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Rom4ik [11]
2 years ago
9

A block of mass 3. 0 kg is hung from a spring, causing it to stretch 12 cm at equilibrium, as shown above. The 3. 0 kg block is

then replaced by a 4. 0 kg block, and the new block is released from the position shown above, at which the spring is unstretched. How far will the 4. 0 kg block fall before its direction is reversed?.
Physics
1 answer:
QveST [7]2 years ago
4 0

Answer: 32 cm

Explanation:

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3 years ago
A clarinetist, setting out for a performance, grabs his 3.070 kg clarinet case (including the clarinet) from the top of the pian
Cerrena [4.2K]

Answer:

the vertical acceleration of the case is 1.46 m/s

Explanation:

Given;

mass of the clarinet case, m = 3.07 kg

upward force applied by the man, F = 25.60 N

Apply Newton's second law of motion;

the upward force on the clarinet case = its weight acting downwards + downward force due to its downward accelaration

F = mg + m(-a)

the acceleration is negative due to downward motion from the top of the piano.

F = mg - ma

ma = mg - F

a = \frac{mg - F}{m} \\\\a = \frac{(3.07 \times 9.8) \ - \ 25.6}{3.07} \\\\a = 1.46 \ m/s^2

Therefore, the vertical acceleration of the case is 1.46 m/s²

4 0
3 years ago
What ideas do you have a bout why different places have lower or higher rates of skin cancer?
AnnZ [28]

Answer: Some places like Florida and California are both very hot and dry places, the sun damages your skin more in sunny places then in colder more breezy places.

Explanation:

4 0
2 years ago
In the model of the hydrogen atom due to Niels Bohr, the electron moves around the proton at a speed of 3.3 × 106 m/s in a circl
irga5000 [103]

Answer:

1.5048\times 10^{-23}\ Am^2

Explanation:

q = Charge of proton = 1.6\times 10^{-19}\ C

r = Radius of circle = 5.7\times 10^{-11}\ m

v = Velocity of proton = 3.3\times 10^6\ m/s

Magnetic moment is given by

M=\frac{1}{2}qrv\\\Rightarrow M=\frac{1}{2}1.6\times 10^{-19}\times 5.7\times 10^{-11}\times 3.3\times 10^6\\\Rightarrow M=1.5048\times 10^{-23}\ Am^2

The magnetic moment associated with this motion is 1.5048\times 10^{-23}\ Am^2

5 0
3 years ago
A bag of sugar weighs 3.50 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration isone-sixth
Alex73 [517]

Answer:

F_{Earth}= 15.57 N

F_{Moon}= 2.60 N

F_{Uranus}= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

Explanation:

The weight of the sugar bag on Earth is:

g=9.81 m/s²

m=3.50 lb=1.59 kg

F_{Earth}=m·g=1.59 kg×9.81 m/s²= 15.57 N

The weight of the sugar bag on the Moon is:

g=9.81 m/s²÷6= 1.635 m/s²

F_{Moon}=m·g=1.59 kg× 1.635 m/s²= 2.60 N

The weight of the sugar bag on the Uranus is:

g=9.81 m/s²×1.09=10.69 m/s²

F_{Uranus}=m·g=1.59 kg×10.69 m/s²= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

5 0
3 years ago
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