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3 years ago

Select the correct comparison between science and technology.

Physics

2 answers:

sweet [91]3 years ago

8 0

Answer: c

Explanation:

mamaluj [8]3 years ago

6 0

Answer:

Explanation:

Science influences society through its knowledge and world view. ... Technology influences society through its products and processes

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Orbiting satellites use geothermal energy panels. True False

Anna007 [38]

False, geothermal means from underground. They use normal solar panels. Hope this helps.

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3 years ago

A 7.4 A current is set up in a circuit for 7.8 min by a rechargeable battery with a 3.0 V emf. By how much is the chemical energ

bearhunter [10]

Answer:

10389.6 J

Explanation:

Power is the rate of doing work with respect to time, its S.I unit is in watts but it can also be expressed in J/s. Power is calculated using the formula:

$Power=\frac{energy}{time}$

Power is also the rate at which energy is used per second.

Energy is the capacity to do work and it is measured in joules (J).

Power = current × voltage = 7.4 A × 3 V = 22.2 W

$Energy=power*time\\\\time=7.8\ min=7.8*60\ s=468\ s\\\\Energy=22.2*468\\\\Energy=10389.6\ J$

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3 years ago

The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21

frutty [35]

Answer:

a) The velocity of the projectile at 2 seconds after launch is 1.9 meters per second. The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) The projectile reaches maximum height 2.192 seconds after launch.

c) The maximum height of the projectile is 26.584 meters above ground.

d) The projectile will hit the ground at 4.523 seconds after launch.

e) The velocity of the projectile right before hitting the ground in -22.871 meters per second.

Explanation:

Complete statement of problem is: The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21.5 meters per second is $h(t) = 3+21.5\cdot t-4.9\cdot t^{2}$ after t seconds. (Round your answers to two decimal places.) (a) Find the velocity after 2 seconds and after 4 seconds, (b) When does the projectile reach its maximum height? (c) What is the maximum height? (d) When does it hit the ground? (e) With what velocity does it hits the ground?

a) From Physics and Differential Calculus we remember that velocity is the first derivative of height. Hence, we need to differentiate the height function in time:

$v(t) = 21.5-9.8\cdot t$ (Eq. 1)

Where $v(t)$ is the velocity function, measured in meters per second.

Now we evaluate this function at given times:

t = 2 s.

$v(2) = 21.5-9.8\cdot (2)$

$v(2) = 1.9\,\frac{m}{s}$

The velocity of the projectile at 2 seconds after launch is 1.9 meters per second.

t = 4 s.

$v(4) = 21.5-9.8\cdot (4)$

$v(4) = -17.7\,\frac{m}{s}$

The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) Maximum height is reached when velocity of projectile is zero. We equalize velocity to zero and solve the expression for $t$ :

$21.5-9.81\cdot t = 0$

$t = 2.192\,s$

The projectile reaches maximum height 2.192 seconds after launch.

c) Maximum height is calculated by evaluating height function at the time found in b). That is:

$h(2.192) = 3+21.5\cdot (2.192)-4.9\cdot (2.192)^{2}$

$h (2.192) = 26.584\,m$

The maximum height of the projectile is 26.584 meters above ground.

d) In this case, we need to equalize the height function to zero and solve for $t$ . That is:

$3+21.5\cdot t-4.9\cdot t^{2} = 0$

Roots are found by means of Quadratic Formula:

$t_{1}\approx 4.523\,s$ and $t_{2}\approx -0.135\,s$

Only the first root offers a physically reasonable solution. Therefore, the projectile will hit the ground at 4.523 seconds after launch.

e) This can be found by evaluating velocity function at the time found in d):

$v(4.523) = 21.5-9.81\cdot (4.523)$

$v(4.523) = -22.871\,\frac{m}{s}$

The velocity of the projectile right before hitting the ground in -22.871 meters per second.

5 0

4 years ago

A centrifuge has an angular velocity of 3,000 rpm, what is the acceleration (in unit of the earth gravity) at a point with a rad

Anna71 [15]

Answer:

$a_{r} = 1006.382g \,\frac{m}{s^{2}}$

Explanation:

Let suppose that centrifuge is rotating at constant angular speed, which means that resultant acceleration is equal to radial acceleration at given radius, whose formula is:

$a_{r} = \omega^{2}\cdot R$

Where:

$\omega$ - Angular speed, measured in radians per second.

$R$ - Radius of rotation, measured in meters.

The angular speed is first determined:

$\omega = \frac{\pi}{30}\cdot \dot n$

Where $\dot n$ is the angular speed, measured in revolutions per minute.

If $\dot n = 3000\,rpm$ , the angular speed measured in radians per second is:

$\omega = \frac{\pi}{30}\cdot (3000\,rpm)$

$\omega \approx 314.159\,\frac{rad}{s}$

Now, if $\omega = 314.159\,\frac{rad}{s}$ and $R = 0.1\,m$ , the resultant acceleration is then:

$a_{r} = \left(314.159\,\frac{rad}{s} \right)^{2}\cdot (0.1\,m)$

$a_{r} = 9869.588\,\frac{m}{s^{2}}$

If gravitational acceleration is equal to 9.807 meters per square second, then the radial acceleration is equivalent to 1006.382 times the gravitational acceleration. That is:

$a_{r} = 1006.382g \,\frac{m}{s^{2}}$

6 0

3 years ago

Jenny and Betty are having a great time at Busch Gardens riding the Ubanga Banga bumper cars. Jenny, who is traveling southward

lutik1710 [3]

They both exert an equal amount of force onto each other

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4 years ago