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babymother [125]

3 years ago

9

A bartender slides a beer mug at 1.6 m/s towards a customer at the end of a frictionless bar that is 1.1 m tall. The customer ma

kes a grab for the mug and misses, and the mug sails off the end of the bar.
(a) How far away from the end of the bar does the mug hit the floor?
m

(b) What are the speed and direction of the mug at impact?

speed m/s
direction ° below the horizontal

2 answers:

Paraphin [41]3 years ago

8 0

Answer:

(a) x = 0.758 m

(b) v = 4.9 m/s, α = 70.97° below the horizontal

Explanation:

The movement is semi-parabolic because the initial velocity is horizontal.

The equations of semiparabolic motion are:

x movement : uniform line movement

x= vx*t Equation (1)

Where:

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity in m/s

y movement: free fall motion

y = (1/2)g*t² Equation (2)

vy= g*t Equation (3)

Where:

y: vertical position in meters (m)

t : time in seconds (s)

vy: vertical velocity in m/s

g: acceleration due to gravity in m/s²

Data

Vx = 1.6 m/s

y = 1.1 m

g = 9.8 m/s²

Time it takes for the beer mug to hit the floor

We replace in the formula (2)

y = (1/2)g*t²

1.1 = (1/2) (9.8)*t²

t² =(2*1.1)/ (9.8)

$t = \sqrt{\frac{2*1.1}{9.8} }$

t= 0.474 s

Horizontal distance the mug reaches

We replace vx= 1.6 m/s and t= 0.474 s in the formula (1):

x = vx*t = (1.6) *(0.474)

x = 0.758 m

Speed (v) and direction (α) of the mug at impact :

In the Equation (3): vy= g*t = 9.8* 0.474 = 4.645 m/s (downward)

$v = \sqrt{v_{x}^{2}+v_{y}^{2} }$

$v = \sqrt{1.6^{2}+4.645^{2} }$

v = 4.9 m/s

$\alpha = tan^{-1} (\frac{v_{y} }{v_{x} } )$

$\alpha = tan^{-1} (\frac{-4.645 }{1.6} )$

α = - 70.97°

α = below the horizontal

Drupady [299]3 years ago

6 0

Answer:

(a): the mug hits the floor 0.752m away from the end of the bar.

(b): the speed of the mug at impact are:

V= 4.87 m/s

direction= 70.82º below the horizontal.

Explanation:

Vx= 1.6 m/s

Vy=?

h= 1.1 m

g= 9.8 m/s²

t is the fall time

$t=\sqrt{\frac{2*h}{g} }$

t=0.47 sec

Vy= g*t

Vy= 4.6 m/s

$V=\sqrt{Vx^{2} +Vy^{2}$

V= 4.87 m/s

α= tan⁻¹(Vy/Vx)

α= -70.82º

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A cyclist traveling at constant speed of 12m/s when he passes a stationary bus.The bus starts moving just as the cyclist passes

Bogdan [553]

Answer:

A.) 8 seconds

B.) 16 seconds

C.) 48 m

Explanation:

Given that a cyclist traveling at constant speed of 12 m/s

and the bus accelerates uniformly at 1.5ms²

A.) The bus has the following parameters

Acceleration a = 1.5 m/s^2

Initial velocity U = 0. Since the bus is starting from rest.

Final velocity V = 12 m/s

Use equation one of linear motion.

V = U + at

Substitute V, U and a into the formula

12 = 0 + 1.5t

1.5t = 12

t = 12/1.5

t = 8 seconds

Therefore, the bus reach the same speed as the cyclist at 8 seconds.

B.) For the cyclist moving at constant speed, acceleration a = 0. Using second equation of motion

h = Ut + 1/2at^2

Since a = 0, the equation is reduced to:

h = Ut.

Also, for the bus,

h = Ut + 1/2at^2

Equate the two equations since the h is the same

Ut = Ut + 1/2at^2

Substitute all the parameters into the formula

12t = 0 + 1/2 × 1.5t^2

12t = 0.75t^2

0.75t = 12

t = 12/0.75

t = 16 seconds

Therefore, the bus takes 16 seconds to catch the cyclist

C.) Use third equation of linear motion.

V^2 = U^2 + 2as

Where s = distance

Substitute V, U and a into the formula

12^2 = 0 + 2 × 1.5 S

144 = 3S

S = 144/3

S = 48 m

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irinina [24]

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aalyn [17]

Answer:

514.27 ( wavelength )

the color is green

602.93 nm ( orange color )

the observation is that there is a change of visible color

Explanation:

A) wavelength of visible light that is most strongly reflected from a point on a soap

refraction n = 1.33

wall thickness (t) = 290 nm

2nt = (2m +1 ) ∝/2 -----equation 1

note when m = 0

therefore ∝ = 4nt/ 1 = 4 * 1.33 * 290 = 1542.8nm we will discard this

when m = 1

equation 1 becomes

∝ = 4nt/3 =( 4 * 1.33 * 290) / 3 = 1542.8 / 3 = 514.27 ( wavelength )

the color is green

B) the wavelength when the wall thickness is 340 nm

∝ = 4nt / 2m +1

where m = 1

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3 years ago

A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule

ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

$p_i = 0$

The final total momentum is instead:

$p_f = m_a v_a + m_c v_c$

where

$m_a = 125 kg$ is the mass of the astronaut

$v_a = 2.50 m/s$ is the velocity of the astronaut

$m_c = 1900 kg$ is the mass of the capsule

$v_c$ is the velocity of the capsule

Since the total momentum must be conserved, we have

$p_i = p_f = 0$

so

$m_a v_a + m_c v_c=0$

Solving the equation for $v_c$ , we find

$v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s$

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

$F \Delta t = \Delta p$

The change in momentum of the astronaut is

$\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s$

And the duration of the push is

$\Delta t = 0.600 s$

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

$F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N$

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

$K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J$

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

$K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J$

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