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Leona [35]
2 years ago
12

3.accuracy is gauged by comparing the measured value of a known standard to its true value. assuming the mass of the water repre

sents a standard for the true volume of water, which piece of glassware has the highest degree of accuracy
Physics
1 answer:
tangare [24]2 years ago
4 0

The glassware with the highest degree of accuracy is 250 mL beaker.

<h3>What is accuracy?</h3>

When in an experiment, a value is measured 5 times, then if the values measured are same for most of the time or like three times out of five, it said to be accurate. The phenomenon is accuracy.

Accuracy compares the experimental value to the theoretical value.

Accepted density of water = 0.99 g/mL

1. 10 mL cylinder

Mass of water = 6.76 g

Volume of water = 6.8 mL

Density = mass /Volume

density = 6.76/6.8 = 0.99412 g/mL

Percentage error = [Observed value - Accepted value/ Accepted value ] x 100

% error = [0.99412 - 0.99 / 0.99 ] x 100

% error = 0.416 %

2. 50 mL cylinder

Mass of water = 24 g

Volume of water = 24.2 mL

density = 24/24.2 = 0.9917 g/mL

Percentage error = [Observed value - Accepted value/ Accepted value ] x 100

% error = [0.9917 - 0.99 / 0.99 ] x 100

% error = 0.172 %

3. 25 mL cylinder

Mass of water = 17 g

Volume of water = 17.1 mL

density = 17/17.1 = 0.99415 g/mL

Percentage error = [Observed value - Accepted value/ Accepted value ] x 100

% error = [0.99415 - 0.99 / 0.99 ] x 100

% error = 0.419 %

4. 250 mL beaker

Mass of water = 35 g

Volume of water = 35.3 mL

density = 35/35.3 = 0.9915 g/mL

Percentage error = [Observed value - Accepted value/ Accepted value ] x 100

% error = [0.9915 - 0.99 / 0.99 ] x 100

% error = 0.152 %

Lesser the percentage error, higher is the accuracy.

Thus, 250 mL beaker has the highest degree of accuracy.

Learn more about accuracy.

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An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f
IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}

Similarly, The equation of mption:

x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})

replacing our assumed values:

where v_=0 \ and \ g= 9.81

x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})

x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

1000= 0.981t-0.981(1-e^{-(10)t}) \ m

By diregarding e^{-(10)t} \ m

1000= 0.981t-0.981

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec

7 0
3 years ago
Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slid
nataly862011 [7]

Answer:

A)

the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B)

data that were not necessary to the solution are;

a) mass of truck and b) mass of load

Explanation:

Given that;

mass of load m_{LS} = 10000 kg

mass of flat bed m_{FB} = 20000 kg

initial speed of truck v_{0} = 12 m/s

coefficient of friction between the load sits and flat bed μs = 0.5

A) the minimum stopping distance for which the load will not slide forward relative to the truck.

Now, using the expression

Fs,max = μs F_{N}     -------------let this be equation 1

where F_{N} = normal force = mg

so

Fs,max = μs mg

ma_{max} = μs mg

divide through by mass

a_{max} = μs g    ---------- let this be equation 2

in equation 2, we substitute in our values

a_{max} = 0.5 × 9.8 m/s²

a_{max} = 4.9 m/s²

now, from the third equation of motion

v² = u² + 2as

v_{f}² = v_{0}² + 2aΔx

where v_{f} is final velocity ( 0 m/s )

a is acceleration( - 4.9 m/s² )

so we substitute

(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx

0 = 144 m²/s² - 9.8 m/s²Δx

9.8 m/s²Δx = 144 m²/s²

Δx = 144 m²/s² /  9.8 m/s²

Δx = 14 m

Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B) data that were not necessary to the solution are;

a) mass of truck and b) mass of load

3 0
3 years ago
A window air conditioner uses 1010 W of electricity and has a coefficient of performance of 3.88. For simplicity, let's assume t
Naya [18.7K]

Answer:

Explanation:

1.) Q=MC_v\delta T\\=\rho \times v \times C_v \times \delta T\\=1.2 \times 7 \times 7 \times 3 \times 720 \times (324-295)\\=3683232J

2.) COP

= \frac{Q_{abs}}{W}=3.88\\\\work=949286.6J

3.) Rate of heat absorption

   COP\times Power\\

=3.88\times 1010watt=3918.8watt

Therefore, cooling in minutes=\frac{\frac{3683232}{3918.8}}{60}=15.66mins

COP cannot cycle= \frac{295}{324-295}\\\\=10.17=\frac{Q_{abs}}{power}

Rate of heat absorption =(1010\times 10.17)watt=10271.7min

cooling time in minutes with cannot cycle=\frac{\frac{3683232}{10271.7}}{60}=5.98mins

3 0
3 years ago
If you do 200 joules of work using a force of 50 newtons, over what distance was the force applied ?
harina [27]
By using the formula for work, we can derive the force

W = Fd
F= \frac{W}{d} = \frac{200}{50} = 4 N

If you found that helpful, please rate this as "Brainiest"
5 0
3 years ago
Read 2 more answers
1. A bus starting from rest moves
skelet666 [1.2K]

Answer:

Speed acquired (Final velocity) = 12 m/s

Distance travelled = 720 meter

Explanation:

Given:

Uniform acceleration = 0.1 m/s²

Initial velocity = 0 m/s

Time taken = 2 minutes = 2 x 60 = 120 second

Find:

(a)  Speed acquired (Final velocity)

(b) Distance travelled

Computation:

v = u + at

v = 0 + 0.1(120)

Speed acquired (Final velocity) = 12 m/s

Distance travelled = ut + (1/2)(a)(t²)

Distance travelled = (0)(120) + (1/2)(0.1)(120²)

Distance travelled = (1/2)(0.1)(14400)

Distance travelled = 720 meter

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3 years ago
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