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NeX [460]
3 years ago
7

in a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallw

ay. If the 3.5 kg book is pushed from rest through a distance of 0.96 m by the horizontal 22 N force from the broom and then has a speed of 1.36 m/s, what is the coefficient of kinetic friction between the book and floor?
Physics
1 answer:
madam [21]3 years ago
3 0

Answer:

μ=0.151

Explanation:

Given that

m= 3.5 Kg

d= 0.96 m

F= 22 N

v= 1.36 m/s

Lets take coefficient of kinetic friction = μ

Friction force Fr=μ m g

Lets take acceleration of block is a m/s²

F- Fr = m a

22 -  μ x 3.5 x 10 = 3.5 a         ( take g =10 m/s²)

a= 6.28 - 35μ  m/s²

The final speed of the block is v

v= 1.36 m/s

We know that

v²= u²+ 2 a d

u= 0 m/s given that

1.36² = 2 x a x 0.96

a= 0.963 m/s²

a= 6.28 - 35μ  m/s²

6.28 - 35μ = 0.963

μ=0.151

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La mascota de felipe esta jugando en el parque despues de un tiempo esta agotado de tanto correr. Felipe conoce que el perro con
MatroZZZ [7]

Responder:

6.704 m / s

Explicación:

Se dice que el trabajo se realiza cuando la fuerza aplicada a un objeto hace que el objeto se mueva. Primero necesitamos calcular la distancia recorrida por el perro usando la fórmula del trabajo realizado.

Trabajo realizado = Fuerza × distancia

Distancia = Trabajo realizado / Fuerza

Distancia = W / mg

S = 176/8 × 9,81

S = 176 / 78,48

S = 2,24 m

Dada la velocidad inicial u = 3.6km / h

Convertir a m / s

= 3.6km × 1000m / 1h × 3600

= 3600/3600

= 1 m / s

u = 1 m / s

Usando la ecuación de movimiento

v² = u² + 2gS para obtener la velocidad final v:

v² = 1² + 2 (9,81) (2,24)

v² = 1 + 43,9488

v² = 44,9488

v = √44,9488

v = 6,704 m / s

Por tanto, la rapidez final del perro es de 6,704 m / s

6 0
3 years ago
A bobsled zips down an ice track, starting from rest at the top of a hill with a vertical height of 150m. Disregarding friction,
olchik [2.2K]
<span>The velocity would be 54.2 m/s We would use the equation 1/2mv^2top+mghtop = 1/2mv^2bottom+mghbottom where m is the mass of the bobsled(which can be ignored), vtop/bottom is the velocity of the bobsled at the top or bottom, g is gravity, and htop/bottom is the height of the bobsled at the top or bottom of the hill. Since the velocity of the bobsled at the top of the hill and height at the bottom of the hill are zero, 1/2mv^2top and mghbottom will equal zero. The equation will be mghtop=1/2mv^2bottom. Thus we would solve for v.</span>
5 0
3 years ago
Read 2 more answers
At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f
jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp =  change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s,  0.65 kg m/s)

The magnitude of this vector is calculated as follows:

|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

4 0
3 years ago
On February 15, 2013, Asteroid 2012 DA14 passed within 17,200 miles [mi] of the surface of the Earth at a relative speed of 7.8
xz_007 [3.2K]

Answer:

The total amount of energy that would have been released if the asteroid hit earth = The kinetic energy of the asteroid = 1.29 × 10¹⁵ J = 1.29 PetaJoules = 1.29 PJ

1 PJ = 10¹⁵ J

Explanation:

Kinetic energy = mv²/2

velocity of the asteroid is given as 7.8 km/s = 7800 m/s

To obtain the mass, we get it from the specific gravity and diameter information given.

Density = specific gravity × 1000 = 3 × 1000 = 3000 kg/m³

But density = mass/volume

So, mass = density × volume.

Taking the informed assumption that the asteroid is a sphere,

Volume = 4πr³/3

Diameter = 30 m, r = D/2 = 15 m

Volume = 4π(15)³/3 = 14137.2 m³

Mass of the asteroid = density × volume = 3000 × 14137.2 = 42411501 kg = 4.24 × 10⁷ kg

Kinetic energy of the asteroid = mv²/2 = (4.24 × 10⁷)(7800²)/2 = 1.29 × 10¹⁵ J

6 0
3 years ago
What two forces act on a falling object ?
goldenfox [79]

Answer:

The two forces acting on the object are weight due to gravity pulling the object towards earth, and drag resisting this motion. When the object is first released, drag is small as velocity is low, so the resultant force is down. This means the object accelerates towards earth.

5 0
3 years ago
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