Which statement shows how to correctly convert from the mass of a compound in grams to the amount of that compound in moles?

The SI unit for energy is Newton true or false

ipn [44]

Explanation:

false....it is joules

..........

6 0

2 years ago

A ball is thrown from a rooftop with an initial downward velocity of magnitude vo = 2.9 m/s. The rooftop is a distance above the

Step2247 [10]

Answer:

a) The velocity of the ball when it hits the ground is -20.5 m/s.

b) To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

Explanation:

I´ve found the complete question on the web:

A ball is thrown from a rooftop with an initial downward velocity of magnitude v0=2.9 m/s. The rooftop is a distance above the ground, h= 21 m. In this problem use a coordinate system in which upwards is positive.

(a) Find the vertical component of the velocity with which the ball hits the ground.

(b) If we wanted the ball's final speed to be exactly 27, 3 m/s from what height, h (in meters), would we need to throw it with the same initial velocity?

The equation of the height and velocity of the ball at any time "t" are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the ball at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the ball at a time "t".

First, let´s find the time it takes the ball to reach the ground (the time at which h = 0)

h = h0 + v0 · t + 1/2 · g · t²

0 = 21 m - 2.9 m/s · t - 1/2 · 9.8 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 1.8 s ( the other solution of the quadratic equation is rejected because it is negative).

Now, using the equation of velocity, let´s find the velocity of the ball at

t = 1.8 s:

v = v0 + g · t

v = -2.9 m/s - 9.8 m/s² · 1.8 s

v = -20.5 m/s

The velocity of the ball when it hits the ground is -20.5 m/s.

b) Now we have the final velocity and have to find the initial height. Using the equation of velocity we can obtain the time it takes the ball to acquire that velocity:

v = v0 + g · t

-27.3 m/s = -2.9 m/s - 9.8 m/s² · t

(-27.3 m/s + 2.9 m/s) / (-9.8 m/s²) = t

t = 2.5 s

The ball has to reach the ground in 2.5 s to acquire a velocity of 27.3 m/s.

Using the equation of height, we can obtain the initial height:

h = h0 + v0 · t + 1/2 · g · t²

0 = h0 -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

-h0 = -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

h0 = 38 m

To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

6 0

4 years ago

Cual sera el caudal que lleva un rio cuando se desplaza a 200 litros cada 40 segundos

alisha [4.7K]

Answer:

Q = 5 L/s

Explanation:

To find the flow you use the following formula (para calcular el caudal usted utiliza la siguiente formula):

$Q=\frac{V}{t}$

V: Volume (volumen) = 200L

t: time (tiempo) = 40 s

you replace the values of the parameters to calculate Q (usted reemplaza los valores de los parámteros V y t para calcular el caudal):

$Q=\frac{200L}{40s}=5\frac{L}{s}$

Hence, the flow is 5 L/s (por lo tanto, el caudal es de 5L/s)

4 0

3 years ago

The joule (J) is a unit of energy. Recall that energy may be converted between many different forms such as mechanical energy, t

REY [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The workdone is $W = -177.275J$

Explanation:

From the question we are told that

The initial Volume is $Vi = 0.160 L$

The final volume is $V_f = 0.510L$

The external pressure is $P = 5.00 \ atm$

Generally the change in volume is

$\Delta V = V_f - V_i$

Substituting values we have

$\Delta V = 0.510 -0.160$

$= 0.350L$

Generally workdone is mathematically represented as

$W = -P \Delta V$

W is negative because the working is done on the environment by the system which is indicated by volume increase

Substituting values

$W = - 5* 0.350$

$= -1.75 \ L \ \cdot atm$

Now $1 \ L \cdot atm = 101.3J$

Therefore $W = -1.75* 101.3$

$= -177.275J$

7 0

3 years ago

Read 2 more answers

A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart

serious [3.7K]

To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

$E_{initial} = E_{final}$

$PE_{initial}+KE_{initial} = PE_{final}+KE_{final}$

Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

$KE_{final} = (PE_{initial}+KE_{initial} )-PE_{final}$

$KE_{final} = (5100+4200)-5700$

$KE_{final} = 3600MJ$

Therefore the final kinetic energy is 3600MJ

5 0

3 years ago