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?

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)

v² = 160.3

v = 12.7m/s

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

3 0

3 years ago