All categories

3 years ago

Question 2 parts a b c and d please

Physics

2 answers:

soldier1979 [14.2K]3 years ago

8 0

A). ..... C

b). ..... it's the same at every point

c). ..... C

d). ..... it's the same at every point

fgiga [73]3 years ago

6 0

The vertical velocity is the greatest at C, where it is closest to the ground.
The projectiles horizontal velocity stays constant through the entire thing, so it is the same for A, B, and C.
The projectiles speed is the greatest at C, because both its vertical and horizontal components of its velocity is greater than or equal to that of the other two points.

You might be interested in

A solid cylinder of mass M = 45 kg, radius R = 0.44 m and uniform density is pivoted on a frictionless axle coaxial with its sym

user100 [1]

Answer:

w_f = 1.0345 rad/s

Explanation:

Given:

- The mass of the solid cylinder M = 45 kg

- Radius of the cylinder R = 0.44 m

- The mass of the particle m = 3.6 kg

- The initial speed of cylinder w_i = 0 rad/s

- The initial speed of particle V_pi = 3.3 m/s

- Mass moment of inertia of cylinder I_c = 0.5*M*R^2

- Mass moment of inertia of a particle around an axis I_p = mR^2

Find:

- What is the magnitude of its angular velocity after the collision?

Solution:

- Consider the mass and the cylinder as a system. We will apply the conservation of angular momentum on the system.

L_i = L_f

- Initially, the particle is at edge at a distance R from center of cylinder axis with a velocity V_pi = 3.3 m/s contributing to the initial angular momentum of the system by:

L_(p,i) = m*V_pi*R

L_(p,i) = 3.6*3.3*0.44

L_(p,i) = 5.2272 kgm^2 /s

- While the cylinder was initially stationary w_i = 0:

L_(c,i) = I*w_i

L_(c,i) = 0.5*M*R^2*0

L_(c,i) = 0 kgm^2 /s

The initial momentum of the system is L_i:

L_i = L_(p,i) + L_(c,i)

L_i = 5.2272 + 0

L_i = 5.2272 kg-m^2/s

- After, the particle attaches itself to the cylinder, the mass and its distribution around the axis has been disturbed - requires an equivalent Inertia for the entire one body I_equivalent. The final angular momentum of the particle is as follows:

L_(p,f) = I_p*w_f

- Similarly, for the cylinder:

L_(c,f) = I_c*w_f

- Note, the final angular velocity w_f are same for both particle and cylinder. Every particle on a singular incompressible (rigid) body rotates at the same angular velocity around a fixed axis.

L_f = L_(p,f) + L_(c,f)

L_f = I_p*w_f + I_c*w_f

L_f = w_f*(I_p + I_c)

-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.

L_i = L_f

5.2272 = w_f*(I_p + I_c)

w_f = 5.2272/ R^2*(m + 0.5M)

Plug in values:

w_f = 5.2272/ 0.44^2*(3.6 + 0.5*45)

w_f = 5.2272/ 5.05296

w_f = 1.0345 rad/s

5 0

3 years ago

An open pipe of length 0.39 m vibrates in the third harmonic with a frequency of 1400 Hz what is the distance from the center of

Svetach [21]

Length of the pipe = 0.39 m

Number of harmonics = 3

Now there are 3 loops so here we can say

$3\times \frac{\lambda}{2} = 0.39$

$\lambda = 0.26 m$

now here at the center of the pipe it will form Node

we need to find the distance of nearest antinode

So distance between node and its nearest antinode will be

$d = \frac{\lambda}{4}$

$d = \frac{0.26}{4} = 0.065 m = 6.5 cm$

So the distance will be 6.5 cm

3 0

3 years ago

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression x= 5.0 cos

lakkis [162]

Answer:

a) x = 4.33 m , b) w = 2 rad / s , f = 0.318 Hz , c) a = - 17.31 cm / s²,

d) T = 3.15 s, e) A = 5.0 cm

Explanation:

In this exercise on simple harmonic motion we are given the expression for motion

x = 5 cos (2t + π / 6)

they ask us for t = 0

a) the position of the particle

x = 5 cos (π / 6)

x = 4.33 m

remember angles are in radians

b) The general form of the equation is

x = A cos (w t + Ф)

when comparing the two equations

w = 2 rad / s

angular velocity and frequency are related

w = 2π f

f = w / 2π

f = 2 / 2pi

f = 0.318 Hz

c) the acceleration is defined by

a == d²x / dt²

a = - A w² cos (wt + Ф)

for t = 0 , we substitute

a = - 5,0 2² cos (π / 6)

a = - 17.31 cm / s²

d) El period is

T = 1/f

T= 1/0.318

T = 3.15 s

e) the amplitude

A = 5.0 cm

3 0

3 years ago

An archer pulls a bowstring back a distance of 20 cm with an average force of 75 N. The arrow has a mass of 20.0 g. When he rele

Masja [62]

To solve this problem we will apply the concepts related to the work theorem for which it is defined as the product of Force and distance. In turn, we will use the energy conservation theorem for which the applied work must be equivalent to the total kinetic energy on the body.

The work is defined as

$W = Fd$