What is your operational definition for a fast reaction time?

dalvyx [7]

) (2.68 x 10¯5) x (4.40 x 10¯8)

The calculator display gives 1.1792 x 10¯12. Rounded off to three significant figures gives 1.18 x 10¯12 as the answer.

2) (2.95 x 107) ÷ (6.28 x 1015)

The calculator display gives 4.6975 x 10¯9. Rounded off to three significant figures gives 4.70 x 10¯9 as the answer.

3) (8.41 x 106) x (5.02 x 1012)

The calculator display gives 4.2218 x 1019. Rounded off to three significant figures gives 4.22 x 1019 as the answer.

When done as a division, the answer to this problem is 1.68 x 10¯6.

4) (9.21 x 10¯4) ÷ (7.60 x 105)

The calculator display gives 1.2118 x 10¯9. Rounded off to three significant figures gives 1.21 x 10¯9 as the answer.

1) (2.68 x 10¯5) x (4.40 x 10¯8)

The calculator display gives 1.1792 x 10¯12. Rounded off to three significant figures gives 1.18 x 10¯12 as the answer.

2) (2.95 x 107) ÷ (6.28 x 1015)

The calculator display gives 4.6975 x 10¯9. Rounded off to three significant figures gives 4.70 x 10¯9 as the answer.

3) (8.41 x 106) x (5.02 x 1012)

The calculator display gives 4.2218 x 1019. Rounded off to three significant figures gives 4.22 x 1019 as the answer.

When done as a division, the answer to this problem is 1.68 x 10¯6.

4) (9.21 x 10¯4) ÷ (7.60 x 105)

The calculator display gives 1.2118 x 10¯9. Rounded off to three significant figures gives 1.21 x 10¯9 as the answer

7 0

2 years ago

With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you releas

xxTIMURxx [149]

Answer:

The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

Let the initial velocity be 'v'.

Given:

Mass of the ball (m) = 130 g = 0.130 kg [ 1 g = 0.001 kg]

Initial height of the ball (h₁) = 1.4 mm = 0.0014 m [ 1 mm = 0.001 m]

Final height of the ball (h₂) = 15 mm = 0.015 m

Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

$\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity$

As it just touches the 15 mm high roof, the final velocity will be zero. So,

$v_f=0\ m/s$ .

Now, the change in kinetic energy is equal to:

$\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2$

Change in gravitational potential energy = Final PE - Initial PE

So,

$\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J$ [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

$0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s$

Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

4 0

2 years ago

If a substance is found to be reactive flammable soluble and explosive what observation is also a physical property

Ainat [17]

Answer:

soluble

Explanation:

Chemical properties only manifest when a chemical reaction occurs. Being reactive, flammable and explosive are chemical properties, because they involve chemical reactions: the substances are changed; the chemical bonds of some substances, called reactants, are broken, and the chemical bonds are created, forming other substances, called products.

Solubility is a physical property because during dissolution no new substances are formed. You can prove it when the solvent evaporates leaving behind the same original substance.

The the observation that the substance is soluble is describing a physical property.

3 0

3 years ago

Physics question, answer completely with work

Alenkasestr [34]

The force is 2.0 N east

Explanation:

The impulse exerted by a force is defined as the product between the force itself and the time interval during which the force is applied. Mathematically, it is equal to the change in momentum experienced by the object on which the force is acting:

$I=F\Delta t = \Delta p$