The geese fly 23 m/s to the south when they migrate for the winter. Identify if this example is speed or velocity.

A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac

RoseWind [281]

Answer:

a) $\Delta U_{g} = 12.945\,J$ , b) $\Delta U_{k} = 12.945\,J$ , c) $k = 2930.059\,\frac{N}{m}$

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

$\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)$

$\Delta U_{g} = 12.945\,J$

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

$\Delta U_{k} = 12.945\,J$

c) The spring constant of the gun is:

$\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}$

$k = \frac{2\cdot \Delta U_{k}}{x^{2}}$

$k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}$

$k = 2930.059\,\frac{N}{m}$

4 0

2 years ago

A quantity that has direction as well as size is a scalar. True or False

lisov135 [29]

Answer:

A quantity that does not depend on the direction is called a scalar quantity. Vector quantities have two characteristics, a magnitude, and a direction. Scalar quantities have only a magnitude. When comparing two vector quantities of the same type, you have to compare both the magnitude and the direction.

Scalar quantities only have magnitude (size). Scalar quantities include distance...

A quantity that is specified by both size and direction is a vector. Displacement includes both size and direction and is an example of a vector. However, distance is a physical quantity that does not include a direction and isn't a vector.

Explanation:

hope this helps...

3 0

3 years ago

Read 2 more answers

PLEASE ANYONE SOLVE THIS NOW FAST PLEASE IM IN A HURRY

m_a_m_a [10]

Hello,

Solution for A:

Force = 3.00N

Mass = 0.50 Kgs

Time = 1.50 Seconds

According to newton's second law of motion;

Force = Mass times Acceleration(a)

3.00 = 0.50 * a

a = 3.00/0.50 = 6.00 m/s^2

We know that acceleration = Velocity / time

So Velocity = time * acceleration = 1.50 * 6 = 9.00 m/s^2

Solution for B:

The net force = 4.00N - 3.00N = 1.00N to the left

Force = 1.00N

Mass = 0.50Kg

Time = 3.00 Seconds

Again; F = MA (Where F is force, M is mass and A is acceleration)

1.00N = 0.5 * A

A = 1/0.5 = 2 m/s^2

Velocity = Acceleration * Time = 2 * 3 = 6 m/s

3 0

3 years ago

Calculate the standard electrode potential difference (e°) of the daniell cell (at 1 bar) if temperature is 473.15 k.

anzhelika [568]

Missing data in the text of the exercise: The molar concentration of Zinc is 10 times the molar concentration of copper.

Solution:

1) First of all, let's calculate the standard electrode potential difference at standard temperature. This is given by:
$E^0=E_{cat}^0-E_{an}^0$
where $E_{cat}^0$ is the standard potential at the cathode, while $E_{an}^0$ is the standard potential at the anode. For a Daniel Cell, at the cathode we have copper: $E_{Cu}^0=+0.34 V$ , while at the anode we have zinc: $E_{Zn}^0=-0.76 V$ . Therefore, at standard temperature the electrode potential difference of the Daniel Cell is
$E^0=+0.34 V-(-0.76 V)=+1.1 V$

2) To calculate $E^0$ at any temperature T, we should use Nerst equation:
$E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}$
where
$R=8.31 J/(K mol)$
$T=473.15 K$ is the temperature in our problem
$z=2$ is the number of electrons transferred in the cell's reaction
$F=9.65\cdot 10^4 C/mol$ is the Faraday's constant
$[Zn]$ and $[Cu]$ are the molar concentrations of zinc and in copper, and in our problem we have $[Zn]=10[Cu]$ .
Using all these data inside the equation, and using $E^0=+1.1 V$ , in the end we find:
$E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}=+1.053 V$

8 0

3 years ago

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 6.0 rev/s in 11.0 s. At th

sleet_krkn [62]

Answer

given,

ω₁ = 0 rev/s

ω₂ = 6 rev/s

t = 11 s

Using equation of rotational motion

The angular acceleration is

ωf - ωi = α t

11 α = 6 - 0

= 0.545 rev/s²

The angular displacement

θ₁= ωi t + (1/2) α t²

θ₁= 0 + (1/2) (0.545)(11)^2

θ₁= 33 rev

case 2

ω₁ = 6 rev/s

ω₂ = 0 rev/s

t = 14 s

Using equation of rotational motion

The angular acceleration is

ωf - ωi = α t

14 α = 0 - 6

= - 0.428 rev/s²

The angular displacement

θ₂= ωi t + (1/2) α t²

θ₂= 6 x 14 + (1/2) (-0.428)(14)^2

θ₂= 42 rev

total revolution in 25 s is equal to

θ = θ₁ + θ₂

θ = 33 + 42

θ = 75 rev

3 0

3 years ago