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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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4. A rock is thrown from the edge of the top of a 100 m tall building at some unknown angle above the horizontal. The rock strik
nikklg [1K]

Answer:

Explanation:

Let the velocity of projectile be v and angle of throw be θ.

The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m

considering its vertical displacement

h = - ut +1/2 g t²

100 = - vsinθ x 5 + .5 x 9.8 x 5²

5vsinθ =  222.5

vsinθ = 44.5

It covers 160 horizontally in 5 s

vcosθ x 5 = 160

v cosθ = 32

squaring and adding

v²sin²θ +v² cos²θ = 44.4² + 32²

v² = 1971.36 + 1024

v = 54.73 m /s

5 0
3 years ago
Read 2 more answers
Time management skills include
dexar [7]
D. Budgeting time, avoiding stress, and prioritizing.
7 0
3 years ago
2 ways in which the effects of friction can be minimised in the experiment​
Vlad [161]

Answer:

Friction can be minimized by using lubricants like oil and grease and by using ball bearing between machine parts. A substance that is introduced between two surfaces in contact, to reduce friction, is called a lubricant. Fluid friction can be minimized by giving suitable shapes to the objects moving in the fluids.

Explanation:

hope it helps

7 0
2 years ago
What is the least force you could exert on the refrigerator to move it?
Ulleksa [173]
 <span>If the refrigerator weights 1365 and you are not exerting any vertical force on it, then the normal force is also 1365N. so Fn=1365 

Fsf = Static frictional force = (coefficient of static friction) * (Normal force) 

So the least for you could exert to move it is equal to the Fsf. 
Fsf = (0.49)(1365N)</span><span>
</span>
8 0
3 years ago
Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ
bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; H_0 = 51km/s/Mly

Age of the universe; t = \ ?

We know that, the reciprocal of the Hubble's constant ( H_0 ) gives an estimate of the age of the universe ( t ). It is expressed as:

Age\ of\ Universe; t = \frac{1}{H_0}

Now,

Hubble's constant; H_0 = 51km/s/Mly

We know that;

1\ light\ years = 9.46*10^{15}m

so

1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m

Therefore;

H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

Now, we input this Hubble's constant value into our equation;

Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0
3 years ago
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