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scZoUnD [109]
3 years ago
12

in physics lab, a cube slides down a frictionless incline as shown in the figure below, and elastically strikes another cube at

the bottom that is only one-half its mass. if the incline is 30 cm high and the table is 90cm off the floor, where does each cube land [hint: both leave the incline moving horizontally]

Physics
1 answer:
Tema [17]3 years ago
3 0
<span>In the physics lab, a cube slides down a frictionless incline as shown in the figure below, check the image for the complete solution:

</span>

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The order of elements in the periodic table is based on the atomic number.<br> true or false
kirill [66]
True the elements are ordered in the atomic number
4 0
3 years ago
A 4.30 g bullet moving at 943 m/s strikes a 730 g wooden block at rest on a frictionless surface. The bullet emerges, traveling
Ymorist [56]

Answer:

(a)2.7 m/s

(b) 5.52 m/s

Explanation:

The total of the system would be conserved as no external force is acting on it.

Initial momentum = final momentum

⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)

⇒ 730 ×v = (4054.9 - 2081.2) =1973.7

⇒v=2.7 m/s

Thus, the resulting speed of the block is 2.7 m/s.

(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.

V_{COM} = \frac{m_b}{m_b+m_{bl}}v_{bi}=\frac{4.30}{4.30+730}\times 943 m/s = 5.52 m/s

Thus, the speed of the bullet-block center of mass is 5.52 m/s.

4 0
4 years ago
2. A carpenter tosses a shingle off a 9.4 m high roof, giving it an initial horizontal
Aleksandr [31]

Sounds like the shingle/ball is thrown from the roof horizontally, so that the distance it travels <em>x</em> after time <em>t</em> horizontally is

<em>x</em> = (7.2 m/s) <em>t</em>

The object's height <em>y</em> at time <em>t</em> is

<em>y</em> = 9.4 m - 1/2 <em>gt</em>²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, and its vertical velocity is

<em>v</em> = -<em>gt</em>

(a) The object hits the ground when <em>y</em> = 0:

0 = 9.4 m - 1/2 <em>gt</em>²

<em>t</em>² = 2 * (9.4 m) / (9.80 m/s²)

<em>t</em> ≈ 1.92 s

at which time the object's vertical velocity is

<em>v</em> = -<em>g</em> (1.92 s) = -18.8 m/s ≈ -19 m/s

(b) See part (a); it takes the object about 1.9 s to reach the ground.

(c) The object travels a horizontal distance of

<em>x</em> = (7.2 m/s) * (1.92 s) ≈ 13.8 m ≈ 14 m

8 0
3 years ago
4.
Akimi4 [234]

Answer:

c

Explanation:

plz make me brainliest i have answered

5 0
3 years ago
A vw beetle goes from 0 to 60 mi/h with an acceleration of 2.35 m/s^2.
Vlad1618 [11]
Part a.
u = 0, the initial velocity
v = 60 mi/h, the final velocity
a = 2.35 m/s², the acceleration.

Note that
1 m = 1609.34 m.
Therefore
v = (60 mi/h)*(1609.34 m/mi)*(1/3600 h/s) = 26.822 m/s
Use the formula
v = u + at
(26.822 m/s) = (2.35 m/s²)*(t s)
t = 26.822/2.35 = 11.4 s

Answer: 11.4 s

Part b.
We already determined that v = 60 mi/h = 26.822 m/s.
t = 0.6 s
Therefore
(26.822 m/s) = (a m/s²)*(0.6 s)
a = 26.822/0.6 = 44.7 m/s²

Answer:  44.7 m/s²
6 0
3 years ago
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