Which scientist determined that electrons had predicted zones orbiting the nucleus?

A cart, which has a mass of m - 2.50 kg, is sitting at the top of an inclined plane which is 3.30 meters long and which meets th

SashulF [63]

Answer:

Part(i) the time taken for this cart to reach the bottom of the inclined plane is 1.457 s

Part(ii) the velocity of the cart when it reaches the bottom of the inclined plane is 4.531 m/s

Part(iii) the kinetic energy of the cart when it reaches the bottom of the inclined plane is 25.663 J

Explanation:

Given;

mass of the cart = 2.5 kg

angle of inclination, β = 18.5⁰

length of inclined plane = 3.3m

Part(i) the time taken for this cart to reach the bottom of the inclined plane

s = ut + ¹/₂×at²

initial vertical velocity, u = 0

s = 3.3 m

s = ¹/₂×at²

$t = \sqrt{\frac{2s}{a} }$

acceleration, of the cart, a = gsinβ

a = 9.8sin(18.5) = 3.11 m/s²

$t = \sqrt{\frac{2X3.3}{3.11 }}= 1.457 s$

Part(ii) the velocity of the cart when it reaches the bottom of the inclined plane

V = a×t

V = 3.11 × 1.457 = 4.531 m/s

Part(iii) the kinetic energy of the cart when it reaches the bottom of the inclined plane

KE = ¹/₂MV²

= ¹/₂ × 2.5× (4.531)²

= 25.663 J

6 0

4 years ago

To practice Tactics Box 9.1 Calculating the Work Done by a Constant Force. Recall that the work W done by a constant force F⃗ at

insens350 [35]

Answer:

The vector magnitudes F and r are always postive, so the sign o W is determined entirely by the angle e between the force and the displacement.Submit Figure 1 off 1 part C

3 0

3 years ago

Read 2 more answers

What is the name given to group 18 on the periodic table?

zhannawk [14.2K]

Answer:

Noble gases

also known as group 0

7 0

3 years ago

Relate the output of energy from a heat engine to the energy put into the heat engine considering the second law of thermodynami

Vinil7 [7]

Answer:

Considering the II law of thermodynamics

From the figure

Out put of energy:

Heat supplied from the source/ reservoir (Q₁) - Heat rejected to the surroundings from the system (Q) = Q₁ - Q₂. Also known as Net work done on the system.

Input of energy:

Amount of heat energy supplied to the system from the source (Q₁ ).

Efficiency (H.E) = η = Output÷ Input

η = (Q₁ - Q₂) ÷ Q₁

OR η = Wnet ÷ Q₁ ; since Wnet = (Q₁ - Q₂)

3 0

3 years ago

A baseball is thrown directly upward from ground level with a velocity of +15 m/s. What are the two times when the ball is 10 m

erastovalidia [21]

Answer:

time is 0.5660 s

and time is - 3.62431 s

Explanation:

velocity u = 15 m/s

height s = 10 m

acceleration due to gravity g = –9.8 m/s²

to find out

time

solution

we will apply here distance equation that is

s = ut - 1/2× gt² ...........1

here put all these value and get time t

here s is height and g is -9.8

so

s = ut - 1/2× gt²

10 = 15t - 1/2× (-9.8)t²

10 = 15t + 4.9t²

solve it we get t

t = 0.56630 and -3.62431

so time is 0.5660 s

and time is - 3.62431 s

8 0

4 years ago