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nevsk [136]
3 years ago
11

A block of wood has a length of 4 cm a width of 5 cm and a height of 10 cm what is the volume of the wood?

Physics
1 answer:
Ghella [55]3 years ago
6 0
The correct answer is B
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What type of image can be larger or smaller than the object?
Stolb23 [73]
It’s D. An enlargement (hope this helps!)
4 0
3 years ago
PLZ HELP WILL GIVE BRAINLIEST
xxTIMURxx [149]

Answer:

12.7m/s

Explanation:

Given parameters:

Mass of the diver = 77kg

Height  = 8.18m

Unknown:

Final velocity  = ?

Solution:

To solve this problem, we use one of the motion equations.

            v²  = u² + 2gh

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

h is the height

             v² = 0² + (2 x 9.8 x 8.18)

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7 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
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

\theta = \frac{1.22 \lambda }{D}

Here,

D is diameter of the eye

D = \frac{1.22 (539nm)}{5.11 mm}

D= 1.287*10^{-4}m

The angle that relates the distance between the lights and the distance to the lamp is given by,

Sin\theta = \frac{d}{L}

For small angle, sin\theta = \theta

sin \theta = \frac{d}{L}

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle sin \theta = \theta

Therefore,

L = \frac{d}{sin\theta}

L = \frac{0.691m}{1.287*10^{-4}}

L = 5367m

Therefore the distance is 5.367km.

4 0
4 years ago
How is calculating amplitude​
kykrilka [37]
It is same as calculating maths for math
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3 years ago
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