All categories

3 years ago

A baseball player slides into third base with an initial speed of 4.0 m/s. If the coefficient of

2 answers:

6 0

The frictional force is given by F = μmg

where μ is the coeficient of friction.

Work done by frictional force = Fd = μmgd

Kinetic energy "lost" = 1/2 mv²

Since we know that W = ΔK,

so, Fd = μmgd = 1/2 mv²

The m's cancel μgd = v² / 2

d = v² / 2μg

d = 4² / 2(0.46)(9.81)

d = 16 / 2(0.46)(9.81)

d = 1.77

Player slides 1.77 m .

musickatia [10]3 years ago

3 0

The frictional force is given by F = μmg

where μ is the coeficient of friction. 

Work done by frictional force = Fd = μmgd 

Kinetic energy "lost" = 1/2 mv² 

Fd = μmgd = 1/2 mv² 

The m's cancel μgd = v² / 2 

d = v² / 2μg 

d = 8² / 2(0.41)(9.8) 

d = 32 / (0.41)(9.8) 

d = 7.96 

Player slides 8 m . 

Note. In your other example μ = 0.46 and v = 4 m/s 

d = v² / 2μg 

= 4² / 2(0.46)(9.8) 

= 8 / (0.46)(9.8) 

= 1.77 or 1.8 m.

Hope i Helped :D

You might be interested in

What is imperceptible human motion?

castortr0y [4]

Unable to be noticed or not felt because of feeling slight

3 0

3 years ago

Why is a specimen smaller than 200 nm not visible with a light microscope?

kifflom [539]

Because the specimen is very small with a light microscope

8 0

3 years ago

9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the

mojhsa [17]

Answer:

Moment about SHOULDER ∑ τ = 3.17 N / m,

Moment respect to ELBOW Στ= 2.80 N m

Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

∑ τ = I α

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

∑ τ = I α

I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

I_forearm = ⅓ m L²

the moment of inertia of the mass in the hand, let's approach as punctual

I_mass = m L²

we substitute

∑ τ = (⅓ m L² + M L²) α

let's calculate

∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

∑ τ = 3.17 N / m

Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

Στ = I α

I = I_ forearm + I_mass

I = ⅓ m (L-0.03)² + M (L-0.03)²

let's calculate

i = ⅓ 2.3 0.47² + 0.5 0.47²

I = 0.2798 Kg m²

Στ = 0.2798 10

Στ= 2.80 N m

3 0

3 years ago

What is an ellipse?

Rzqust [24]

Answer:An ellipse is a closed curve consisting of points whose distances from each of two fixed points (foci) all add up to the same value .

Explanation:

6 0

2 years ago

Read 2 more answers

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m

ikadub [295]

(a) $-39.4^{\circ}$

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

$p_y =0$

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

$0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2$

where

m is the mass of each ball

$v_1= 4.60 m/s$ is the velocity of the cue ball after the collision

$v_2 = 3.40 m/s$ is the velocity of the second ball after the collision

$\phi_1=28.0^{\circ}$ is the angle of the cue ball with the x-axis

$\phi_2$ is the angle of the second ball

Solving for $\phi_2$ , we find the angle between the direction of motion of the second ball and the original direction of motion:

$sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}$

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

$p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m$

The initial total momentum along the x-direction as

$p_x = m u$

where

m is the mass of the cue ball

$u$ is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

$mu = 6.69 m\\u = 6.69 m/s$

7 0

3 years ago