How is the axe an example of a wedge (think about simple machines)?

What is the SI unit for moment

natali 33 [55]

Answer:

Newton Metre

Explanation:

8 0

3 years ago

Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R.

Dennis_Churaev [7]

Answer:

$\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}$

Explanation:

The rotational kinetic energy when the cylinder is with the rope is:

$E_k=\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^2$

where we used the fact that both rope and cylinder hast the same w. This E_k must conserve, that is, E_k must equal E_k when the rope leaves the cylinder. Hence, the final w is given by:

$E_{k1}=E_{k2}\\\\\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^{2}=\frac{1}{2}I_c\omega^2\\\\\omega=\sqrt{\omega_0^2(\frac{I_c+I_r}{I_c})}$ (1)

For Ic and Ir we can assume that the rope is a ring of the same radius of the cylinder. Then, we have:

$I_c=\frac{1}{2}MR^2\\\\I_r=mR^2$

Finally, by replacing in (1):

$\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}$

hope this helps!!

7 0

3 years ago

A 0.046 kg golf ball hit by a driver can accelerate from rest to 67 m/s in 1 ms while the driver is in contact with the ball. Ho

Flauer [41]

Answer:

Average force = 67 mn

Explanation:

Given:

Initial velocity u = 0 m/s

Final velocity v = 67 m/s

Time t = 1 ms = 0.001 sec.

Computation:

Using Momentum theory

Change in momentum = F × Δt

(v-u)/t = F × Δt

F × 0.001 = (67 - 0)/0.001

F= 67,000,000

Average force = 67 mn

3 0

3 years ago

A 85-kg astronaut is stranded from his space shuttle. He throws a 1-kg hammer away from the shuttle with a velocity of 17 m/s. H

algol [13]

Answer:

0.2 m/s

Explanation:

given,

mass of astronaut, M = 85 Kg

mass of hammer, m = 1 Kg

velocity of hammer , v =17 m/s

speed of astronaut, v' = ?

initial speed of the astronaut and the hammer be equal to zero = ?

Using conservation of momentum

(M + m) V = M v' + m v

(M + m) x 0 = 85 x v' + 1 x 17

85 v' = -17

v' = -0.2 m/s

negative sign represent the astronaut is moving in opposite direction of hammer.

Hence, the speed of the astronaut is equal to 0.2 m/s

4 0

3 years ago

What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.

valkas [14]

<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)

Where;

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M=6.39(10)^{23}kg$ is the mass of Mars

$a=3.4(10)^{6}m$ is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}$ (2)

$T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}$ (3)

$T=\sqrt{11581157.44 s^{2}}$ (4)

Finally:

$T=3403.1099s=56.718min$ This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0

3 years ago