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NARA [144]
3 years ago
14

PLS HELP THIS IS WORTH A LOT OF POINTS ON MY TEST. WILL GIVE BRAINLIEST

Physics
2 answers:
vova2212 [387]3 years ago
8 0
  • Initial velocity=0m/s
  • Acceleration=g=10m/s^2
  • Time=t=6s

  • Height be h

\\ \rm\Rrightarrow h=\dfrac{1}{2}gt^2

\\ \rm\Rrightarrow h=\dfrac{1}{2}(10)(6)^2

\\ \rm\Rrightarrow h=5(36)

\\ \rm\Rrightarrow h=180m

Ede4ka [16]3 years ago
7 0

Answer:

Explanation:

By conservation of energy, speed of the ball going up = speed of ball coming down with the ball stops at the top.

Because the gravity acceleration is constant, by symmetry, half of total time, 6/2 = 3s, is for going up and the last 3s for coming down.

Consider the last 3s when the ball drops from top to bottom, the initial velocity = 0 and acceleration = 10m/s^2

distance traveled = initial velocity * time + 1/2 * acceleration * time^2

= 0*3 + 1/2*10*3^2

= 5*9

= 45m

So maximum height of the ball is 45m.

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Se apunta un rifle horizontalmente con mira a un blanco pequeño que está a 200m en el suelo. La velocidad inicial de la bala es
vfiekz [6]

Answer:

Lo importante a tener en cuenta sobre esta pregunta es que la velocidad horizontal de la bala no hace ninguna diferencia en cuanto al tiempo que tarda en caer al suelo.

Debido a que el arma no ha aplicado ninguna fuerza vertical a la bala, la única fuerza que afecta la bala es la gravedad. Esto significa que la bala tarda tanto en caer al suelo como lo haría si se cayera, a pesar de que ahora viaja una gran distancia horizontal en la duración.

Para encontrar el tiempo de viaje antes de tocar el suelo, tenemos 3 valores:

-El desplazamiento desde el suelo que la bala debe viajar, s = 1.5m

-La aceleración que experimenta la bala. Como la gravedad está acelerando la bala hacia abajo, a = g = ~ 9.81m / s ^ 2

-La velocidad inicial de la bala verticalmente. Como la bala es estacionaria verticalmente (solo viaja horizontalmente al inicio), u = 0m

Examinamos nuestras ecuaciones de movimiento, comúnmente conocidas como ecuaciones SUVAT. Es posible que necesite aprender estos para su examen, pero algunas tablas de examen los proporcionan.

Debido a que tenemos s, u y a, y estamos buscando el tiempo t, la ecuación relevante es

s = ut + 0.5 (en ^ 2)

Completando nuestros valores tenemos:

1.5 = 0t + 0.5 (9.81 x t ^ 2)

1.5 = 4.905 x t ^ 2

Divide 1.5 entre 4.905 para encontrar t ^ 2

t ^ 2 = 0.3058 ...

Simplemente encontramos la raíz cuadrada de t ^ 2 para encontrar t, el tiempo que tarda la bala en llegar al suelo:

t = 0.553s (3 cifras significativas)

Para encontrar la distancia horizontal, d, que la bala ha viajado antes de tocar el suelo, podemos usar la ecuación que vincula el desplazamiento s con cierta velocidad v durante un tiempo t:

s = vt

La velocidad horizontal de la bala, v = 430

El tiempo antes de que la bala toque el suelo, t = 0.553

Entonces d = vt = 430 * 0.553 = 238m (3 cifras significativas)

3 0
4 years ago
A person with a mass of 40 kg is sitting on a box. What is the value of the normal force
VMariaS [17]

normal force=mass*gravitational force

normal force=40*0

normal force=40

8 0
3 years ago
a speed swimmer love to race around the parks pine which is 25m around if she can swim 20 laps in 7200s what is her average spee
aleksley [76]

Answer:

the average speed of the swimmer is 0.069 m/s.

Explanation:

Given;

complete distance around the park pine, d = 25 m

total lap completed, = 20 laps

time of laps completion, t = 7200 s

The total distance completed by the swimmer = 20 x 25 = 500 m

The average speed of the swimmer = distance / time

                                                            = (500 m) / (7200 s)

                                                            = 0.069 m/s.

Therefore, the average speed of the swimmer is 0.069 m/s.

8 0
3 years ago
In
son4ous [18]
(Direction) for the fact that it will continue having the momentum at the constant speed in which the engines turned off.
6 0
3 years ago
4. A trolley of mass 2kg rests next to a trolley of mass 3 kg on a flat
nydimaria [60]

Answer:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. The total momentum of the trolleys after separation is zero

c. The momentum of the 2 kg trolley after separation is 12 kg·m/s

d. The momentum of the 3 kg trolley is -12 kg·m/s

e. The velocity of the 3 kg trolley = -4 m/s

Explanation:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0

c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s

d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s

e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley

∴ The momentum of the 3 kg trolley = 3 kg × Velocity of the 3 kg trolley = -12 kg·m/s

The velocity of the 3 kg trolley = -12 kg·m/s/(3 kg) = -4 m/s

3 0
3 years ago
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