measured quantities in which the number of reported digits reflects the precision in the measurement- more digits, more precision; less digits, less precision. You just studied 14 terms!

Question #2- Units are important because without proper measurement and units to express them, we can never express physical laws precisely just from qualitative reasoning. Units are incredibly important to physics. Two of the most important reasons are the following: (1) they help us. to avoid making mistakes in computation, and (2) they serve as a check on computations once they are completed. In the first case, you can avoid adding 3m and 25cm and coming up with the wrong answer.

Explanation: Hope this helps please mark brainliest!

4 0

3 years ago

Read 2 more answers

One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit

nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

$\lambda$ = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

$\theta=\dfrac{1.22\lambda}{D}$

The width is given by

$d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m$

The required width is 0.000003782 m

Minimum resolvable line separation is given by

$\dfrac{0.000003782}{2}=0.000001891\ m$

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when $\lambda=157\ nm$

$d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m$

The new minimum resolvable line separation between adjacent lines is

$\dfrac{0.00000239425}{2}=0.000001197125\ m$

6 0

4 years ago

A diver makes 1.0 revolutions on the way from a 9.2-m-high platform to the water. Assuming zero initial vertical velocity, find

____ [38]

Answer:

Average angular velocity ≈ 4.59 rad/s

Explanation:

Using the equation of motion,

H = ut + (1/2)t² ............................ equation 1.

Where H= height, u = initial velocity(m/s), g = acceleration due to gravity(m/s²), t = time(s) u= 0 ∴ ut =0

H =(1/2)gt².................................... equation 2.

making t² the subject of the relation in equation 2,

∴ t² = 2H/g

Where H = 9.2 m, g= 9.8 m/s

∴ t² = ( 2×9.2)/9.8

t = √(2 × 9.2/9.8) = √(18.4/9.8)

t = 1.37 s.

The average angular velocity = θ/t

Where θ = is the number of revolution that the diver makes, t = time

θ = 1 rev.

Since 1 rev = 2π (rad)

t = 1.37 s

Average angular velocity = 2π/t

π = 3.143

Average angular velocity = (2×3.143)/1.37 = 6.286/1.37

Average angular velocity ≈ 4.59 rad/s

8 0

4 years ago

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

juin [17]

Answer:

Explanation:

Given

For first case

launch angle $\theta =45^{\circ}C$

at highest point $h=150 m/s$

$150=u\cos 45$

$u=\frac{150}{\cos 45}=212.132 m/s$

For second case

$\theta _2=37^{\circ}C$

at highest Point velocity is $u\cos \theta _2$

$=212.132\times \cos 37$

$=169.41 m/s$

as there is no acceleration in x direction therefore horizontal velocity is same

7 0

3 years ago

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What is the wavelength that corresponds to a frequency of 6.00x1014 hz?

kipiarov [429]

The basic relationship between wavelength $\lambda$ , frequency f and speed c of an electromagnetic wave is
$\lambda= \frac{c}{f}$
where c is the speed of light. Substituting numbers, we find:
$\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{6.0 \cdot 10^{14} Hz}=5\cdot 10^{-7} m$

4 0

4 years ago