Question 4

A 6 m/s vector pointing North is added to a 2 m/s vector pointing East. What are the magnitude and direction of the resultant?

irina [24]

Answer:

A + B = C Ax = 2 Ay = 0 Bx = 0 By = 6

Ax + Bx = Cx = 2

Ay + By = Cy = 6

C = (2^2 + 6^2)^1/2 = 6.32

Tan Cy / Cx = 6 / 2 = 3

Cy at 71.6 deg

6 0

3 years ago

This experiment is to see if water flows faster out of a smaller can or a larger can.

ddd [48]

Answer:rh

Explanation:fjdj

3 0

3 years ago

Which equation is most likely used to determine the acceleration from a velocity vs:time graph?

tresset_1 [31]

Acceleration, a = (v - u)/t

where v is the final velocity, u is the initial velocity, and t is the time.

This formula on a velocity time graph represents the slope of the graph.

7 0

3 years ago

Read 2 more answers

A uniform horizontal bar of mass m1 and length L is supported by two identical massless strings. String A Both strings are verti

NeX [460]

Answer:

a) T_A = $\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )$ , b) T_B = g [m₂ ( $\frac{x}{d} -1$ ) + m₁ ( $\frac{L}{ 2d} -1$ ) ]

c) x = $d - \frac{m_1}{m_2} \ \frac{L}{2d}$ , d) m₂ = m₁ ( $\frac{ L}{2d} -1$ )

Explanation:

After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act

a) The tension of string A is requested

The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive

∑ τ = 0

T_A d - W₂ x -W₁ L/2 = 0

T_A = $\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )$

b) the tension in string B

we write the expression of the translational equilibrium

∑ F = 0

T_A - W₂ - W₁ - T_B = 0

T_B = T_A -W₂ - W₁

T_ B = $\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )$ - g m₂ - g m₁

T_B = g [m₂ ( $\frac{x}{d} -1$ ) + m₁ ( $\frac{L}{ 2d} -1$ ) ]

c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system

T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0

at the point that begins to rotate T_B = 0

g m₂ (d -x) - g m₁ (0.5 L -d) + 0 = 0

m₂ (d-x) = m₁ (0.5 L- d)

m₂ x = m₂ d - m₁ (0.5 L- d)

x = $d - \frac{m_1}{m_2} \ \frac{L}{2d}$

d) The mass of the block for which it is always in equilibrium

this is the mass for which x = 0

0 = d - \frac{m_1}{m_2} \ \frac{L}{2d}

$\frac{m_1}{m_2} \ (0.5L -d) = d$

$\frac{m_1}{m_2} = \frac{ d}{0.5L-d}$

m₂ = m₁ $\frac{0.5 L -d}{d}$

m₂ = m₁ ( $\frac{ L}{2d} -1$ )

5 0

3 years ago

Your friend thinks that the escape speed should be greater for more massive objects than for less massive objects. Provide an ar

Brrunno [24]

Answer:

Concepts and Principles

1- Kinetic Energy: The kinetic energy of an object is:

K=1/2*m*v^2 (1)

where m is the object's mass and v is its speed relative to the chosen coordinate system.

2- Gravitational potential energy of a system consisting of Earth and any object is:

U_g = -Gm_E*m_o/r*E-o (2)

where m_E is the mass of Earth (5.97x 10^24 kg), m_o is the mass of the object, and G = 6.67 x 10^-11 N m^2/kg^2 is Newton's gravitational constant.

Solution

The argument:

My friend thinks that escape speed should be greater for more massive objects than for less massive objects because the gravitational pull on a more massive object is greater than the gravitational pull for a less massive object and therefore the more massive object needs more speed to escape this gravitational pull.

The counterargument:

We provide a mathematical counterargument. Consider a projectile of mass m, leaving the surface of a planet with escape speed v. The projectile has a kinetic energy K given by Equation (1):

K=1/2*m*v^2 (1)

and a gravitational potential energy Ug given by Equation (2):

Ug = -G*Mm/R

where M is the mass of the planet and R is its radius. When the projectile reaches infinity, it stops and thus has no kinetic energy. It also has no potential energy because an infinite separation between two bodies is our zero-potential-energy configuration. Therefore, its total energy at infinity is zero. Applying the principle of energy consersation, we see that the total energy at the planet's surface must also have been zero:

K+U=0

1/2*m*v^2 + (-G*Mm/R) = 0

1/2*m*v^2 = G*Mm/R

1/2*v^2 = G*M/R

solving for v we get

v = √2G*M/R

so we see v does not depend on the mass of the projectile

8 0

3 years ago