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uysha [10]
3 years ago
8

Question 5 (1 point)

Physics
1 answer:
katovenus [111]3 years ago
8 0

Answer:

The horizontal distance the ball travels is approximately 4.055 meters

Explanation:

The given parameters are;

The height from which the child kicks the ball = 3.5 m

The horizontal speed of the ball = 4.8 m/s

Therefore, we have;

The time it takes the ball to hit the ground is given by the relation;

h = u·t + 1/2·g·t²

Where;

u = The initial vertical velocity of the ball = 0 m/s

t = The time it takes the ball to hit the ground

g = The acceleration due to gravity = 9.81 m/s²

h = The height of the ball = 3.5 m

3.5 = 0 × t + 1/2 × 9.81 × t²

3.5 = 1/2 × 9.81 × t²

∴ t² = 3.5/(1/2 × 9.81)

∴ t = √(3.5/(1/2 × 9.81) = 0.8447 s

t ≈ 0.8447 s

The time the ball takes in flight = t ≈ 0.8447 s

Therefore;

The horizontal distance the ball travels = The horizontal velocity × The time of flight

∴ The horizontal distance the ball travels = 4.8 × 0.8447 ≈ 4.055

The horizontal distance the ball travels ≈ 4.055 meters.

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A horizontal force of magnitude 46.7 N pushes a block of mass 4.35 kg across a floor where the coefficient of kinetic friction i
exis [7]

(a) 138.2 J

Since the applied force is parallel to the displacement of the block, the work done by the force is given by:

W=Fd

where

F = 46.7 N is the magnitude of the force

d = 2.96 m is the displacement of the block

Substituting the numbers into the equation, we find

W=(46.7 N)(2.96 m)=138.2 J

(b) 45.1 J

In order to calculate the total energy dissipated among the floor and the block as thermal energy, we have to calculate the work done by the frictional force, which is

W_f = F_f d = (-\mu mg)d

where\mu=0.635 is the coefficient of friction

m = 4.35 kg is the mass of the block

g = 9.8 m/s^2

d = 2.96 m is the displacement

and the negative sign is due to the fact that the frictional force has opposite direction to the displacement.

Substituting, we find

W_f =-(0.635)(4.35 kg)(9.8 m/s^2)(2.96 m)=-80.1 J

The magnitude of this work is equal to the sum of the thermal energy dissipated between the floor and the block:

W_f = E_{floor}+E_{block}

And since we know

W_f = 80.1 J\\E_{floor}=35.0 J

we find

E_{floor}=W_f-E_{block}=80.1 J-35.0 J=45.1 J

(c) 58.1 J

According to the work-energy theorem, the increase in kinetic energy of the block must be equal to the work done by the applied force minus the work done by friction (which becomes wasted thermal energy):

\Delta K=W-W_f

Substituting

W=138.2 J\\W_f = 80.1 J

We find

\Delta K=138.2 J-80.1 J=58.1 J

7 0
3 years ago
Se colocan tres objetos, muy cerca uno del otro, dos al mismo tiempo. Cuando se juntan los objetos A y B, se repelen. Cuando se
docker41 [41]

Answer:

Los objetos A y C tienen cargas del mismo signo (opcion a)

Explanation:

Hay dos tipos de cargas : cargas positivas y cargas negativas.

La Ley de Coulomb dice que la fuerza electrostática entre dos cargas puntuales es proporcional al producto de las cargas e inversamente proporcional al cuadrado de la distancia que las separa, y tiene la dirección de la línea que las une y se cumple que:

  • La fuerza ejercida sobre una carga apunta hacia la otra cuando las dos tienen distinto signo (fuerza atractiva).
  • El sentido de la fuerza se dirige hacia el lado opuesto de la carga cuando ambas tienen el mismo signo (fuerza repulsiva).

Es decir que las cargas de igual signo se repelen, mientras que las de diferente signo se atraen.

Entonces, si se juntan los objetos A y B y se repelen significa que la carga es del mismo signo.

Cuando se acercan los objetos B y C, se repelen. Entonces significa que posee carga de igual signo.

Por lo que podes concluir que <u><em>los objetos A y C tienen cargas del mismo signo (opcion a)</em></u>

8 0
3 years ago
Which of the following is a substance that is found between the cell membrane and the nucleus, which primarily consists of water
Alex
This would be Cytoplasm , hope this helps 

Hugs

5 0
4 years ago
A +35 µC point charge is placed 46 cm from an identical +35 µC charge. How much work would be required to move a +0.50 µC test c
Ira Lisetskai [31]

Answer:

512.5 mJ

Explanation:

Let the two identical charges be q = +35 µC and distance between them be r₁ = 46 cm. A charge q' = +0.50 µC located mid-point between them is at r₂ = 46 cm/2 = 23 cm = 0.23 m.

The electric potential at this point due to the two charges q is thus

V = kq/r₂ + kq/r₂

= 2kq/r₂

= 2 × 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C/0.23 m

= 630/0.23  × 10³ V

= 2739.13 × 10³ V

= 2.739 MV

When the charge q' is moved 12 cm closer to either of the two charges, its distance from each charge is now r₃ = r₂ + 12 cm = 23 cm + 12 = 35 cm = 0.35 m and r₄ = r₂ - 12 cm = 23 cm - 12 cm = 11 cm = 0.11 cm.

So, the new electric potential at this point is

V' = kq/r₃ + kq/r₄

= kq(1/r₃ + 1/r₄)

= 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C(1/0.35 m + 1/0.11 m)

= 315 × 10³(2.857 + 9.091) V

= 315 × 10³ (11.948) V

= 3763.62 × 10³ V

= 3.764 MV

Now, the work done in moving the charge q' to the point 12 cm from either charge is

W = q'(V' - V)

= 0.5 × 10⁻⁶ C(3.764 MV - 2.739 MV)

= 0.5 × 10⁻⁶ C(1.025 × 10⁶) V

= 0.5125 J

= 512.5 mJ

8 0
3 years ago
Think about routine tasks that astronauts might need to do inside and outside a
xz_007 [3.2K]
<h3><u><em>Answer:</em></u></h3><h3><u><em>Ok, here we go...</em></u></h3><h3><u><em>Explanation:</em></u></h3><h3><u><em>Astronauts perform many tasks as they orbit Earth. The space station is designed to be a permanent orbiting research facility. Its major purpose is to perform world-class science and research that only a microgravity environment can provide. The station crew spends their day working on science experiments that require their input, as well as monitoring those that are controlled from the ground. They also take part in medical experiments to determine how well their bodies are adjusting to living in microgravity for long periods of time.</em></u></h3><h3><u><em></em></u></h3><h3><u><em>Working on the space station also means ensuring the maintenance and health of the orbiting platform. Crew members are constantly checking support systems and cleaning filters, updating computer equipment: doing many of the things homeowners must do to ensure their largest investment stays in good shape. Similarly, the Mission Control Center constantly monitors the space station and sends messages each day through voice or email with new instructions or plans to assist the crew members in their daily routines.</em></u></h3><h3><u><em></em></u></h3>

<u><em></em></u>

from the NASA website, COPYRIGHT Jun 8, 2015

7 0
3 years ago
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