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2 years ago

Define a combination circuit.

Physics

1 answer:

Neporo4naja [7]2 years ago

7 0

Answer:

a circuit that is a blend of series paths and parallel paths. See Figure for a visual explanation. Most circuits are of this variety. Don't be afraid to tackle these circuits as far as the math goes.

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Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projec

sammy [17]

Answer:

1) The maximum jump height is reached at A. $0.337s$

2) The maximum center of mass height off of the ground is B. $1.64m$

3) The time of flight is C. $0.834s$

4) The distance of jump is B. $7.49m$

Explanation:

First of all we need to decompose velocity in its rectangular components, so

$v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s$

1) We use, $v_{fy}=v_{iy}-gt$ , as we clear it for $t$ and using the fact that $v_{fy}=0$ at max height, we obtain $t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s$

2) We can use the formula $y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ for $t=0.337s$ , so

$y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m$

3) We can use the formula $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find total time of fligth, so $0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66$ , as it is a second-grade polynomial, we find that its positive root is $t=0.834s$

4) Finally, we use $x=v_{x}t=8.05m/s(0.834s)=6.71m$ , as it has an additional displacement of $0.77m$ due the leg extension we obtain,

$x=6.71m+0.77m=7.48m$ , aprox $x=7.49m$

5 0

3 years ago

How do you put 1/2000 into scientific notation? *

maksim [4K]

It’s e 2.0 x 10^-4 because it is a fraction

4 0

3 years ago

A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s

gulaghasi [49]

(a) 0.96 m/s

The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:

$T=2\cdot 2.5 s=5.0 s$

The frequency of the waves is the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{5.0 s}=0.20 Hz$

The wavelength instead is just the distance between two consecutive crests, so

$\lambda=4.8 m$

And the wave speed is given by:

$v=\lambda f=(4.8 m)(0.20 Hz)=0.96 m/s$

(b) 0.265 m

The total distance between the highest point of the wave and its lowest point is

d = 0.53 m

The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is

$A=\frac{d}{2}=\frac{0.53 m}{2}=0.265 m$

(c) Amplitude: 0.15 m, wave speed: same as before

In this case, the amplitude of the wave would be lower. In fact,

d = 0.30 m

So the amplitude would be

$A=\frac{d}{2}=\frac{0.30 m}{2}=0.15 m$

Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.

8 0

3 years ago

Using examples, explain why the first and second Newton laws of motion are significant for living organisms.

Triss [41]

Answer:

1) Newton's first law of motion states an object will remain at rest or in uniform will be in uniform motion in a straight line unless a force acts on it

2) Newton's second law states the acceleration of an object is directly proportional to the applied force acting on an object and inversely proportional to the mass of the object

Explanation:

1) With Newton's first law, we are able arrange things within a space and schedule meetings in time knowing that they will remain in place unless an external force changes their positions

2) An example of Newton's second law of motion is that small objects such as a ball are easily accelerated and can be given appreciable acceleration for flight by single, one time contact (such as kicking the ball) while larger objects such as a rock require sustained force application to change their location.

6 0

3 years ago

Two identical loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of

natita [175]

Answer:

The phase difference is 0.659 rad.

Explanation:

Given that,

Distance between two identical loudspeakers d= 1.00 m

Distance between speakers and listener r= 4.00 m

Frequency = 300 Hz

Suppose we need to find the phase difference in radian between the waves from the speakers when they reach the observer