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mina [271]
3 years ago
14

If you touch the two probes together while the DMM is set to resistance, what will happen?

Physics
1 answer:
rosijanka [135]3 years ago
8 0

If the meter's battery is good and the meter is working OK, it should read zero ohms.

Most meters have a little adjustment wheel on the side to SET the reading to exactly zero when the probes are touched together.

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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


5 0
2 years ago
The ratio of output force is the what of a machine
Crazy boy [7]
The ratio of output force to the input force is generally the mechanical advantage of the machine.
7 0
3 years ago
Your roommate leaves a 120W fan running in your apartment.Over the course of an hour,how much thermal energy does the fan add to
zubka84 [21]

Answer:

4.32\cdot 10^5 J

Explanation:

Power is related to energy by the following relationship:

P=\frac{E}{t}

where

P is the power used

E is the energy used

t is the time elapsed

In this problem, we know that

- the power of the fan is P = 120 W

- the fan has been running for one hour, which corresponds to a time of

t = 1 h \cdot (60 min/h)(60 s/min)=3600 s

So we can re-arrange the previous equation to find E, the energy (in the form of thermal energy) released by the fan:

E=Pt=(120 W)(3600 s)=4.32\cdot 10^5 J

3 0
2 years ago
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Which category of galaxy does not have a distinctive shape?
mylen [45]
The category of galaxy which does not have a distinctive shape is D. an irregular galaxy.
A spiral galaxy has a spiral shape, an elliptical galaxy has an elliptical shape, and a barred-spiral galaxy has a barred-spiral shape. The only galaxy type which does not have a constant shape is an irregular galaxy.
5 0
3 years ago
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When a cannon fires a cannonball, the cannon will recoil backward because the:
gulaghasi [49]

C) total linear momentum of the ball and cannon is conserved.

Basically it happens that in the beginning before there is a momentum acting on the two bodies, these are a unique system. Here the total momentum of the System is 0. However, when the positive momentum of the cannonball is added, the system will be immediately affected by a negative momentum which will pull back the cannon. Could this be extrapolated as a condition of Newton's third law.

4 0
2 years ago
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