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lyudmila [28]
3 years ago
13

An accelerometer is a device that uses the extension of a spring to measure acceleration in terms of Earth's gravitational accel

eration (g). What is the approximate acceleration if this accelerometer spring is extended just 0.43 centimeters?. . a) 0.24 g. b) 0.80 g. c) 2.00 g. d) 2.40 g
Physics
2 answers:
Nimfa-mama [501]3 years ago
8 0

0.80g (B) is correct for plato!

ELEN [110]3 years ago
3 0
In your question explains that an accelerometer is a device that uses the extension spring to measure acceleration in terms of Earth's Gravitational accelerations. So if this device is extended just 0.43 centimeters the possible accelerations is letter B. 0.80 g
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A proton and an electron are moving in the +x direction in a magnetic field in the +z
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Answer:

Explanation:

A proton and electron are moving in the positive x direction, this shows that their velocity will be in the positive x direction

V = v•i

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B = B•k

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F = q(v×B)

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Then, for the positive charge, the force will act in the negative direction of the y-axis

B. For electron

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5 0
3 years ago
If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is
viktelen [127]
To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: V_{f}=V_{i}+gt
where
V_{f} is the final velocity 
V_{i} is the initial velocity 
g is the acceleration due to gravity 32 \frac{ft}{s^2}
t is the time 
2. Kinematic equation for distance: d=V_{i}t+ \frac{1}{2} gt^2
where
d is the distance 
V_{i} is the initial velocity 
V_{f} is the final velocity
g is the acceleration due to gravity 32 \frac{ft}{s^2}
t is the time 

First, we are going to convert 21.82 mi/h to ft/s:
21.82 \frac{mi}{h} =31.21 \frac{ft}{s}

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: V_{f}=0. We also know that it initial speed is 31.21 ft/s, so V_{i}=31.21. Lets replace those values in our formula to find t:
V_{f}=V_{i}+gt
0=31.21+(-32)t
-32t=-31.21
t= \frac{-31.21}{-32}
t=0.98seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
d=V_{i}t+ \frac{1}{2} gt^2
d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2
d=15.22ft 

Now we can add the height of the building and the maximum height of the object:
d=160+15.22=175.22ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
d=V_{i}t+ \frac{1}{2} gt^2
175.22=31.21t+ \frac{1}{2} (32)t^2
16t^2+31.21t-175.22=0
t=2.47 or t=-4.43
Since time cannot be negative, t=2.47 is the time it takes the object to fall to the ground. 

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
V_{f}=V_{i}+gt
V_{f}=31.21+(32)(2.47)
V_{f}=110.25 ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s


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Answer:

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