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3 years ago

Can work ever be negative?

Physics

1 answer:

Rom4ik [11]3 years ago

7 0

Answer:

B yes when displacement is negative

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Gold-leaf electroscope uses

Kruka [31]

-identifies an electric charge
-it can identify its polarity (positive or negative) if you compare it to a charge that you already know
-can identify the magnitude of a charge (how big of a charge it is)

7 0

3 years ago

2 (a) A plane from airport A flies 280 km to the east to airport B. The plane then travelled north to airport C, 190 km away. (i

Mekhanik [1.2K]

Answer:

See below

Explanation:

See attached diagram

280 km east then 190 km north

Use Pythagorean theorem to find the resultant displacement

d^2 = 280^2 + 190^2

d = 338.4 km

Angle will be arctan ( 190/280) = 34.16 °

4 0

2 years ago

A sample with a path length of 1 cm absorbs 99.0% of the incident light at a wavelength of 274 nm, measured with respect to an a

Ksju [112]

Answer:

17. NADH has a molar extinction coefficient of 6200 M2 cm at 340 nm. Calculate the molar concentration of NADH required to obtain an absorbance of 0.1 at 340 nm in a 1-cm path length cuvette. 18. A sample with a path length of 1 cm absorbs 99.0% of the incident light at a wavelength of 274 nm, measured with respect to an appropriate solvent blank. Tyrosine is known to be the only chromophore present in the sample that has significant absorption at 274 nm. Calculate the molar concentration of tyrosine in the sample.

Explanation:

8 0

3 years ago

Jenny loves puffed rice cereal. As she is pouring the cereal, two little

Ede4ka [16]

Answer:

4.9 x 10-23 N

Explanation:

I took the test

7 0

3 years ago

Read 2 more answers

Determine the velocity that a car should have while traveling around a frictionless curve of radius 100m and that is banked 20 d

alex41 [277]

Answer:

$v=18.89\frac{m}{s}$

Explanation:

From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:

$\sum F_x:Nsin(20^\circ)=ma_c(1)\\\sum F_y:Ncos(20^\circ)=mg(2)$

Solving N from (2) and replacing in (1):

$N=\frac{mg}{cos(20^\circ)}\\(\frac{mg}{cos(20^\circ)})sin(20^\circ)=ma_c\\gtan(20^\circ)=a_c$

The centripetal acceleration is given by:

$a_c=\frac{v^2}{r}$

Replacing and solving for v:

$gtan(20^\circ)=\frac{v^2}{r}\\v=\sqrt{grtan(20^\circ)}\\v=\sqrt{9.8\frac{m}{s^2}(100m)tan(20^\circ)}\\v=18.89\frac{m}{s}$

4 0

3 years ago