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3 years ago

Anaya Goodloe decides to spike the football after a big touchdown (on Earth) and it

Physics

1 answer:

Llana [10]3 years ago

6 0

Answer:

The football leaves with the velocity, u = 15.68 m/s

Explanation:

Given data,

The football bounces back up off the ground and is airborne for, t = 3.2 s

Let the football bounces back up off the ground in the vertical direction

The formula for time of flight is given by,

t = 2u /g

∴ u = gt / 2

Substituting the values,

u = 9.8 x 3.2 / 2

u = 15.68 m/s

Hence, the football leaves with the velocity, u = 15.68 m/s

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Two cubes, one silver and one iron, have the same mass and temperature. A quantity Q of heat is removed from each cube. Which on

scoundrel [369]

Answer:

C) specific heat capacity

Explanation:

As we know that heat given to the system to change the temperature is given as

$Q = ms \Delta T$

here we know that

Q = thermal energy given to the system

s = specific heat capacity of the system

$\Delta T$ = change in temperature

so here we know that

$\Delta T = \frac{Q}{ms}$

here we know that heat given to the two cubes is of same amount as well as the mass is same but the final temperature is different because of the specific heat capacity (s)

So correct answer would be

C) specific heat capacity

8 0

3 years ago

Which of the following is true of microscopic particles? They can be localized or not localized in space They can explain the ph

Alona [7]

Answer:

true about microscopic particle is that they can explain the phenomena of diffraction and interference.

Explanation

the first thing mentioned in question particles are localized in space according to classical physics particle can be well localized in space because its momentum and position can be determine simultaneously .second their properties can be measured and they can act like wave particle by particle build up a wave. so all of they things mention above only thing true about particles is that they describe the phenomena of diffraction and interference.

5 0

2 years ago

When two point charges are a distance d apart, the magnitude of the electrostatic force between them is F. If the distance betwe

EastWind [94]

Answer:

The magnitude of the electrostatic force decreases by a factor 9

Explanation:

The electrostatic force between two charges is given by:

$F=k\frac{q_1 q_2}{d^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

d is the distance between the two charges

We see that the magnitude of the force is inversely proportional to the square of the distance. If the distance is increased to 3d: d' = 3d, the new electrostatic force would be:

$F'=k\frac{q_1 q_2}{(d')^2}=k\frac{q_1 q_2}{(3d)^2}=\frac{1}{9} k\frac{q_1 q_2}{d^2}=\frac{F}{9}$

So, the electrostatic force decreases by a factor 9.

3 0

2 years ago

Read 2 more answers

Particle A of charge 2.70 10-4 C is at the origin, particle B of charge -6.36 10-4 C is at (4.00 m, 0), and particle C of charge

Serhud [2]

Answer:

FC vector representation

$F_{C} =(19.03i+13.78j)N$

Magnitude of FC

$F_{C}=23.495N$

Vector direction FC

$\beta=35.91$ degrees: angle that forms FC with the horizontal

Explanation:

Conceptual analysis

Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.

The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.

$\alpha = tan^{-1}(\frac{3}{4}) = 36.86$ degrees

To calculate the magnitudes of the forces we apply Coulomb's law:

$F_{AC} = \frac{k*q_{A}*q_{C}}{r_{AC}^2}$ Equation (1): Magnitude of the electric force of the charge qA over the charge qC

$F_{BC} = \frac{k*q_{B}*q_{C}}{r_{BC}^2}$ Equation (2) : Magnitude of the electric force of the charge qB over the charge qC

Known data

$k=8.99*10^9 \frac{N*m^2}{C^2}$

$q_{A}=2.70*10^{-4} C$

$q_{B}=-6.36*10^{-4} C$

$q_{C}=1.04*10^{-4} C$

$r_{AC} =3$

$r_{BC}=\sqrt{4^2+3^2} = 5$

Problem development

In the equations (1) and (2) to calculate FAC Y FBC:

$F_{AC} =8.99*10^9*(2.70*10^{-4}* 1.04*10^{-4})/(3)^2=28.05N$

$F_{BC} =8.99*10^9*(6.36*10^{-4}* 1.04*10^{-4})/(5)^2=23.785N$

Components of the FBC force at x and y:

$F_{BCx}=23.785 *Cos(36.86)=19.03N$

$F_{BCy}=23.785 *Sin(36.86)=14.27N$

Components of the resulting force acting on qC:

$F_{Cx} = F_{ACx}+ F_{BCx}=0+19.03=19.03N$

$F_{Cy} = F_{ACy}+ F_{BCy}=28.05-14.27=13.78N$

FC vector representation

$F_{C} =(19.03i+13.78j)N$

Magnitude of FC

$F_{C}= \sqrt{19.03^2+13.78^2} =23.495N$

Vector direction FC

$\beta = tan^{-1} (\frac{13.78}{19.03})=35.91$ degrees: angle that forms FC with the horizontal

7 0

3 years ago

3. A ball is thrown straight up with an initial velocity of 40 m/s.

grandymaker [24]

Answer:

The velocity at the top of its path will be zero (0)

Explanation:

We can solve this problem or particular situation using the principle of energy conservation.

Which tells us that energy is transformed from kinetic energy to potential energy and vice versa. A reference point should be considered at which the potential energy is zero, and at this point the initial velocity of 40 [m/s] is printed to the ball.

$Ek=Ep\\where:\\Ek=kinetic energy [J]\\Ep=potencial energy [J]$

The potential energy is determined by:

$Ep=m*g*h\\where:\\m=mass of the ball[kg}\\g=gravity[m/s^2]\\h=heigth [m]\\$

The kinetic energy is determined by:

$Ek=\frac{1}{2}*m*v_{0} ^{2} \\where\\v_{0} = initial velocity[m/s]$

$Ek=Ep\\\frac{1}{2} *m*v_{0} ^{2} =m*9.81*h\\h=\frac{40^{2}}{2*9.81} \\h=81.5[m]$

This will be the maximum path but, its velocity at this point will be zero. Because now all the kinetic energy has been transformed in potential energy.

8 0

3 years ago