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Eduardwww [97]
2 years ago
15

13–82. the 8-kg sack slides down the smooth ramp. if it has a speed of 1.5 m> s when y = 0.2 m, determine the normal reaction

the ramp exerts on the sack and the rate of increase in the speed of sack at this instant.
y
*13–84. the 2-lb block is released from rest at a and slides down along the smooth cylindrical surface. if the attached spring has a stiffness k = 2 lb> ft, determine its unstretched length so that it does not allow the block to leave the surface until u = 60°.
a
prob. 13–82
prob. 13–84
2,fuuu 2,frur
Physics
1 answer:
marissa [1.9K]2 years ago
4 0
The second problem requires a figure to be answered. For the first problem
The acceleration of the sack is
1.5² - 0² = 2a(0.2)
a = 5.63 m/s2
The reaction of the ramp is
F = 8 kg (5.63 m/s2)
F = 45 N
Differentiate the kinematic equation involving time to get the rate of increase of the velocity.
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A highly charged piece of metal (with uniform potential throughout) tends to spark at places where the radius of curvature is sm
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Answer:

look it up

Explanation:

8 0
2 years ago
Which statement best compares and contrasts two physical properties of matter?
Stolb23 [73]

Answer:

D

Explanation:

Boiling points and melting points are similar because they both involve the change in a state of a material, but they are different because boiling point involves a change from a liquid to a gas and melting point involves a change from a solid to a liquid.

5 0
2 years ago
The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the ob
givi [52]

Answer:

Charge, Q=1.56\times 10^{-8}\ C

Explanation:

It is given that,

Electric field strength, E = 180000 N/C

Distance from a small object, r = 2.8 cm = 0.028 m

Electric field at a point is given by :

E=\dfrac{kQ}{r^2}

Q is the charge on an object

Q=\dfrac{Er^2}{k}

Q=\dfrac{180000\ N/C\times (0.028\ m)^2}{9\times 10^9\ Nm^2/C^2}

Q=1.56\times 10^{-8}\ C

So, the charge on the object is 1.56\times 10^{-8}\ C. Hence, this is the required solution.

7 0
2 years ago
There are two forces on the 2 kg box in the overhead view of the following figure, but only one is shown. For F1=20N, a= 12 m/s2
maw [93]

Answer:

second force = 32.784

Magnitude =\sqrt{32.784

θ = -90°

Explanation:

a)

Fnet = ma

F1 + F2 = ma

20N + F2 = 2(12 × cos30° + 12 ×sin30°)

F2 = 2 × 12 ( sin 30° + cos 30°)

    = 24 × ( 1 + √3 )÷ 2

  =12 (1 +√3 )

  = 32.784

b) \sqrt{12(1 +\sqrt{3}}

= \sqrt{12 ( 1+ 1.732)}

= \sqrt{12 (2.732)}

= \sqrt{32.784}  

c)

θ = 30° + 180°

θ = 210°

210° - 300°

θ = -90°

8 0
2 years ago
2 boxes connected by a plus sign hold Wave 1 on top and Wave 2 on bottom. The crests of Wave 1 line up with the troughs of Wave
loris [4]

Answer:

No, the resulting wave in the diagram does not demonstrate destructive interference. The resulting wave in the diagram shows a bigger wave than Wave 1 or Wave 2. If it demonstrated destructive interference, it would be a smaller wave or a horizontal line. With destructive interference, waves break down to form a smaller wave, or cancel each other out, resulting in no wave formation.

6 0
2 years ago
Read 2 more answers
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