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Korolek [52]
2 years ago
12

I need help with #24 ASAP .. I have to get it done today.. I need help with #24 right now

Physics
1 answer:
Savatey [412]2 years ago
8 0

Answer:

Explanation: the  temperature range is at 1,400

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What happens to myosin and actin as sarcomeres relax?
Leokris [45]
The myosin heads pull on the actin, bringing them closer together

4 0
3 years ago
1. A 2.5 kg led projector is launched as a projectile off a tall building. At one point, as it
spin [16.1K]

Answer:

Explanation:

I got everything but i. Don't know why but it's eluding me. So let's do everything but that.

a. PE = mgh so

   PE = (2.5)(98)(14) and

   PE = 340 J

b. KE=\frac{1}{2}mv^2 so

   KE=\frac{1}{2}(2.5)(14)^2 and

   KE = 250 J

c. TE = KE + PE so

   TE = 340 + 250 and

   TE = 590 J

d. PE at 8.7 m:

   PE = (2.5)(9.8)(8.7) and

   PE = 210 J

e. The KE at the same height:

   TE = KE + PE and

   590 = KE + 210 so

   KE = 380 J

f. The velocity at that height:

   380=\frac{1}{2}(2.5)v^2 and

   v=\sqrt{\frac{2(380)}{2.5} } so

   v = 17 m/s

g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:

   590 = KE + PE and

   PE = (2.5)(9.8)(11.6) so

   PE = 280 then

   590 = KE + 280 so

   KE = 310 then

   310=\frac{1}{2}(2.5)v^2 and

   v=\sqrt{\frac{2(310)}{2.5} } so

   v = 16 m/s

h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:

   v=v_0+at and

   26 = 0 + 9.8t and

   26 = 9.8t so the time at 26 m/s is

   t = 2.7 seconds. Now we use that in the equation for displacement:

   Δx = v_0t+\frac{1}{2}at^2 and filling in the time the object was at 26 m/s:

   Δx = 0t + \frac{1}{2}(-9.8)2.7)^2 so

   Δx = 36 m

i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.

4 0
2 years ago
A wave has a wavelength of 9 mm and a frequency of 14 hertz. What is its speed?
lesantik [10]
Wavespeed = frequency x wavelength
= 14 x 9
= 126 mm/s
= 0.126 m/s
5 0
3 years ago
A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if
Dmitrij [34]

To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

Planet gravitational force

F_p = 6*10^{22}N

F_p = \frac{GMm}{R^2}

F_p = 9*10^{22}N

Distance between planet and star

r = \frac{R}{2}

Gravitational force is

F = \frac{GMm}{r^2}

Applying the new distance,

F = \frac{GMm}{(\frac{R}{2})^2}

F =  4\frac{GMm}{R^2}

Replacing with the previous force,

F = 4F_p

Replacing our values

F= 4(9*10^{22}N)

F = 36*10^{22}N

Therefore the magnitude of the force on the star due to the planet is  36*10^{22}N

5 0
3 years ago
the electrical resistance of copper wire varies directly with its length and inversely with the square of the diameter of the wi
leonid [27]

Answer:24.47

Explanation:

Given

L_1=30 m

d_1=3 mm

R_1=25 \Omega

L_2=40 m

d_2=3.5 mm

we know Resistance R=\frac{\rho L}{A}

Where R=resistance

\rho =resistivity

L=Length

A=area of cross-section

A=\frac{\pi d^2}{4}

Thus R\propto \frac{L}{d^2}

therefore

R_1\propto \frac{L_1}{d_1^2}--------1

R_2\propto \frac{L_2}{d_2^2}--------2

divide 1 and 2 we get

\frac{R_1}{R_2}=\frac{L_1}{L_2}\times \frac{d_2^2}{d_1^2}

\frac{R_1}{R_2}=\frac{30}{40}\times \frac{3.5^2}{3^2}

R_2=25\times 0.734\times 1.33

R_2=24.47 \Omega

8 0
3 years ago
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