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Tpy6a [65]
2 years ago
15

11. A box with a mass of 1.0 kg is resting on a horizontal surface and the coefficient of friction between the block and the sur

face is 0.20. It is accelerated by attaching a 1.5 kg mass, as shown in the diagram. Assume that the pulley is frictionless and that the cord has negligible mass.
a. What is the acceleration of the box? What is the tension?
Physics
2 answers:
Volgvan2 years ago
6 0

Answer:

A force is applied acting to the right. Assume that friction is negligible. Ikg 0.5N. For each question, one or more features of the system.

Explanation:hope this helps

AlladinOne [14]2 years ago
5 0

Answer:

We know that the force pulling the box in the positive x direction has a magnitude of m g sin 30 . Using Newtons Second Law, F = ma , we just need to solve for a :

ma=mgsin30

a=gsin30

=(10m/s2)(0.500)

=5m/s2

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The mass of Jupiter is 1.9 × 1027 and that of the sun is 1.99 × 1030. The
n200080 [17]

Answer:

F = 4.147 × 10^23

v = 1.31 × 10^4

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Given the following :

mass of Jupiter (m1) = 1.9 × 10^27

Mass of sun (m2) = 1.99 × 10^30

Distance between sun and jupiter (r) = 7.8 × 10^11m

Gravitational force (F) :

(Gm1m2) / r^2

Where ; G = 6.673×10^-11 ( Gravitational constant)

F = [(6.673×10^-11) × (1.9 × 10^27) × (1.99 × 10^30)] / (7.8 × 10^11)^2

F = [25.231 × 10^(-11+27+30)] / (60.84 × 10^22)

F = (25.231 × 10^46) / (60.84 × 10^22)

F = 3.235 × 10^(46 - 22)

F = 0.4147 × 10^24

F = 4.147 × 10^23

Speed of Jupiter (v) :

v = √(Fr) / m1

v = √[(4.147 × 10^23) × (7.8 × 10^11) / (1.9 × 10^27)

v = √32.3466 × 10^(23+11) / 1.9 × 10^27

v = √32.3466× 10^34 / 1.9 × 10^27

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4 0
3 years ago
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nexus9112 [7]

Answer:

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Volume of the mixture = V_1+V_2+V_3

Density of the mixture:

\frac{M}{V_1+V_2+V_3}=15.58 g/cm^3

8 0
3 years ago
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