Can somebody help please !<br><br> a. -8.3 m/s<br> b.-4.2 m/s<br> c.-0.12 m/s<br> d. 0 m/s

if you have a block of steel with a density of 7.8g/cm3 and a block of gold with a density of 19.3g/cm3 each with the same volum

natita [175]

Gold because 19.3 is greater than 7.8

3 0

3 years ago

13) Fred is travelling 3.4 m/s and decelerates at a rate of 0.83 m/s until he comes to a stop

Finger [1]

Answer:

<h2>Tt will take 4.04 seconds</h2>

Explanation:

In this problem/exercise, we are going to apply the newtons first equation of motion to solve for the time taken

Given that

final velocity= 3.4m/s

initia velocity u= 0m/s

deceleration= 0.84 m/s^2

time t= ?

applying

v=u+at

Substituting our given data into the expression above we have

3.4=0+0.84t

3.4=0.84t

divide both sides by 0.84 we have

3.4/0.84= t

t= 4.04 seconds

3 0

3 years ago

An external force of 25 N acts on a system for 10 s. What is the impulse delivered to the system?

Ipatiy [6.2K]

Answer:

I = 250N*s

Explanation:

I = F*t

I = impulse or momentum

F = Force

t = time

I = 25N*10s

I = 250N*s

7 0

3 years ago

A golfer hits a shot to a green that is elevated 3.20 m above the point where the ball is struck. The ball leaves the club at a

dem82 [27]

Answer:

16.17 m/s

Explanation:

h = 3.2 m

u = 18.1 m/s

Angle of projection, θ = 49°

Let H be the maximum height reached by the ball.

The formula for the maximum height is given by

$H=\frac{u^{2}Sin^{2}\theta }{2g}$

$H=\frac{18.1^{2}\times Sin^{2}49 }{2\times 9.8}=9.52 m$

The vertical distance fall down by the ball, h' H - h = 9.52 - 3.2 = 6.32 m

Let v be the velocity of ball with which it strikes the ground.

Use third equation of motion for vertical direction

$v_{y}^{2}=u_{y}^{2}+2gh'$

here, uy = 0

So,

$v_{y}^{2}=2\times 9.8 \times 6.32$

vy = 11.13 m/s

vx = u Cos 49 = 18.1 x 0.656 = 11.87 m/s

The resultant velocity is given by

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}$

$v=\sqrt{11.87^{2}+11.13^{2}}$

v = 16.27 m/s

7 0

4 years ago

What is the physical meaning of Schrodinger's wave function? (b). Calculate the minst three energy levels of an electron in an i

KonstantinChe [14]

Explanation:

(a) The Schrodinger's wave function represent the position of a particle at a particular instant of time. It is also known as the probability amplitude. It is also used to find the location of a particle.

(b) The width of a potential well, $l=5\ A=5\times 10^{10}\ m$

For first energy level, n = 1

Energy in infinite potential well is given by :

$E=\dfrac{n^2h^2}{8ml}$

$E=\dfrac{(1)^2\times (6.63\times 10^{-34})^2}{8\times 9.1\times 10^{-31}\times 5\times 10^{10}}$

E = 0.0120 Joules

For second energy level, n = 2

$E=\dfrac{n^2h^2}{8ml}$

$E=\dfrac{(2)^2\times (6.63\times 10^{-34})^2}{8\times 9.1\times 10^{-31}\times 5\times 10^{10}}$

E = 0.0483 Joules

For third energy level, n = 3

$E=\dfrac{n^2h^2}{8ml}$

$E=\dfrac{(3)^2\times (6.63\times 10^{-34})^2}{8\times 9.1\times 10^{-31}\times 5\times 10^{10}}$

E = 0.108 Joules

Hence, this is the required solution.

7 0

3 years ago

Can somebody help please ! a. -8.3 m/s b.-4.2 m/s c.-0.12 m/s d. 0 m/s