Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 25 kg and the larger bottom crate has a mass of m2 = 91 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is µs = 0.79 and the coefficient of kinetic friction between the two crates is µk = 0.62. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem).1)The rope is pulled with a tension T = 234 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate?2)In the previous situation, what is the magnitude of the frictional force the lower crate exerts on the upper crate?3)What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide?4)The tension is increased in the rope to 1187 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?m/s25)As the upper crate slides, what is the acceleration of the lower crate?

Answers

Answer 1

Answer:

Answer:

Explanation:

Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.

1 ) Common acceleration a = force / total mass

= 234 / ( 25 +91 )

= 2.017 m s⁻².

2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂

= mass x acceleration

= 25 x 2.017

= 50.425 N

The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.

3 ) Maximum friction force that is possible between m₁ and m₂

= μ_s m₁g

= .79 x 25 x 9.8

= 193.55 N

Acceleration of m₁

= 193 .55 / 25

= 7.742 m s⁻²

This is the common acceleration in case of maximum tension required

So tension in rope

= ( 25 +91 ) x 7.742

= 898 N

4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁

= μ_k m₁g

= .62 x 25 x 9.8

= 151.9 N

Acceleration of m₁

= 151.9 / 25

= 6.076 m s⁻².

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How many times will the temperature of oxygen with a mass of 1 kg increase if its volume is increased by 4 times, and the pressure is decreased by 2 times?Round off the answer to the nearest whole number.

Answers

Answer:

9.2 Relating Pressure, Volume,

Figure 1. In 1783, the first (a) hydrogen-filled balloon flight, (b) manned hot air balloon flight, and (c) manned hydrogen-filled balloon flight occurred. When the hydrogen-filled balloon depicted in (a) landed, the frightened villagers of Gonesse reportedly destroyed it with pitchforks and knives. The launch of the latter was reportedly viewed by 400,000 people in Paris.

Explanation:

hope its help :)

nicsfrom #philippines

A large aquarium has portholes of thin transparent plastic with a radius of curvature of 1.95 m and their convex sides facing into the water. A shark hovers in front of a porthole, sizing up the dinner prospects outside the tank.a) If one of the sharks teeth is exactly 46.5 cm from the plastic, how far from the plastic does it appear to be to observers outside the tank? (You can ignore refraction due to the plastic.)b) Does the shark appear to be right side up or upside down?c) If the tooth has an actual length of 5.00 cm, how long does it appear to the observers?

Answers

Answer:

Explanation:

For refraction through a curved surface , the formula is as follows

μ₂ / v - μ₁ / u = (μ₂ -μ₁ )/R , Here μ₂( air) = 1 , μ₁ ( water) = 4/3 , R = 1.95 m

u , object distance = - .465 m

1 / v + 1.333 / .465 = (1 -1.333 )/1.95

1 / v + 2.8667 = - .171

1 / v = - 2.8667 - .171 = - 3.0377

v = - .3292 m

= - 32.92 cm

image will be formed in water.

c ) magnification = μ₁v / μ₂u , μ₁ = 1.33 , μ₂ = 1 , u = 46.5 , v = 32.92 .

= (1.33 x 32.92) / (1 x 46.5)

= .94

size of image of teeth = .94 x 5

= 4.7 cm .

A"boat"is"moving"to"the"right"at"5"m/s"with"respect"to"the"water."A"wave"moving"to"the"left,"opposite"the"motion"of"the"boat."The"waves"have"2.0"m"between"the"top"of"the"crests"and"the"bottom"of"the"troughs."The"period"of"the"wave"is"8.3"s"and"their"wavelength"is"110"m."At"one"instant"the"boat"sits"on"a"crest"of"the"wave,"20"seconds"later,"what"is"the"vertical"displacement"of"the"boat

Answers

Answer:

0.99m

Explanation:

Firs you calculate the relative velocity between the boat and the wave. The velocity of the boat is 5m/s and the velocity of the wave is given by:

$v=\lambda f=\lambda(1)/(T)=(110m)(1)/(8.3s)=13.25(m)/(s)$

the relative velocity is:

$v'=13.25m/s-5m/s=8.25(m)/(s)$

This velocity is used to know which is the distance traveled by the boat after 20 seconds:

$x'=v't=(8.25m/s)(20s)=165m$

Next, you use the general for of a wave:

$f(x,t)=Acos(kx-\omega t)=Acos((2\pi)/(\lambda)x-\omega t)$

you take the amplitude as 2.0/2 = 1.0m.

$\omega=(2\pi)/(T)=(2\pi)/(8.3s)=0.75(rad)/(s)$

by replacing the values of the parameters in f(x,t) you obtain the vertical displacement of the boat:

$f(165,20)=1.0m\ cos((2\pi)/(110m)(165)-(0.75(rad)/(s))(20s))\n\nf(165,20)=0.99m$

The position of a particle is given by the function x=(5t3−8t2+12)m, where t is in s. at what time does the particle reach its minimum velocity?

Answers

The particle reach its minimum velocity at time 1.06 sec.

The function is given as

x=5t^3-8t^2+12

Differentiating the above equation with respect to time, to obtain the velocity

dx/dt=v=15t^2-16t

For maximum and minimum values, put dx/dt=0

15t^2-16t=0

On solving the equation, t=0, 1.06

Therefore at time t=1.06 sec, the particle has the minimum value of velocity.

The particle reaches its minimum velocity at t = 0 s or t = 16/15 s

Further explanation

Acceleration is rate of change of velocity.

$\large {\boxed {a = (v - u)/(t) } }$

$\large {\boxed {d = (v + u)/(2)~t } }$

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

Given:

$x = ( 5t^3 - 8t^2 + 12) ~ m$

To find the velocity function, we will derive the position function above.

$v = (dx)/(dt)$

$v = 5(3)t^(3-1) - 8(2)t^(2-1)$

$v = ( 15t^2 - 16t ) ~ m/s$

Next to calculate the time to reach its minimum speed, then v = 0 m/s

$0 = ( 15t^2 - 16t )$

$0 = t( 15t - 16)$

$\large {\boxed {t = 0 ~s ~ or ~ t = 16/15 ~ s} }$

Learn more

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

Chuck wants to investigate how gas moleculesmove in a container. Which model would be most
helpful to represent this motion?
A. stacking blocks to build a tower
B. freezing water in an ice cube tray
C. bouncing elastic balls off of each other and
the walls of a room
D. placing a closed, water-filled plastic bag in
the sun and watching condensation form

Answers

The answers C the molecules in gas move rapidly and all around they are spread out and bounce off each-other

the period of the satellite is exact 42.391 hours, the earth's mass is 5.98 kg and the radius of th earth is 3958.8 miles, what is the distance of the satellite from the surface of the earth in miles?