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kondaur [170]
3 years ago
12

A sailboat is moving across the water at 3.0 m/s. A gust of wind fills its sails and it accelerates at a constant 2.0 m/s2. At t

he same instant, a motorboat at rest starts its engines and accelerates at 4.0 m/s2. After 3.0 seconds have elapsed, find the velocity of the sailboat.
Physics
1 answer:
Flauer [41]3 years ago
5 0

Answer:

v = 9 m/s

Explanation:

It is given that,

Initial speed of the sailboat, u = 3 m/s

Acceleration of the sailboat, a=2\ m/s^2

Initial speed of the motorboat, u = 0

Acceleration of the motorboat, a=4\ m/s^2

Time elapsed, t = 3 s

To find,

The velocity of the sailboat

Solve,

Let v is the velocity of the sailboat after 3 seconds. By using the equation of kinematics, it can be calculated.

v=u+at

v=3\ m/s+2\ m/s^2\times 3\ s

v = 9 m/s

Therefore, the velocity of the sailboat is 9 m/s.

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Explanation:

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3 years ago
Use the galvanometers to determine the amount and direction of the induced current. Which galvanometer is
docker41 [41]

Answer:

Option B

Explanation:

Looking at the 3 galvanometer readings given above, for galvanometer A, the reading is -2 mA.

For galvanometer B, the reading is 4 mA.

While for galvanometer C, the reading is -5 MA

Thus, option B is correct.

4 0
3 years ago
does increasing the frequency of a wave also increase its wavelength if not how are these quantities related​
Aleks [24]

Answer: Increasing the frequency does not increase the wavelength. They are inversely related.

Explanation:

As wavelength increases, frequency decreases. If you look at a transverse wave and it has a long wavelength, there only a few waves produce. Which means there is less frequency produced. So as wavelength increases, frequency decreases. The other way around can work to. As frequency increases, wavelength decreases. They are inversely related.

8 0
3 years ago
The drawing shows 6 point charges arranged in a rectangle. The value of q is 2.83 uC and the distance d is 0.123 m. Find the tot
vova2212 [387]

the total electric potential at location P, which is at the center of the rectangle is 0V.

The charges placed at the corner of the rectangle are same in magnitude but different in charge. hence the total electric potential will be same in  magnitude but different in charge and will be cancelled. Similarly, all the total electric potential will be cancelled and resultant will be zero.

<h3>What is total electric potential?</h3>
  • The amount of labor required to convey a unit of electric charge from a reference point to a given place in an electric field is known as the electric potential (also known as the electric field potential, potential drop, or the electrostatic potential).
  • More specifically, it is the energy per unit charge for a test charge that is negligibly disruptive to the field under discussion. In order to prevent the test charge from gaining kinetic energy or radiating, the travel across the field is also meant to occur with very little acceleration.
  • The electric potential at the reference location is, by definition, zero units. Any point may be used as the reference point, but typically it is earth or a point at infinity.

To learn more about total electric potential with the given link

brainly.com/question/14776328

#SPJ4

3 0
2 years ago
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
3 years ago
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