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2 years ago

2. Apply Mathematics If the amplitude of the 6 PM wave increases to 0.6 m, how many times greater would the energy become?

Physics

1 answer:

RoseWind [281]2 years ago

7 0

Answer:

8 and 093 over 0.6 then 303

Explanation:

cross over the numbers and get the answer

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Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, th

alex41 [277]

Answer: 321 J

Explanation:

Given

Mass of the box $m=3\ kg$

Force applied is $F=25\ N$

Displacement of the box is $s=15\ m$

Velocity acquired by the box is $v=6\ m/s$

acceleration associated with it is $a=\dfrac{F}{m}$

$\Rightarrow a=\dfrac{25}{3}\ m/s^2$

Work done by force is $W=F\cdot s$

$W=25\times 15\\W=375\ J$

change in kinetic energy is $\Delta K$

$\Rightarrow \Delta K=\dfrac{1}{2}m(v^2-0)\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 6^2\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 36\\\\\Rightarrow \Delta K=54\ J$

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy

$\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J$

Therefore, the magnitude of work done by friction is $321\ J$

3 0

3 years ago

What is the resultant force on the human cannonball in the

Svetllana [295]

Answer:

heart

Explanation:

4 0

3 years ago

URGENT

Gemiola [76]

Answer:

5.25 J

Explanation:

W = PE = (f)(x)

PE = 35N*0.15m

PE = 5.25 N*m

1 N*m = 1 J

PE = 5.25 J

7 0

2 years ago

An object is accelerating at 2m/s^2. What is the net force if the mass is 125kg?

natita [175]

Answer:

250 N

Explanation:

Newton's second law:

∑F = ma

∑F = (125 kg) (2 m/s²)

∑F = 250 N

3 0

3 years ago

What is the value of the boltzmann constant of 1.7 moles of gas at 290 K that has an average kinetic energy of 1.4 MJ? A. 7x10^-

fredd [130]

Answer:

The value of of the Boltzmann constant is $3.14\times 10^{-23} J/K$ .

Explanation:

Generally the formula for average kinetic energy of a molecule :

$K.E=\frac{3}{2}kT$

where,

k = Boltzmann’s constant

T = temperature = 290 K

Given:

Number of molecules in 1.7 moles:

$1.7\times 6.022\times 10^{23} molecules=1.0237\times 10^{24} molecules$

Average kinetic energy of 1.7 moles = 1.4 MJ = $1.4\times 10^6 J$

Average kinetic energy of 1 molecule = $\frac{1.4\times 10^6 J}{1.023\times 10^{24}} =1.3675\times 10^{-18} J$

$1.3675\times 10^{-18} J=\frac{3}{2}\times k\times (290K)$

$k=3.14\times 10^{-23} J/K$

8 0

3 years ago