Arrange the following substances from the lightest to the heaviest:

A target in a shooting gallery consists of a vertical square wooden board, 0.250 m on a side and with mass 0.750 kg, that pivots

Alenkasestr [34]

Here in this case since there is no torque about the hinge axis for the system of bullet and block then we can say that angular momentum of this system will remain conserved

$L_i = L_f$

$mv \frac{L}{2} = (I_1 + I_2)\omega$

here we will have

L = 0.250 m

v = 385 m/s

m = 1.90 gram

now moment of inertia of the plate will be

$I_1 = \frac{ML^2}{3}$

$I_1 = \frac{0.750 (0.250)^2}{3} = 0.0156 kg m^2$

$I_2 = m(\frac{L}{2})^2 = 0.0019(0.125)^2 = 2.97 \times 10^{-5} kg m^2$

now from above equation

$0.0019 (385)(0.125) = (0.0156 + 2.97 \times 10^{-5})\omega$

$\omega = 5.85 rad/s$

8 0

2 years ago

While traveling from Boston to Hartford, Person A drives at a constant speed of 55 mph for the entire trip. Person B drives at 6

Ne4ueva [31]

Answer:

B will take 1.034 times the time of A from Boston to Hartford.

Explanation:

Let the distance from Boston to Hartford be S.

Person A drives at a constant speed of 55 mph for the entire trip,

Time taken by person A

$t_A=\frac{S}{55}$

Person B drives at 65 mph for half the distance and then drives 45 mph for the second half of the distance.

Time taken by person B

$t_B=\frac{\frac{S}{2}}{65}+\frac{\frac{S}{2}}{45}=\frac{S}{130}+\frac{S}{90}=\frac{220S}{130\times 90}=\frac{11S}{585}$

Ratio of time of arrival of B to A

$\frac{t_B}{t_A}=\frac{\frac{11S}{585}}{\frac{S}{55}}=\frac{121}{117}=1.034$

B will take 1.034 times the time of A from Boston to Hartford.

8 0

2 years ago

A puck rests on a horizontal frictionless plane. A string is wound around the puck and pulled on with constant force. What fract

Ivan

Answer:

Explanation:

Let v be the linear velocity , ω be the angular velocity and I be the moment of inertia of the the puck.

Kinetic energy ( linear ) = 1/2 mv²

Rotational kinetic energy = 1/2 I ω²

I = 1/2 m r² ( m and r be the mass and radius of the puck )

Rotational kinetic energy = 1/2 x1/2 m r² ω²

= 1/4 m v² ( v = r ω )

Total energy

= Kinetic energy ( linear ) + Rotational kinetic energy

= 1/2 mv² + 1/4 m v²

= 3/4 mv²

rotational K E / Total K E = 1/4 m v² / 3/4 mv²

= 1 /3

So 1 /3 rd of total energy is rotational K E.

3 0

2 years ago

Which of the following four circuit diagrams best represents the experiment described in this problem?

Valentin [98]

We don't see any circuit diagrams.

This worries us for a few seconds, until we realize that we don't know anything about the experiment described in the problem either, so we don't have to worry about it at all.

6 0

2 years ago

A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f

konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

$A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m$

(c) the maximum speed attained by the object during its motion

$E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s$

8 0

2 years ago