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Leto [7]
2 years ago
9

Which one of the following statements is not true of free falling object

Physics
1 answer:
Leya [2.2K]2 years ago
7 0

Answer:

FORCE as for my answer....

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The ability to adjust to an artificially displaced or even inverted visual field is called
egoroff_w [7]

Answer:

méthode if séparation muddy water

3 0
2 years ago
Un avión vuela a una velocidad de 900 km/h. Si tarda en viajar desde Canarias hasta la península 180 s ¿qué distancia recorre en
wel

Answer:

El avión recorrió 45 km en los 180 s.

Explanation:

La relación entre velocidad, distancia y tiempo se da de la siguiente manera;

Velocidad= \dfrac{Distancia}{Hora}

Por lo cual los parámetros dados son los siguientes;

Velocidad = 900 km/h = 250 m / s

Tiempo = 180 s

Estamos obligados a calcular la distancia recorrida

De la ecuación para la velocidad dada arriba, tenemos;

Distancia recorrida = Velocidad pf viaje × Tiempo de viaje

Distancia recorrida = 900 km/h × 180 s = 900

Distancia recorrida = 900 km/h × 1 h/60 min × 1 min/60 s × 180 s = 45 km

Por lo tanto, el avión viajó 45 km en 180 s.

8 0
3 years ago
A single circular loop of wire of radius 0.75 m carries a constant current of 3.0 A. The loop may be rotated about an axis that
NISA [10]

Answer:

B = 0.8 T

Explanation:

It is given that,

Radius of circular loop, r = 0.75 m

Current in the loop, I = 3 A

The loop may be rotated about an axis that passes through the center and lies in the plane of the loop.

When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 Nm.

We need to find the magnitude of the uniform magnetic field exerting this torque on the loop. Torque acting on the loop is given by :

\tau=NIAB\sin\theta

B is magnetic field

B=\dfrac{\tau}{NIA\sin\theta}\\\\B=\dfrac{1.8}{1\times \pi \times (0.75)^2\times 3\times \sin(25)}\\\\B=0.8\ T

So, the magnitude of the uniform magnetic field exerting this torque on the loop is 0.8 T.

6 0
3 years ago
HELP BRANLIEST
Vsevolod [243]

Explanation:

- Newton's first law of motion:

"An object at rest (or in uniform motion) remains at rest (or in uniform motion) unless acted upon an unbalanced force

In this situation, we can apply Newton's first law to the keys of the keyboard that are not hit by the fingers of the man. In fact, as no force act on the keys, they remain at rest.

- Newton's second law of motion:

"The acceleration experienced by an object is proportional to the net force exerted on the object; mathematically:

F=ma

where F is the net force, m is the mass of the object, and a its acceleration"

In this case, we can apply Newton's second law to the keys of the keyboard that are hit by the man: in fact, as they are hit, they experience a downward force, and therefore they experience a downward acceleration.

"Newton's third law of motion:

"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"

Here We can apply Newton's third law to the pair of objects finger-key: in fact, as the finger apply a force on the key (action force), then the key exerts a force back on the finger (reaction force), equal and opposite.

3 0
3 years ago
A car is traveling at the bottom of a 9.00-meter-radius circular hill with a constant speed v. The moment the car is at the bott
Rina8888 [55]

Answer:

Explanation:

reading of scale = reaction force of surface R

centripetal force = R - mg = m v² / R , m is mass , v is velocity and R is radius of the circular path .

R = mg + m v² / R

given ,

m v² / R = .80 mg

v² = .80 x g x R

= .8 x 9.8 x 9 = 70.56

v = 8.4 m /s

3 0
2 years ago
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