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3 years ago

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Which of the following methods is most likely to be used to determine the absolute age of animal bones that are less than 45,000

years old? A. Radiocarbon B. Potassium- argon dating C. Uranium- lead dating D. Relative dating

1 answer:

Leni [432]3 years ago

6 0

Radiocarbono la opcion a

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The discovery of which force most directly impacted our society by allowing the development of nuclear weapons

zhuklara [117]

nuclear ... truman, hiroshima, nagasaki

4 0

3 years ago

What is the resulting velocity of the launcher if the net force on the launcher is equal to the reaction force?

xz_007 [3.2K]

Answer:

according to this question best answer is C

5 0

3 years ago

A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

a)

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

b)

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

c)

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in the linear speed

$\Delta t$ is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

$v=8322 m/s$

Therefore, its linear speed is not changing, so the change in linear speed is zero:

$\Delta v=0$

And therefore, the tangential acceleration is zero as well:

$a_t=\frac{0}{\Delta t}=0 m/s^2$

5 0

3 years ago

Tech A says that some electric actuators are positioned by an A/C ECU which checks the air flow with sensors. Tech B says that e

Svetradugi [14.3K]

Answer:

Tech B

Explanation:

The control of electric actuators may be through through a two-wire or a five-wire circuit through the driver circuit by bidirectionally controlling via the motor wire polarity. The door position is determined by counting the actuator commutator pulses by the control module in the actuator with 2-wire, while the actuator with 5-wire uses potentiometer feedback.

The A/C ECU is the Air Condition Engine Control Unit.

7 0

3 years ago

A simple gaussmeter for measuring horizontal magnetic fields consists of a stiff 52 cm wire that hangs from a conducting pivot s

vivado [14]

Answer:

8.1345°

Explanation:

We apply $\sum F_x=0$ to the wire to obtain:

$\sum F_x=-F_m+mg\ sin \theta=0$

#The magnitude of the magnetic force acting on the wire is given by:

$F_m=IlB\ sin \phi, \phi=90\textdegree\\\\F_m=IlB$

#Substitute for $F_m$ to obtain:

$-IlB+mg\ sin \ \theta=0\ \ \ \ \ \ \ \ ...i$

Solve for $\theta$ :

$\theta=sin^{-1}[\frac{IlB}{mg}]$

We the substitute the numerical values to calculate the equilibrium angular displacement:

$\theta= sin^{-1}[\frac{0.2A\times0.52m\times 0.040T}{0.003\ kg\times9.8m/s^2}]\\\\=8.1345\textdegree$

Hence, the equilibrium angular displacement of the wire from vertical if the horizontal magnetic field is 8.1345°

3 0

3 years ago