At what point is the KINETIC ENERGY at its highest?

A block is at rest on the incline shown in the figure. The coefficients of static and ki- netic friction are μs = 0.62 and μk =

GREYUIT [131]

The frictional force is 218.6 N

Explanation:

The block in the problem is at rest along the inclined surface: this means that the net force acting along the direction parallel to the incline must be zero.

There are two forces acting along this direction:

- The component of the weight parallel to the incline, downward along the plane, of magnitude

$mg sin \theta$

where

m = 46 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=29^{\circ}$ is the angle of the incline

- The (static) frictional force, acting upward, of magnitude $F_f$

Since the block is in equilibrium, we can write

$mg sin \theta - F_f = 0$

And substituting, we find the force of friction:

$F_f = mg sin \theta = (46)(9.8)(sin 29^{\circ})=218.6 N$

Learn more about frictional force along an inclined plane:

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7 0

3 years ago

A metal wire has a resistance of 14.00 Ω at a temperature of 25.0°C. If the same wire has a resistance of 14.55 Ω at 90.0°C, wha

aliina [53]

Answer:

13.52 Ω

Explanation:

coefficient of thermal resistance be α

R₀ , R₂₅ , R₉₀ and R₋₃₂ be resistances at 0 , 25 , 90 , and - 32 degree

R₂₅ = R₀ + α x 25

R₉₀ = R₀ + α x 90

R₉₀ - R₂₅ = 65 x α

α = (R₉₀ - R₂₅ )/ 65

= (14.55 - 14) / 65

= .55 / 65 Ω per °C,

R₂₅ = R₀ + α x 25

14 = R₀ + (.55 / 65 )x 25

= R₀ + .2115

R₀ = 13.7885 Ω

R₋₃₂ = R₀ - α x 32

= 13.7885 -( .55 / 65) x 32

= 13.7885 - .27077

= 13.51773 Ω

= 13.52 Ω

7 0

3 years ago

Read 2 more answers

Two electrons are at rest and separated by a distance of 4.32 × 10-10 m. When they are released they accelerate away from each o

sasho [114]

Answer:

Speed of electron when their separation increased by a factor of 4.10 is 9.41 x 10⁵ m/s .

Explanation:

The electric potential energy is given by the relation :

$U = \frac{kq_{1}q_{2} }{r}$

Here q₁ and q₂ are the two charge particles and r is the distance between them and k is electric constant.

In this case, there are two electrons which are separated by the distance 4.32 x 10⁻¹⁰ m.

Let e be the electron charge and r₁ be the distance between them. Then, the initial electric potential energy is :

$U_{1} = \frac{ke^{2} }{r_{1} }$

Now, the distance between the electrons increases by the factor of 4.10. Let r₂ be the new distance between them i.e. r₂ = 4.10 r₁.

Thus, the new electric potential energy is :

$U_{2} = \frac{ke^{2} }{r_{2} }=\frac{ke^{2} }{4.10r_{1} }$

Applying law of conservation of energy :

ΔU = ΔK

Here ΔU is change in electric potential energy and ΔK is change in kinetic energy.

( U₁ - U₂ ) = ( K₂ - K₁ )

Here K₂ and K₁ are initial and final kinetic energy of electron.

Since, the electron initially is at rest, so its initial kinetic energy is zero. Thus, the above equation becomes:

K₂ = U₁ - U₂

$\frac{1}{2}mv^{2}=\frac{ke^{2} }{r_{1} }- \frac{ke^{2} }{4.10r_{1} }$

Here m and v are the mass and final speed of electron respectively.

$v^{2}=\frac{2}{m} \frac{ke^{2} }{r_{1} }(1- \frac{1 }{4.10 })$

Substitute 9.1 x 10⁻³¹ kg for m, 9 x 10⁹ N m² C⁻² for k, 1.6 x 10⁻¹⁹ C for e and 4.32 x 10⁻¹⁰ m for r₁ in the above equation.

$v^{2}=\frac{2}{9.1\times10^{-31} } \frac{9\times10^{9}\times(1.6\times10^{-19})^{2} }{4.32\times10^{-10} }(1- \frac{1 }{4.10 })$

$v^{2}=8.86\times10^{11}$

v = 9.41 x 10⁵ m/s

5 0

3 years ago

A stone is thrown vertically downward with an initial speed of 12.0 m/s from the top of a building. The stone takes 1.54 s to re

mr_godi [17]

(a) The stone travels a vertical distance y of

y = (12.0 m/s) t + 1/2 g t ²

where g = 9.80 m/s² is the acceleration due to gravity. Note that this equation assume the downward direction to be positive, and that y = 0 corresponds to the height from which the stone is thrown.

So if it reaches the ground in t = 1.54 s, then the height of the building y is

y = (12.0 m/s) (1.54 s) + 1/2 (9.80 m/s²) (1.54 s)² ≈ 30.1 m

(b) The stone's (downward) velocity v at time t is

v = 12.0 m/s + g t

so that after t = 1.54 s, its velocity is

v = 12.0 m/s + (9.80 m/s²) (1.54 s) ≈ 27.1 m/s

(and of course, speed is the magnitude of velocity)

8 0

3 years ago

One type of slingshot can be made from a length of rope and a leather pocket for holding the stone. The stone can be thrown by w

grigory [225]

Answer:

15.66 rad/s

Explanation:

The vertical motion and horizontal motion are independent of each other.

t = √ ( 2 s/ g) where t = time for the ball to reach the ground and s is the height of the cliff = 18.0 m

t = √ ( 36 / 9.81 ) = 1.916 secs

horizontal distance travel = ut where u is the horizontal velocity of the stone = 30 × r (radius)

tangential velocity V = angular velocity ( ω) × radius

distance traveled = ω × r × t = 30 × r

radius cancelled on both side

ω = 30 / 1.9156 = 15.66 rad/s

4 0

3 years ago