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tangare [24]
4 years ago
12

A mass of 4.8 kg is dropped from a height of 4.84 meters above a vertical spring anchored at its lower end to the floor. If the

spring has a height of 82 cm and a constant of 24 N/cm, how far, to the nearest tenth of a cm, is the spring compressed?
Physics
1 answer:
krek1111 [17]4 years ago
5 0

Answer:

45.6 cm

Explanation:

Let x (m) be the length that the spring is compressed. We know that when we drop the mass from 4.84 m above and compress the springi, ts gravitational energy shall be converted to spring potential energy due to the law of energy conservation

E_g = E_p

mgh = 0.5kx^2

where h = 4.84 + x is the distance from the dropping point the the compressed point, and k = 24N/cm = 2400N/m is the spring constant, g = 9.81 m/s2 is the gravitational acceleration constant. And m = 4.8 kg is the object mass.

4.8*9.81(4.84 + x) = 0.5 * 2400 * x^2

47.088x + 227.906 = 1200x^2

1200x^2 - 47.088x - 227.906 = 0

x = 0.456m or 45.6 cm

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A cosmic catastrophic event occurred that caused the tilt of the Earth's axis relative to its plane of orbit to increase from 23
Gnom [1K]

Answer: The elimination of seasonal variations

Explanation:

Since the cosmic catastrophic event which occurred led to the tilt of the Earth's axis relative to the plane of orbit to increase from 23.5° to 90°, the most obvious effect of this change would be the elimination of seasonal variations.

It should be noted that seasonal variation refers to the variation in a time series that's within a year which is repeated. The cause of seasonal variation can include rainfall, temperature, etc.

7 0
3 years ago
If 56.5 m3 of a gas are collected at a pressure of 455 mm Hg, what volume will the gas occupy if the pressure is changed to 632
ExtremeBDS [4]

Assuming ideal conditions, Boyle's law says that

<em>P₁ V₁ </em>= <em>P₂</em> <em>V₂</em>

where <em>P₁ </em>and <em>V₁</em> are the initial pressure and temperature, respectively, and <em>P₂</em> and <em>V₂</em> are the final pressure and temperature.

So you have

(455 mm Hg) (56.5 m³) = (632 mm Hg) <em>V₂</em>

==>   <em>V₂</em> = (455 mm Hg) (56.5 m³) / (632 mm Hg) ≈ 40.7 m³

4 0
3 years ago
A 4.00 µf capacitor is connected to a 12.0 v battery.
atroni [7]
(a) The capacitance of the capacitor is:
C=4 \mu F=4 \cdot 10^{-6}F
and the voltage applied across its plates is
V=12.0 V

The relationship between the charge Q on each plate of the capacitor, the capacitance and the voltage is:
C= \frac{Q}{V}
and re-arranging it we find the charge stored in the capacitor:
Q=CV=(4 \cdot 10^{-6} F)(12.0 V)=4.8 \cdot 10^{-5} C

(b) The electrical potential energy stored in a capacitor is given by
U= \frac{1}{2}CV^2
where C is the capacitance and V is the voltage. The new voltage is 
V=1.50 V
so the energy stored in the capacitor is
U= \frac{1}{2}(4 \cdot 10^{-6} F)(1.50 V)^2=4.5 \cdot 10^{-6} J
3 0
3 years ago
Read 2 more answers
Moon rocks resemble rocks from which of the following layers of the earth?
Vedmedyk [2.9K]
Mantle I think idrk cuz of erosion
7 0
3 years ago
A cosmic-ray electron moves at perpendicular to earth’s magnetic field at an altitude where the field strength is. what is the r
Harlamova29_29 [7]

4.266 m is the radius of the circular path the electron follows.

Given

Speed of electron (v) = 7.5 × 10⁶ m/s

Earth's Magnetic Field (B) = 1 × 10⁻⁵ T

We already know that

Mass of electron (m) = 9.1 × 10⁻³¹ kg

Charge on electron (q) = 1.6 × 10⁻¹⁹ C

According to the formula

Radius of circular path(r) = mass on electron × speed/ Charge × Magnetic field

Radius of circular path(r) = m × v/q × B

Put the values into the formula

r = 9.1 × 10⁻³¹ × 7.5 × 10⁶/ 1.6 × 10⁻¹⁹ × 10⁻⁵

On solving, we get

r = 4.266 m

Hence, 4.266 m is the radius of the circular path the electron follows.

Learn more about magnetic field here brainly.com/question/26257705

#SPJ4

6 0
1 year ago
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