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SVEN [57.7K]
2 years ago
5

Convert 2.4 milimetres into metre​

Physics
2 answers:
AleksAgata [21]2 years ago
4 0
0.0024

Milimetres are before centimetres and centimetres are before metres
olchik [2.2K]2 years ago
3 0

Answer: 2.4 millimeters = 0.0024 meters

Explanation: A millimeter is 1/1000 of a meter. By diving 2.4 by 1000, you get 0.0024.

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A compact car, mass 660 kg, is moving at 1.00 ✕ 102 km/h toward the east. The driver of the compact car suddenly applies the bra
Maslowich

Answer:59.43\times 10^3 kg-m/s

Explanation:

Given

mass of car m=660 kg

Initial velocity of car =102 km/h\approx 28.33 m/s towards east

Time taken to stop t=2.1 s

Force exerted F_{avg}=4.8\times 10^3 N

change in momentum is given  by impulse imparted to the car

Impulse(J)=-F\cdot t

J=-4.8\times 10^3\times 2.1

J=-59.49\times 10^3 kg-m/s

negative Sign indicates that impulse is imparted opposite to the direction of motion

magnitude of momentum J=59.49\times 10^3

6 0
3 years ago
Read 2 more answers
If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wave
kotegsom [21]

Answer:

0.82 mm

Explanation:

The formula for calculation an n^{th} bright fringe from the central maxima is given as:

y_n=\frac{n \lambda D}{d}

so for the distance of the second-order fringe when wavelength \lambda_1 = 745-nm can be calculated as:

y_2 = \frac{n \lambda_1 D}{d}

where;

n = 2

\lambda_1 = 745-nm

D = 1.0 m

d = 0.54 mm

substituting the parameters in the above equation; we have:

y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y_2 = 0.00276 m

y_2 = 2.76 × 10 ⁻³ m

The distance of the second order fringe when the wavelength \lambda_2 = 660-nm is as follows:

y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y^'}_2 = 1.94 × 10 ⁻³ m

So, the distance apart the two fringe can now be calculated as:

\delta y = y_2-y^{'}_2

\delta y = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

\delta y = 10 ⁻³ (2.76 - 1.94)

\delta y = 10 ⁻³ (0.82)

\delta y = 0.82 × 10 ⁻³ m

\delta y =  0.82 × 10 ⁻³ m (\frac{1.0mm}{10^{-3}m} )

\delta y = 0.82 mm

Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm

6 0
3 years ago
At 20°c, the resistance of a sample of nickel is 525 ohms. what is the resistance when the sample is heated to 70°C?​
Ipatiy [6.2K]

The resistance of the sample is 682.5\Omega

Explanation:

The relationship between resistance of a material and temperature is given by the equation

R(T)=R_0(1+\alpha (T-T_0))

where

R_0 is the resistance at the temperature T_0

\alpha is the temperature coefficient of resistance

For the sample of nickel in this problem, we have:

R_0 = 525 \Omega when the temperature is T_0 = 20^{\circ}C

While the temperature coefficient of resistance of nickel is

\alpha = 0.006/^{\circ}C

Therefore, the resistance of the sample when its temperature is

T=70^{\circ}C

is

R=(525)(1+0.006(70-20))=682.5 \Omega

Learn more about resistance:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

3 0
3 years ago
Rate at which wave travels or propagates
Softa [21]

Answer:

v =fλ v = f λ where v

Explanation:

4 0
3 years ago
A variable that should not change throughout the experiment is a(n):
Natalka [10]
The answer is control variable
7 0
3 years ago
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