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3 years ago

I WILL MARK BRAINLIST!!!!

1 answer:

7 0

Are you sure it’s mechanical energy? I haven’t taken a physics class in a while but mechanical energy doesn’t sound right.

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Question 1 (1 point)

KATRIN_1 [288]

Pretty sure it is weather :))

7 0

3 years ago

A wave traveling in water has a frequency of 500.0 Hz and a wavelength of 3.00 m. What is the speed of the wave?

Sergeu [11.5K]

Answer:

1500 m/s

Explanation:

Recall that for a wave,

Speed = frequency x wavelength

here we are given frequency = 500 Hz and wavelength = 3m

simply substitute into above equation

Speed = 500 Hz x 3m

= 1500 m/s

6 0

3 years ago

A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi

Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

$d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L$

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

$x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L$

Finally, using the Parallel Axis Theorem, we calculate I_B:

$I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4} ML^{2} =\frac{1}{3} ML^{2}$

5 0

3 years ago

A sample of nitrogen gas has a volume of 5.0 ml at a pressure of 1.50 atm. what is the pressure exerted by the gas if the volume

olchik [2.2K]

A sample of nitrogen gas has a volume of 5.0 ml at a pressure of 1.50 atm. what is the pressure exerted by the gas if the volume increases to 30.0 ml, at constant temperature is 0.25atm.

On constant temperature, the pressure and volume relation become constant before and after the change in quantitities have occurred.

According to Boyle's Law,

P₁V₁ = P₂V₂

where, P₁ is pressure exerted by the gas initially

V₁ is the volume of gas initially

P₂ is pressure exerted by the gas finally

V₂ is the volume of gas finally

Given,

P₁ = 1.5 atm

V₁ = 5 ml

V₂ = 30 ml

P₂ =?

On substituting the given values in the above equation:

P₁V₁ = P₂V₂

1.5 atm × 5 ml = P₂ × 30 ml

P₂ = 0.25 atm

Hence, pressure exerted by the gas is 0.25atm.

Learn more about Boyle's Law here, brainly.com/question/1437490

#SPJ4

8 0

2 years ago

Interactive Solution 9.1 presents a model for solving this problem. The wheel of a car has a radius of 0.300 m. The engine of th

Murrr4er [49]

Answer:

740 N

Explanation:

We are given that

Radius,r=0.3 m

Torque, $\tau=222 Nm$

We have to find the magnitude of the static frictional force.

According to question

Torque by engine=Torque by static friction

$222=f\times r$

$f=\frac{222}{r}$

$f=\frac{222}{0.3}$

$f=740 N$

Hence, the magnitude of static frictional force=740 N

8 0

3 years ago