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3 years ago

Why is it possible to remove a paper without moving the object on it.(apply newton's law)

Physics

1 answer:

klasskru [66]3 years ago

7 0

This is possible due to inertia of motion. which is nothing but newton's first law.

according to this law , an object tries to retain its state of motion or rest unless acted upon by an external force.

consider an object placed on a paper, initially both the object and paper are at rest. to pull the paper , we apply force on the paper and paper gains velocity. but the object keeps its motion of rest and hence the paper can be removed without moving the object.

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The two cars collide at right angles in the intersection of two icy roads. Car A has a mass of 1965 kg and car B has a mass of 1

Sunny_sXe [5.5K]

Answer:

U2 = 47.38m/s = initial velocity of B before impact

Explanation:

An example of the diagram is shown in the attached file because of missing angle of direction in the question

Mass A, B are mass of cars

A = 1965

B =1245

U1 = initial velocity of A = 52km/hr

U2 = initial velocity of B

V = common final velocity of two cars

BU2 = (A + B)*V sin ¤ ...eq1 y plane

AU1 = (A + B) *V cos ¤ ....equ 2plane

From equ 2

V = AU1/(A + B)*cos ¤

Substitute V into equation 1

We have

U2 = (AU1/B)tan ¤ where ¤ = angle of direction which is taken to be 30°

Substitute all parameters to get

U2 = (1965/1245)*52 * tan 30°

U2 = 47.38m/s

8 0

3 years ago

According to Ohm’s law, which combination of units is the same as the unit for resistance? volt ÷ ampere ampere × volt volt + am

Maksim231197 [3]

Answer:

volt ÷ ampere

Explanation:

The mathematical form of Ohms law is given by :

V = IR

Where V is voltage

I is current

R is resistance

$R=\dfrac{V}{I}$

The unit of voltage is volt and that of current is ampere

Unit of resistance :

$R=\dfrac{\text{volt}}{\text{ampere}}$

So, volt ÷ ampere is the same as the unit of resistance. Hence, the correct option is (a).

6 0

2 years ago

Read 2 more answers

A golf club hits a stationary 0.05kg golf ball with and average force of 5.0 x 10^3 newtons accelerating the ball at 44 meters p

maxonik [38]

Answer: The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

Explanation:

Force applied on the golf ball = $5.0\times 10^3 N$

Mass of the ball = 0.05 kg

Velocity with which ball is accelerating = 44 m/s

Time period over which forece applied = t

$f=ma=\frac{m\times v}{t}$

$t=\frac{0.05 kg\times 44m/s}{5.0\times 10^3 N}=4.4\times 10^{-4} seconds$

$Impulse=(force)\times (time)=f\times t = 5.0\times 10^3\times 4.4\times 10^{-4} seconds=2.2$ Newton seconds

The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

7 0

3 years ago

Una prenda de 320gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene 40 cm y gira con una frecuencia de 4

Nitella [24]

Answer:

Período del tambor: $T = 0.25\,s$ , fuerza sobre la prenda: $F \approx 80.852\,N$ , velocidad lineal del tambor: $v \approx 10.053\,\frac{m}{s}$ , velocidad angular del tambor: $\omega \approx 25.133\,\frac{rad}{s}$ .

Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle a) el período, b) la velocidad angular, c) la fuerza con la que gira la prenda y d) la velocidad lineal de la lavadora."

El tambor gira a velocidad angular constante ( $\omega$ ), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga ( $a$ ), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor ( $T$ ), en segundos: