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Mamont248 [21]
2 years ago
10

A glider of mass m1 on a frictionless horizontal track is connected to an object of mass m2 by a massless string. The glider acc

elerates to the right, the object accelerates downward, and the string rotates the pulley. The pulley has a moment of inertia I. What is the relationship among T1 (the tension in the horizontal part of the string), T2 (the tension in the vertical part of the string), and the weight m2g of the object
Physics
1 answer:
inysia [295]2 years ago
5 0

Answer:

m2g -> T2 -> T1

Explanation:

m2g -> T2 -> T1

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All scientists try to base their conclusions on _____.
seraphim [82]
Either theory or evidence
8 0
2 years ago
Read 2 more answers
8) A flat circular loop having one turn and radius 5.0 cm is positioned with its plane perpendicular to a uniform 0.60-T magneti
Marrrta [24]

Answer:

EMF induced in the loop is 9.4 V

Explanation:

As we know that initial magnetic flux of the loop is given as

\phi_1 = B.A

\phi_1 = (0.60)(\pi (0.05)^2)

\phi_1 = 4.7 \times 10^{-3} Wb

As soon as the area of the loop becomes zero the final magnetic flux of the loop is ZERO

Now as per faraday's law of electromagnetic induction the EMF is induced due to rate of change in magnetic flux

so we will have

EMF = \frac{\Delta \phi}{\Delta t}

so we will have

EMF = \frac{4.7 \times 10^{-3} - 0}{0.50 \times 10^{-3}}

EMF = 9.4 V

7 0
2 years ago
A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?
ankoles [38]
Given required solution

M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
S=VT
=5m/s*2.4s
=12m
so W=12*100
W=1200J
4 0
1 year ago
Two infinite parallel surfaces carry uniform charge densities of 0.20 nC/m2 and -0.60 nC/m2. What is the magnitude of the electr
svlad2 [7]

Answer:

Electric field, E = 45.19 N/C

Explanation:

It is given that,

Surface charge density of first surface, \sigma_1=0.2\ nC/m^2=0.2\times 10^{-9}\ C/m^2

Surface charge density of second surface, \sigma=-0.6\ nC/m^2=-0.6\times 10^{-9}\ C/m^2

The electric field at a point between the two surfaces is given by :

E=\dfrac{\sigma}{2\epsilon_o}

E=\dfrac{\sigma1-\sigma_2}{2\epsilon_o}

E=\dfrac{0.2\times 10^{-9}-(-0.6\times 10^{-9})}{2\times 8.85\times 10^{-12}}

E = 45.19 N/C

So, the magnitude of the electric field at a point between the two surfaces is 45.19 N/C. Hence, this is the required solution.

6 0
1 year ago
A metal cube with sides of length a=1cm is moving at velocity v0→=1m/sj^ across a uniform magnetic field B0→=5Tk^. The cube is o
Nitella [24]

Answer:

the magnitude of the electric field is 1.25 N/C

Explanation:

The induced emf in the cube ε = LB.v where B = magnitude of electric field = 5 T , L = length of side of cube = 1 cm = 0.01 m and v = velocity of cube = 1 m/s

ε = LB.v = 0.01 m × 5 T × 1 m/s = 0.05 V

Also, induced emf in the cube ε = ∫E.ds around the loop of the cube where E = electric field in metal cube

ε = ∫E.ds

ε = Eds since E is always parallel to the side of the cube

= E∫ds  ∫ds = 4L since we have 4 sides

= E(4L)

= 4EL

So,4EL = 0.05 V

E = 0.05 V/4L

= 0.05 V/(4 × 0.01 m)

= 0.05 V/0.04 m

= 1.25 V/m

= 1.25 N/C

So, the magnitude of the electric field is 1.25 N/C

7 0
1 year ago
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