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german
3 years ago
8

Two blocks of masses 1.0 kg and 2.0 kg, respectively, are pushed by a constant applied force F across a horizontal frictionless

table with constant acceleration such that the blocks remain in contact with each other, as shown above. The 1.0 kg block pushes the 2.0 kg block with a force of 2.0 N. The acceleration of the two blocks is
0

1.0 m/s2

1.5 m/s2

2.0 m/s2

3.0 m/s2
Physics
1 answer:
BaLLatris [955]3 years ago
6 0

Answer:

1.0 m/s^2

Explanation: happy to help :)

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A 7.3 cm diameter loop of wire is initially oriented so that its plane is perpendicular to a magnetic field of 0.61 T pointing u
kenny6666 [7]

Answer:

induced emf =  28.65 mV

Explanation:

given data

diameter = 7.3 cm

magnetic field = 0.61

time period = 0.13 s

to find out

magnitude of the induced emf

solution

we know radius is diameter / 2

radius = 7.3 / 2

radius = 3.65 m

so induced emf is dπ/dt  = Adb/dt

induced emf =  A × ΔB / Δt

induced emf =  πr² × ΔB / Δt

induced emf =  π (0..65)² × ( 0.61 - (-0.28))  / 0.13

induced emf =  0.0286538 V

so induced emf =  28.65 mV

3 0
3 years ago
6. Convert -180° to radians
icang [17]

Answer:

-3.141593

Explanation:

3 0
3 years ago
Read 2 more answers
A ball is thrown down vertically with an initial speed of 31 ft/s from a height of 40 ft. (a) What is its speed just before it s
ICE Princess25 [194]

Answer:

a. 41.96ft/s

b. 1.096s

Explanation:

a. v²=u²+2gs

v²=31²+2×10×40

V=41.96ft/s

b. t=(v-u) /g

t=(41.96-31)/10

t=1.096s

5 0
3 years ago
A Foucault pendulum consists of a brass sphere with a diameter of 31.0 cm suspended from a steel cable 11.0 m long (both measure
kozerog [31]

Answer:

43.7 °C

Explanation:

\alpha_b = Coefficient of linear expansion of brass = 18\times 10^{-6}\ ^{\circ}C

\alpha_s = Coefficient of linear expansion of steel = 11\times 10^{-6}\ ^{\circ}C

L_{0b} = Initial length of brass = 31 cm

L_{0s} = Initial length of steel = 11 m

\Delta L = Total change in length = 3 mm

Total change in length would be

\Delta L=\Delta L_b+\Delta L_s\\\Rightarrow \Delta L=L_{0b}\alpha_b\Delta T+L_{0s}\alpha_b\Delta T\\\Rightarrow \Delta T=\frac{\Delta L}{L_{0b}\alpha_b+L_{0s}\alpha_b}\\\Rightarrow \Delta T=\frac{0.003}{0.31\times 18\times 10^{-6}+11\times 10^{-6}\times 11}\\\Rightarrow \Delta T=23.7\ ^{\circ}C

\Delta T=23.7\\\Rightarrow T_f-T_i=23.7\\\Rightarrow T_f=23.7+T_i\\\Rightarrow T_f=23.7+20\\\Rightarrow T_f=43.7\ ^{\circ}C

The final temperature is 43.7 °C

6 0
4 years ago
This problem has been solved!See the answerCommunication with submerged submarines via radio waves is difficult because seawater
mash [69]

Answer:

4000 km

Explanation:

as we know velocity of electromagnetic wave is c

c = 3 * 10^8 m/s

frequency  given (f) = 76 Hz

wavelength ?

using

c =fλ

λ =   \frac{c}{f}

λ =  \frac{3 * 10^8}{76}  ≈4000 km

 so final answer λ = 4000km

8 0
3 years ago
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