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3 years ago

28.25 mL, three signicant digits

2 answers:

3 0

Answer:3) variable affinities (stickiness) for something it is running past. Physical ... -measurement number (significant digits) unit (such as inches) -Significant ... Mass 1 oz. 28.25 g. Relations Between English and Metric Units Mass 1 dram. 1.772 g ... -graduated cylinder has an error of about 1% (± 0.1 mL in 10 mL). -Volumetric

Explanation:

Nutka1998 [239]3 years ago

3 0

Answer:

1. 28.25 mL, three significant digits.

2. 54.074 mL, three significant digits

3. 600.006 km, four significant digits

4. 1356 kg + 4.2 kg + 19.891 kg

Explanation:

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Which Of The Following Research Methods Are Widely Used By Psychologists?

goldenfox [79]

Answer: As you said D.

Explanation:

4 0

3 years ago

A cyclist going downhill is accelerating at 1. 2 m/s2. If the final velocity of the cyclist is 16 m/s after 10 seconds, what is

mel-nik [20]

Answer:

$\boxed {\boxed {\sf v_i= 4 \ m/s}}$

Explanation:

We are asked to find the cyclist's initial velocity. We are given the acceleration, final velocity, and time, so we will use the following kinematic equation.

$v_f= v_i + at$

The cyclist is acceleration at 1.2 meters per second squared. After 10 seconds, the velocity is 16 meters per second.

$v_f$ = 16 m/s
a= 1.2 m/s²
t= 10 s

Substitute the values into the formula.

$16 \ m/s = v_i + (1.2 \ m/s^2)(10 \ s)$

Multiply.

$16 \ m/s = v_i + (1.2 \ m/s^2 * 10 \ s)$

$16 \ m/s = v_i + 12 \ m/s$

We are solving for the initial velocity, so we must isolate the variable $v_i$ . Subtract 12 meters per second from both sides of the equation.

$16 \ m/s - 12 \ m/s = v_i + 12 \ m/s -12 \ m/s$

$4 \ m/s = v_i$

The cyclist's initial velocity is <u>4 meters per second.</u>

6 0

2 years ago

What type of energy does the box have after it is done being pulled?

Ksivusya [100]

Assuming that the box is moving when it is being pulled, Work is done on the box.

So work is the Force times the distance

W=Fd

But what is work actually ? When something moves due to force over some change in distance, it have energy.

But where does this energy come from ? Does it magically appear ? The energy comes from the applied force onto the box.

So the energy have been transferred. And it’s like that throughout the universe

Now to save time, I’ll just tell you the answer: kinetic energy

:)

3 0

3 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$