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2 years ago

In the illustration, which two simple machines are being used to enable the student to reach the door?

Physics

2 answers:

lisov135 [29]2 years ago

4 0

The answer will be D the wheel and axle her wheel chair and the ramp is the inclined plane.

vodka [1.7K]2 years ago

3 0

D. wheel-and-axle AND an incline plane

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PLEASE HELP ME I FONT HAVE THAT MUCH TIME LEFT!!! THE ALL CAPS ARE TO CATCH YOUR ATTENTION SO NOW THAT I’VE GOT IT, PLEASE HELP

podryga [215]

Answer:

hmm if it were up to me i would say gravity potential energy and sorry I don't really have a third one hope this helps though.

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2 years ago

Read 2 more answers

A charge of q= +15 uC moves in a Northeast direction with a speed 5 m/s, 25 degrees East of a magnetic field, pointing North, wi

grin007 [14]

Explanation:

It is given that,

Magnitude of charge, $q=15\ \mu C=15\times 10^{-6}\ C$

It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.

Magnetic field, $B=0.08\ j$

Velocity, $v=(5\ cos25)i+(5\ sin25)j$

$v=[(4.53)i+(2.11)j]\ m/s$

We need to find the magnitude of force on the charge. Magnetic force is given by :

$F=q(v\times B)$

$F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]$

<em>Since</em>, $i\times j=k\ and\ j\times j=0$

$F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]$

$F=0.00000543\ kN$

$F=5.43\times 10^{-6}\ kN$

So, the force acting on the charge is $5.43\times 10^{-6}\ kN$ and is moving in positive z axis. Hence, this is the required solution.

6 0

3 years ago

What will happen if you light up a candle that is in a beaker?

salantis [7]

It will melt so it is c

8 0

2 years ago

When a force of 450N pushes on a 20kg box as

ehidna [41]

Answer:

ma+mgsinh0+f=F∴(25)(0.75)+(25)(10)sinh0+μkN=F∴18.75+(250)(0.6h)+μk(mgcosh0=F⟹18.75+150+μk((25)(10)(0.76))=500∴168.75+μk(190)=500⟹μk(190)=331.25⟹μk=1.74

Explanation:

7 0

2 years ago

Consider an e. coli to be a cylinder with a diameter of 1 micrometer (um) and with a length of 2 micrometer (um).

julia-pushkina [17]

Answer:

$A=6.28\ \mu m^2$

Part 1

$V=1.57 \ \mu m^3$

Part 2

$A=6.28\ \mu m^2$

Explanation:

Given that

Diameter,d=1 μm

Length ,l=2 μm

As we know that volume of cylinder given as

$V=\pi r^2l$

$V=\pi \times 0.5^2\times 2 \ \mu m^3$

$V=1.57 \ \mu m^3$

Surface area,A

A=π d l

$A=\pi \times 1 \times 2\ \mu m^2$

$A=6.28\ \mu m^2$

Part 1

$V=1.57 \ \mu m^3$

Part 2

$A=6.28\ \mu m^2$

4 0

3 years ago