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Vladimir79 [104]
3 years ago
12

A 2 300-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 3.00 m before coming into contact

with the top of the beam, and it drives the beam 15.6 cm farther into the ground before coming to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.
Physics
1 answer:
Mashcka [7]3 years ago
6 0

Answer:

433903.8 N

Explanation:

From work energy theorem, the total work done is equivalent to change in kinetic energy. Kinetic energy is depedant on speed and since pile drive finally comes to rest, then final velocity is zero. Also, its initial velocity is not given. Therefore, the sum of work due to gravity and beam equals to zero. Work due to gravity is product of mass of pile driver, acceleration due to gravity and height while work due to beam is a product of force and distance. Substituting the given values then

2300*9.81*3+(F*0.156)=0

F=-433903.84615384615384615384615384615384615 N

Approximately, the magnitude of force is 433903.8 N and it acts upwards

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A very powerful vacuum cleaner which has a hose of circular cross section can lift a brick of mass 12 kg when the hose is placed
valkas [14]

Answer:

A). 1.9 cm

Explanation:

m = Mass of brick = 12 kg

g = Acceleration due to gravity = 9.81 m/s²

r = Radius of hose

A = Area = \pi r^2

F = Force = mg

Let us assume that the pressure required to lift the brick would be atmospheric pressure

P=\dfrac{F}{A}\\\Rightarrow P=\dfrac{F}{\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{F}{\pi P}}\\\Rightarrow r=\sqrt{\dfrac{12\times 9.81}{\pi\times 101325}}\\\Rightarrow r=0.01923\ m=1.9\ cm

The radius of the hose should be 1.9 cm

6 0
3 years ago
Which phase of cell division results in the formation of 4 new haploid cells
Alexandra [31]
The answer to your question is Meiosis.

Hope this helps! God bless
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7 0
4 years ago
A weather balloon is partially inflated with helium gas to a volume of 2.0 m³. The pressure was measured at 101 kPa and the temp
Yanka [14]

Answer:

4.7 m³

Explanation:

We'll use the gas law P1 • V1 / T1 = P2 • V2 / T2

* Givens :

P1 = 101 kPa , V1 = 2 m³ , T1 = 300.15 K , P2 = 40 kPa , T2 = 283.15 K

( We must always convert the temperature unit to Kelvin "K")

* What we want to find :

V2 = ?

* Solution :

101 × 2 / 300.15 = 40 × V2 / 283.15

V2 × 40 / 283.15 ≈ 0.67

V2 = 0.67 × 283.15 / 40

V2 ≈ 4.7 m³

7 0
2 years ago
If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front o
tangare [24]
The correct answer is letter A. 6 millimeters. <span>If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front of the lens, the height of the image is 6 millimeters.
</span>
Solution:
18 / x = 12 / 4 
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7 0
3 years ago
The nuclei of large atoms, such as uranium, with 9292 protons, can be modeled as spherically symmetric spheres of charge. The ra
Scrat [10]

Answer:

Part 1 E = 2.42 * 10^{21}  N/C

Part 2 E = 1.3 * 10^{13}  N/C

Part 3 E = 0

Explanation:

Given

Number of protons = 92

Radius of nucleus r_n = 7.4 * 10^{-15} m

Distance of the electrons r_1 = 1.0 * 10^ {-10} m

Part 1

Electric field produced by  just outside its surface

E  = \frac{q}{4\pi*E_0* r_n^2 } \\E  = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(7.4* 10^{-15})^2} \\E = 2.42 * 10^{21}  N/C

Part 2

Electric field produced by  just outside its surface

E  = \frac{q}{4\pi*E_0* r_n^2 } \\E  = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(1* 10^{-15})^2} \\E = 1.3 * 10^{13}  N/C

Part 3

The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other

hence, the solution is

Part 1 E = 2.42 * 10^{21}  N/C

Part 2 E = 1.3 * 10^{13}  N/C

Part 3 E = 0

7 0
3 years ago
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