What force would be needed to accelerate a 0.040-kg golf ball at 20.0 m/s?

The minute hand of a wall clock measures 16 cm from its tip to the axis about which it rotates. The magnitude and angle of the d

olya-2409 [2.1K]

Answer:

Explanation:

Given

Minute hand length =16 cm

Time at a quarter after the hour to half past i.e. 1 hr 45 min

Angle covered by minute hand in 1 hr is 360 and in 45 minutes 270

$|r|=\frac{3\times 2\pi r}{4}=75.408 cm$

$Angle =270^{\circ}$

(c)For the next half hour

Effectively it has covered 2 revolution and a quarter

$|r|=\frac{2\pi r}{4}=25.136 cm$

angle turned $=90^{\circ}$

(f)Hour after that

After an hour it again comes back to its original position thus displacement is same =25.136

Angle turned will also be same i.e. $90 ^{\circ}$

7 0

2 years ago

Choose the correct statement of Kirchhoff's voltage law.

frutty [35]

Answer:

A) If one travels around a closed path adding the voltages for which one enters the negative reference and subtracting the voltages for which one enters the positive reference, the total is zero.

Explanation:

Kirchhoff's voltage law deals with the conservation of energy when the current flows in a closed-loop path.

It states that the algebraic sum of the voltages around any closed loop in a circuit is always zero.

In other words, the algebraic sum of all the potential differences through a loop must be equal to zero.

3 0

2 years ago

A person drops a stone down a well and hears the echo 8.9 s later. if it takes 0.9 s for the echo to travel up the well, approxi

Temka [501]

Total time in between the dropping of the stone and hearing of the echo = 8.9 s

Time taken by the sound to reach the person = 0.9 s

Time taken by the stone to reach the bottom of the well = 8.9 - 0.9 = 8 seconds

Initial speed (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s^2

Time taken (t) = 8 seconds

Let the depth of the well be h.

Using the second equation of motion:

$h = ut + \frac{1}{2}\times a \times t^2$

$h = 0 \times 8 + \frac{1}{2} \times 9.8 \times 8^2$

h = 313.6 m

Hence, the depth of the well is 313.6 m

4 0

2 years ago

1. Determina el momento que produce una fuerza de 7 N tangente a una rueda de un metro de diámetro, sabiendo que el punto de apl

Rudik [331]

Answer:

τ= F r into the blade

Explanation:

The moment of a force is defined by

τ = F x r

where the bold indicates vectors

Let us write in the expression in magnitude

τ = F r sin θ

in our case the force is tangent to the wheel therefore the angle between F and the radius is 90º, and the sin 90 = 1

τ= F r

The direction of τ can be used by the rule of the right hand, the fingers curve in the direction of the torque when advancing from the force to the radius and the thumb points in the direction of the torque.

In this case, for a clockwise rotation, the fingers are curved in the direction and the thumb points into the blade, this is the direction of the τ.

TRASLATE

El momento de una fura es definido por

τ = F x r

donde la negrillas indican vectores

Escribamos en ta expresión en magnitud

τ = F r sin θ

en nuestro caso la fuerza es tangente a la rueda por lo tanto el angulo entre F y el radios es 90º, y el sin 90=1

τ = F r

la dirección de tau la podemos usar la regla de la mano derecha, los dedos curva en la dirección del torque al avanzar dese la fuerza al radio y el pulgar apunta en la dirección del torque.

En este caso para un giro en sentido horario los dedos se curvan ente sentido y el pulgar apunta hacia dentro de lla hoja, esta es la dirección del troque

5 0

2 years ago

The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.

Reika [66]

Answer:

$tan \theta = \mu_s$

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

$mg sin \theta - \mu_s R =0$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, $\theta$ the angle of the slope

$\mu_s R$ is the frictional force, with $\mu_s$ being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

$R-mg cos \theta = 0$

where

R is the normal reaction

$mg cos \theta$ is the component of the weight perpendicular to the slope

Solving for R,

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Re-arranging the equation,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of $\mu_s$ , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0

2 years ago