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3 years ago

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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

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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The mass of a sports car is 1000 kg. The shape of the car is such that the aerodynamic drag coefficient is 0.260 and the frontal

alisha [4.7K]

Answer:

The initial acceleration is 0,221 m/s^2.

Explanation:

$Fcar = Fairfriction\\m*a = 0.5*A*c*density*v^2\\a = 0.5*A*c*density*v^2 / 1000 kg = 0.5*2.10m^2*0.260*1.295kg/m^3*(25 m/s)^2 / 1000 kg = 0,221 m/s^2$

8 0

3 years ago

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The velocity function (in meters per second) is given for a particle moving along a line. v(t) = 5t − 9, 0 ≤ t ≤ 3 (a) Find the

Reil [10]

Answer:

a) The displacement is -4.5 m.

b) The traveled distance is 11.7 m.

Explanation:

Hi there!

a)The velocity of the particle is the derivative of the displacement function, x(t):

v(t) = dx/dt = 5t - 9

Separating varibles:

dx = (5t - 9)dt

Integrating both sides from x = x0 to x and from t = 0 to t.

x - x0 = 1/2 · 5t² - 9t

x = 1/2 · 5t² - 9t + x0

If we place the origin of the system of reference at x = x0, the displacement at t = 3 will be x(3):

x(3) = 1/2 · 5 · (3)² - 9(3) + 0

x(3) = -4.5

The displacement at t = 3 s is -4.5 m. It means that the particle is located 4.5 m to the left from the origin of the system of reference at t = 3 s.

b) When the velocity is negative, the particle moves to the left. Let´s find the time at which the velocity is less than zero:

v = 5t - 9

0 > 5t - 9

9/5 > t

1.8 s > t

Then until t = 1.8 s, the particle moves to the left from the origin of the reference system.

Let´s find the position of the particle at that time:

x = 1/2 · 5t² - 9t

x = 1/2 · 5(1.8 s)² - 9(1.8 s)

x = -8.1 m

From t = 0 to t = 1.8 s the traveled distance is 8.1 m. After 1.8 s, the particle has positive velocity. It means that the particle is moving to the right, towards the origin. If at t = 3 the position of the particle is -4.5 m, then the traveled distance from x = -8.1 m to x = -4.5 m is (8.1 m - 4.5 m) 3.6 m.

Then, the total traveled distance is (8.1 m + 3.6 m) 11.7 m.

4 0

3 years ago

Work done= ________ transferred

mamaluj [8]

Answer:

Work done= Energy transferred

Explanation:

Work is the transfer of energy. In physics we say that work is done on an object when you transfer energy to that object. If you put energy into an object, then you do work on that object (mass).

4 0

3 years ago

A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) Ho

Crank

Answer:

a)N = 3.125 * 10¹¹

b) I(avg) = 2.5 × 10⁻⁵A

c)P(avg) = 1250W

d)P = 2.5 × 10⁷W

Explanation:

Given that,

pulse current is 0.50 A

duration of pulse Δt = 0.1 × 10⁻⁶s

a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles

N = Δq/e

charge is given by Δq = IΔt

so,

N = IΔt / e

$N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1$

N = 3.125 * 10¹¹

b) Q = nqt

where q is the charge of 1puse

n = number of pulse

the average current is given as I(avg) = Q/t

I(avg) = nq

I(avg) = nIΔt

= (500)(0.5)(0.1 × 10⁻⁶)

= 2.5 × 10⁻⁵A

C) If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,

eV = K

V = K/e

the power is given by

P = IV

P(avg) = I(avg)K / e

$P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}$

= 1250W

d) Final peak=

P= Ik/e

= $= P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W$

P = 2.5 × 10⁷W

5 0

3 years ago

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Based on the velocity-time graph given, the acceleration of the object is..

fredd [130]

It’s gonna have to b since it’s decreasing

4 0

3 years ago