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dezoksy [38]
3 years ago
15

The precision of a laboratory instrument is ± 0.05 g. The accepted value for your measurement is 7.92 g. Which measurements are

in the accepted range? Check all that apply.
7.85 g

7.89 g

7.91 g

7.97 g

7.99 g
Physics
2 answers:
skelet666 [1.2K]3 years ago
3 0

Answer:

7.89 7.91

Explanation:

The ranges of measurement lie between 7.92-0.05 and 7.92+0.05

7.87g and 7.97g

STALIN [3.7K]3 years ago
3 0

Answer: 7.89g, 7.91g and 7.97 g

Explanation:

Precision has to do with the maximum value of allowance that can be accepted when added to the original measurement of an object.

Since the accepted value of measurement is 7.92g with precision of ± 0.05 g

The accepted range will be either 7.92+0.5 and 7.92 - 0.5 to give 7.97g and 7.87g

This two values are the maximum and minimum value that can be accepted respectively. Any value that falls in between can also be accepted.

7.87g<=x<=7.97 (acceptable range)

Therefore only 7.89g, 7.91g and 7.97g falls within this range.

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Pani-rosa [81]

Given :

Mass of block , M = 20 kg .

Force applied , F = 80 N .

Acceleration of block , a=2.5\ m/s^2 .

To Find :

The coefficient is Kinetic force friction between the block and the table .

Solution :

We know , Force equation on block is given by :

F_{net }=F-\mu_k mg \\\\ma = F-\mu_k mg \\\\20\times 2.5 = 80 -\mu_k \times 20 \times 10\\\\\mu_k\times 200=30\\\\\mu_k=\dfrac{30}{200}\\\\\ mu_k=0.15

Therefore , coefficient is Kinetic force friction between the block and the table is 0.15 .

Hence , this is the required solution .

5 0
4 years ago
A parachutist with a camera, both descending at a speed of 10.8 m/s, releases that camera at an altitude of 50 m. In this proble
Vadim26 [7]

Answer:

The velocity of the camera is 33.11 m/s.

Explanation:

Given that,

Speed = 10.8 m/s

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Suppose determine the velocity of the camera just before it hits the ground?

We need to calculate the velocity of the camera

Using equation of motion

v^2=u^2+2gs

Where, v = final velocity of camera

u = initial speed of camera

s = distance

Put the value into the formula

v^2=(-10.8)^2+2\times(-9.8)\times(-50)

v=\sqrt{1096.64}

v=33.11\ m/s

The direction will be downward so it is the negative velocity.

Hence, The velocity of the camera is 33.11 m/s.

8 0
3 years ago
An unwary football player collides with a padded goalpost while running at a velocity of 5.70 m/s and comes to a full stop after
ollegr [7]
The relationship between the initial velocity, final velociy, distance, and deceleration can be expressed in the following equation.
          
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The value of a (which is the deceleration) is 0.06 m/s². Thus, the answer is that the deceleration value is approximately 0.06 m/s².
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In the interaction between a hammer and the nail it hits, is a force exerted on the nail? On the hammer? How many forces occur i
Vikki [24]
When you hit a hammer on the head of a nail, all the momentum associated with hammer is transferred into nail. And due to this, the surface of nail gets deformed or if you touch it you can feel the rise in temperature. When you hit a hammer on nail, hammer exerts force on nail and by Newton's 3rd law, nail will also exert a force on the hammer perpendicular to the surface of the hammer of same magnitude but its direction will be reversed. Thus, only 2 forces acts in this interaction and they are action-reaction pairs. 
4 0
3 years ago
A cylinder with a piston contains 0.250 mol of oxygen at2.40x105 Pa and 355 K. The oxygen may be treated as an ideal gas.The gas
Nikitich [7]

Answer:

= 285 Joules

Explanation:

a) answer can be found out in attachment

(b) The temperature for the isothermal compression is the same as the temp at the end of the isobaric expansion. Since pressure is held constant but volume doubles, we use the ideal gas law:

p V = nR T          to see that the temperature also doubles.

.So...   temp for isothermal compression =   355×2 = 710 K

.(c)   The max pressure occurs at the top point. At this point, the volume is back to the original value but the temperature is twice the original value. So the pressure at this point is twice the original, or

max pressure = 2×240000 Pa =  480000 Pa  =   4.80 x 10^5 Pa

(d) total work done by the piston = workdone during isothermal compression - work done during expansion =

=  nRT ln(V initial / V final)-p (V initial - V final)

=   nRT ln(2) - nR(T final - T initial)  

= 0.250× 8.314 ×710×ln(2)-0.250×8.314× (710 - 355)

=     285 Joules

7 0
4 years ago
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