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photoshop1234 [79]

3 years ago

13

A 2.0 kg 2.0kg2, point, 0, start text, k, g, end text cart moving right at 5.0 m s 5.0 s m 5, point, 0, start fraction, st

art text, m, end text, divided by, start text, s, end text, end fraction on a frictionless track collides with a 3.0 kg 3.0kg3, point, 0, start text, k, g, end text cart initially at rest. The 2.0 kg 2.0kg2, point, 0, start text, k, g, end text cart has a final speed of 1.0 m s 1.0 s m 1, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to the left. What is the final speed of the 3.0 kg 3.0kg3, point, 0, start text, k, g, end text cart?

2 answers:

butalik [34]3 years ago

7 0

Yes así le tienes que hacer para aprender

Elena L [17]3 years ago

3 0

Answer: The answer is 4.0 m/s.

Explanation: khan acadmey.

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If the volume of one drop is 0.031 mL according to Stu Dent’s measurement, approximately what volume would 22 drops be? Answer w

olasank [31]

Answer:

Volume of 22 drop will be 0.68 ml

Explanation:

We have given volume of one drop = 0.031 ml

We know that 1 liter = 1000 ml

So $1ml=10^{-3}L$

So 0.031 ml will be equal to $0.031\times 10^{-3}L$

We have to find the volume of 22 drop

For finding volume of 22 drop we have to multiply volume of one drop by 22

So volume of 22 drop will be $=22\times 0.031\times 10^{-3}=0.682\times 10^{-3}L=0.68mL$

So volume of 22 drop will be 0.68 ml

7 0

4 years ago

A proton is launched from an infinite plane of charge with surface charge density -1.10×10-6 C/m2. If the proton has an initial

Katyanochek1 [597]

The distance covered by the proton is 48.4 m

Explanation:

The electric field produced by an electrically charged infinite plane is given by

$E=\frac{\sigma}{2\epsilon_0}$

where in this case,

$\sigma = -1.10\cdot 10^{-6} C/m^2$ is the surface charge density

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

Substituting,

$E=\frac{-1.10\cdot 10^{-6}}{2(8.85\cdot 10^{-12})}=-5.65\cdot 10^4N/C$

And the direction is towards the plane (because the charge is negative).

The electric force on the proton due to this field is

$F=qE$

where

$q=1.6\cdot 10^{-19}C$ is the proton charge

Substituting,

$F=(1.6\cdot 10^{-19})(-5.65\cdot 10^4)=-9.0\cdot 10^{-15} N$

where the direction is toward the plane.

Now we can calculate the proton's acceleration using Newton's second law:

$a=\frac{F}{m}$

where

$m=1.67\cdot 10^{-27}kg$ is the proton mass

Substituting,

$a=\frac{-9.0\cdot 10^{-15}}{1.67\cdot 10^{-27}}=-5.4\cdot 10^{-12} m/s^2$

Now we can finally apply the following suvat equation for accelerated motion to find the distance travelled by the proton:

$v^2-u^2=2as$

where

v = 0 is the final velocity

$u=2.40\cdot 10^7 m/s$ is the initial velocity

a is the acceleration

s is the distance covered

And solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{0-(2.4\cdot 10^7)^2}{2(-5.4\cdot 10^{12})}=53.3 m$

Therefore, the closest answer is 48.4 m.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

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#LearnwithBrainly

5 0

3 years ago

There is a 50 g sample of Ra-229. It has a half-life of 4 minutes.

Sloan [31]

Via half-life equation we have:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h} }$

Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:

$A_{final}=50(\frac{1}{2})^\frac{12}{4}=50(\frac{1}{2})^{3}=50(\frac{1}{8})=6.25g$

There is 6.25 grams left of Ra-229 after 12 minutes.

4 0

4 years ago

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A pipe closed at both ends can have standing wav

uranmaximum [27]

Answer:

68.8 Hz

137.6 Hz, 206.4 Hz

Explanation:

L = Length of tube = 2.5 m

v = Velocity of sound in air = 344 m/s

Distance between nodes is given by

$L=\dfrac{\lambda}{2}+m\dfrac{\lambda}{2}\\\Rightarrow \dfrac{\lambda(n+1)}{2}=L\\\Rightarrow \lambda=\dfrac{2L}{n+1}$

Where n = 0, 1, 2, 3, ...

Making n+1 = n

$\lambda=\dfrac{2L}{n}$

where n = 1, 2, 3 .....

For fundamental frequency n = 1

$\lambda=\dfrac{2\times 2.5}{1}\\\Rightarrow \lambda=5\ m$

Frequency is given by

$f=\dfrac{v}{\lambda}\\\Rightarrow f=\dfrac{344}{5}\\\Rightarrow f=68.8\ Hz$

The fundamental frequency is 68.8 Hz

First overtone

$2f=2\times 68.8=137.6\ Hz$

Second overtone

$3f=3\times 68.8=206.4\ Hz$

The overtones are 137.6 Hz, 206.4 Hz

4 0

3 years ago

Battery of 12 V supplies charge of 1000 C to a device work battery

Y_Kistochka [10]

Answer:

W = 12 kJ

Explanation:

We have,

Voltage of a battery is 12 V

Charge supplied by the battery is 1000 C

It is assumed to find the work is done by the battery in moving the charge. The work done is given in terms of voltage as :

$W=qV\\\\W=1000\times 12\\\\W=12000\ J$

or

W = 12 kJ

So, the work of 12 kJ is done by battery in moving charges.

5 0

4 years ago