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tresset_1 [31]
3 years ago
14

Which piece of equipment would give the MOST accurate measurement of 45 mL of a liquid? A) A 100 mL graduated cylinder B) A 50 m

L graduated cylinder C) A 100 mL beaker D) A 50 mL flask
Physics
2 answers:
user100 [1]3 years ago
5 0

Its b for sure


Explanation : because it has graduations for max. 50ml



Plese mark as brainliest


Ludmilka [50]3 years ago
3 0

b is the answer i think


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A diver who has a mass of 68 kg climbs to a diving platform that is 7.5 m above the surface of a pool. How much gravitational po
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The flow of electrons through a circuit is measured in which of the following units? A. electrical pressure B. amperes C. volts
damaskus [11]

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quantity of charge, measured in Coulombs, or in Ampere-seconds.

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8 0
3 years ago
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

5 0
3 years ago
This is “Fusion Reactions”.<br> Please answer number 8. Thank you.
Angelina_Jolie [31]

Answer:

²₁H + ³₂He —> ⁴₂He + ¹₁H

Explanation:

From the question given above,

²₁H + ³₂He —> __ + ¹₁H

Let ⁿₐX be the unknown.

Thus the equation becomes:

²₁H + ³₂He —> ⁿₐX + ¹₁H

We shall determine, n, a and X. This can be obtained as follow:

For n:

2 + 3 = n + 1

5 = n + 1

Collect like terms

n = 5 – 1

n = 4

For a:

1 + 2 = a + 1

3 = a + 1

Collect like terms

a = 3 – 1

a = 2

For X:

n = 4

a = 2

X =?

ⁿₐX => ⁴₂X => ⁴₂He

Thus, the balanced equation is

²₁H + ³₂He —> ⁴₂He + ¹₁H

8 0
3 years ago
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