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AysviL [449]
4 years ago
15

Two 0.50-kg carts are pushed toward each other from starting positions at either end of a 6.0-mlow-friction track. Each cart is

pushed with a force of 5.0 N , and that force is exerted for a distance of 1.0 m.
Part A

What is the work done on the two-cart system?

Part B

What is the change in kinetic energy of the system?

Part C

What is the kinetic energy of the center of mass of the system?
Physics
1 answer:
Rzqust [24]4 years ago
3 0

Answer:

A) W=10J

B) \Delta K=10J

C) K_{CM}=0J

Explanation:

Since the problem is in one dimension, we can consider that one cart (1) is pushed in the positive direction while the other (2) is pushed in the negative one, so the work done on the system will be:

W=F_1.d_1+F_2.d_2=(5N)(1m)+(-5N)(-1m)=10J

Using the work-energy theorem W_{Net}=\Delta K, and since all work done on the system is the one calculated before, we have \Delta K=10J

The net force on the system is F_{Net}=0N, so there is no acceleration of the center of mass and it's kinetic energy is the initial one, K_{CM}=0J.

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A cork has weight mg and density 25% of water’s density. A string is tied around the cork and attached to the bottom of a water-
yuradex [85]

Answer:

Explanation:

Given

Density of Cork \rho _c=0.25\rho _{water}

Considering V be the volume of Cork

Buoyant Force will be acting Upward and Weight is acting Downward along with T

Since density of water is more than cork therefore Cork will try to escape out of water but due to tension it will not

we can write as

\rho _{water}Vg-\rho _Cvg-T=0

where T=tension

Thus Tension T is

T=Vg(\rho _{water}-\rho _c)

Taking \rho _c common

T=\rho _cVg(\frac{\rho _{water}}{\rho _c}-1)    

T=mg(\frac{\rho _{water}}{\rho _c}-1)

T=mg(4-1)

T=3mg

4 0
3 years ago
If your correct, you'll be marked as brainliest. Please help!!
erik [133]
1st Blank: Synthetic 
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8 0
3 years ago
A wire of length 6cm makes an angle of 20° with a 3 mT
Crazy boy [7]

Answer:

Approximately 7.3 \times 10^{-3}\; \rm A (approximately 7.3\; \rm mA) assuming that the magnetic field and the wire are both horizontal.

Explanation:

Let \theta denote the angle between the wire and the magnetic field.

Let B denote the magnitude of the magnetic field.

Let l denote the length of the wire.

Let I denote the current in this wire.

The magnetic force on the wire would be:

F = B \cdot l \cdot I \cdot \sin(\theta).

Because of the \sin(\theta) term, the magnetic force on the wire is maximized when the wire is perpendicular to the magnetic field (such that the angle between them is 90^\circ.)

In this question:

  • \theta = 20^\circ (or, equivalently, (\pi / 9) radians, if the calculator is in radian mode.)
  • B = 3\; \rm mT = 3 \times 10^{-3}\; \rm T.
  • l = 6\; \rm cm = 6 \times 10^{-2}\;\rm m.
  • F = 1.5\times 10^{-4}\; \rm N.

Rearrange the equation F = l \cdot I \cdot \sin(\theta) to find an expression for I, the current in this wire.

\begin{aligned} I &= \frac{F}{l \cdot \sin(\theta)} \\ &= \frac{3\times 10^{-3}\; \rm T}{6 \times 10^{-2}\; \rm m \times \sin \left(20^{\circ}\right)} \\ &\approx 7.3 \times 10^{-3}\; \rm A = 7.3 \; \rm mA\end{aligned}.

5 0
3 years ago
In the absence of an electric field, a radioactive beam strikes a fluorescent screen at a single point. When an electric field i
leonid [27]

Answer: Beta, alpha and gamma ray

Explanation: The component being deflected to the positive side is the beta radiation because it is negatively charged and thus is attracted by the positive terminal of the Electric Field.

The component being deflected to the negative side is the alpha radiation, it's is positively charged and thus being attracted by the negative part of the Electric Field.

The component that went straight down without deflection is the gamma radiation. It is neutral and possess no charge and thus is not deflected.

4 0
3 years ago
Read 2 more answers
A segment of wire of total length 3.0 m carries a 15-A current and is formed into a semicircle. Determine the magnitude of the m
miskamm [114]

Answer:

4.9x10^-6T

Explanation:

See attached file

6 0
3 years ago
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