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algol [13]
4 years ago
9

Could anyone can help me for this question I don’t understand how I can do?!! Please it really important!!

Physics
1 answer:
stiv31 [10]4 years ago
6 0
Im sry i cant help u
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It is a stretch of the atmosphere ranging from the upper mesosphere to the lower parts of the thermosphere. It’s useful to us in radio communication.
8 0
3 years ago
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A 26-kg sled is on a snow-covered slope. The coefficients of friction between the sled’s runners and the snow are µs = 0.096 and µ
sweet-ann [11.9K]

Answer:

Explanation:

Given

mass of sled =26 kg

coefficient of static friction \mu _s=0.096

coefficient of kinetic friction \mu _k=0.072

In order to move sled from rest we need to provide a force greater than static friction which is given by

f_s=\mu mg=0.096\times 26\times 9.8=24.46 N

After Moving Sled kinetic friction comes in to play which is less than static friction

f_k=\mu _kmg=0.072\times 26\times 9.8=18.34 N

therefore minimum force to keep moving sledge at constant velocity is 18.34 N

3 0
3 years ago
Examine the cross-sectional slice of a stem of a mint plant. Which statement most specifically describes mint?
Maslowich

The available options are:

Mint is a dicot.

Mint is a monocot.

Mint is an angiosperm.

Mint is a bulb plant.

Answer:

Mint is a dicot.

Explanation:

Given the fact that Mint is considered to be a member of Lamiaceae, an angiosperm plant which is characterized by typically having leaves that consist of reticulate vacation and appears like veins in structure. It also has a seed that contains two cotyledons.

Hence, it is considered a DICOT PLANT due to these characteristics. The botanical name of Mint is referred to as Mentha arvensis.

7 0
3 years ago
The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at posit
const2013 [10]

Answer:

fr = ½ m v₀²/x

Explanation:

This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.

The best way to solve this exercise is to use the energy work theorem

            W = ΔK

Where work is defined as the product of force by distance

           W = fr x cos 180

The angle is because the friction force opposes the movement

          Δk =K_{f} –K₀

          ΔK = 0 - ½ m v₀²

We substitute

         - fr x = - ½ m v₀²      

           fr = ½ m v₀²/x

8 0
3 years ago
A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th
GREYUIT [131]

Answer:

1.26\cdot 10^7 m/s

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

F=qvB

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

qvB=m\frac{v^2}{r}

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

v=\frac{qBr}{m}

where in this problem we have:

q=1.6\cdot 10^{-19}C is the magnitude of the charge of the electron

B=208 G=208\cdot 10^{-4}T is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

r=3.45 mm = 3.45\cdot 10^{-3} m

m=9.11\cdot 10^{-31} kg is the mass of the electron

So, the electron's speed is

v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s

6 0
3 years ago
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