All categories

2 years ago

Hello all you people who need points use me to get points your welcome :)

Physics

2 answers:

Colt1911 [192]2 years ago

7 0

Answer:

thxs so much

Explanation:

frozen [14]2 years ago

5 0

Answer:

Thank you so much!!!!

Explanation:

I really need this points

You might be interested in

How to convert acceleration to velocity.

Nana76 [90]

You can't. Velocity and acceleration measure two different things, so their units are incompatible. It's like asking, "How many meters does this book weigh?"

Maybe you mean "find" acceleration using given velocities, or a velocity function?

4 0

2 years ago

The wave length of violet light rounded to the nearest nanometer is a __ nm

nataly862011 [7]

Wavelength= speed / frequency

so.....3× 10^8 / 7.26×10^14
= .413× 10^(-6)

in scientific notation= 4.13×10^(-7)

in nanometer = 413 nm

6 0

2 years ago

a DJ uses a 110-volt outlet to plug in a strobe light. if the current flowing through the light is 0.050 amps, how much resistan

tigry1 [53]

Ohms Law: V = IR

V is the voltage in volts

I is the current in amps

R is the resistance in Ohms

Rearrange: R = V/I

R = (110)/(0.050)

R = 2200

There are 2200 Ohms of resistance in the circuit.

5 0

3 years ago

A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until t

postnew [5]

Answer:

Explanation:

Given

mass of saturated liquid water $m=1.4\ kg$

at $200^{\circ}$ specific volume is $\nu =0.001157\ m^3\kg$ (From Table A-4,Saturated water Temperature table)

$V_1=m\nu _1$

$V_1=1.4\times 0.001157$

$V_1=1.6198\times 10^{-3}\ m^3$

Final Volume $V_2=4V_1$

$V_2=4\times (1.6198\times 10^{-3})$

$V_2=6.4792\times 10^{-3}\ m^3$

Specific volume at this stage

$\nu _2=\frac{V_2}{m}$

$\nu _2=\frac{6.4792\times 10^{-3}}{1.4}$

$\nu _2=0.004628\ m^3/kg$

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

$T_2=T_1^{*}+\frac{T_2^{*}-T_1^{*}}{\alpha _2^{*}-\alpha _1^{*}}\times (\alpha _2-\alpha _1^{*})$

$T_2=370^{\circ}+\frac{373.95-370}{0.003106-0.004953}\times (0.004628-0.004953)$

$T_2=370.7^{\circ} C$

4 0

3 years ago

A uniform 2.50m ladder of mass 7.30kg is leaning against a vertical wall while making an angle of 51.0degree with the floor. A w

Zigmanuir [339]

Answer:

19.95 J

Explanation:

The center of mass of the ladder is initially at a height of:

$h_1=\frac{L}{2}sin\theta$

The center of mass of the ladder ends at a height of:

$h_2= \frac{L}{2}sin90$ =L/2

So, the work done is equal to the change in potential energy which is:

W = PE = $mg(h_2-h_1)$

now $h_2-h_1= 1-sin\theta$

therefore

W = [mgL/2]×[1 - sin(theta)]

W = [(7.30)(9.81)(2.50)/2]×[1-sin(51°)]

solving this we get

W = 19.95 J

8 0

2 years ago