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A cat jumps in a horizontal direction with an initial speed of 2.0 m/s from the

ID: 1624733 • Letter: A

Question

A cat jumps in a horizontal direction with an initial speed of 2.0 m/s from the edge of a table and lands 2.0 m away from the table. What is the height of the table? 1.2 m 2.6 m 4.9 m 9.8 m 19.6 m 160 calories are added 10 grams of ice at -16^degree C. How many grams of ice are changed to water? (Latent heat of fusion of water is 80 cal/g; Specific heat of ice is 0.5 cal/g-^degree C.) 4.0 3.0 2.0 1.0 none A student sits on a rotating stool holding two 3.0-kg objects. When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation and he rotates with an angular speed of 0.60 rad/s. The moment of inertia of the student plus stool is 4.0 kg-m^2 and is assumed to be constant. The student then of the in objects horizontally to 0.2 m from the rotation axis. What is the new angular speed (in rad/s) of the student? 0.42 0.76 0.98 1.2 1.4 Mercury rests in an open-tube manometer as shown. Atmospheric pressure P_o is 750 Torr. Density of mercury is 13.6 g/cm^3. What is the pressure (in Torr) of the gas in the tank? 910 890 770 730 590 An aluminum can of mass 0.7 kg contains water at 20^degree C. Iron pellets with a total mass of 3 kg at 80^degree C are dropped into the can and the can and its contents come to equilibrium at 30^degree C. Specific heat of water = 4190 J/(kg middot^degree C); Specific heat of aluminum = 910 J/(kg middot^degree C); Specific heat of iron = 470 J(kg middot^degree C). How many kilograms of water are in the can? 0.95 1.26 1.53 1.78 2.37 A 30 N sled is pushed up a frictionless 30^degree inclined Plane by a force of 200 N applied parallel to the ground. The sled is moved 4 m along the plane. Find the work done the 200-N force. -120 J 120 J 400 J 693 J 800 J

Explanation / Answer

As you asked multiple questions, so l am just giving answer of first three questions.

7.

given that,

vx = 2 m/s

x = 2 m

x = vx*t

t = 2 / 2 = 1 s

Apply kinetic equation of motion for vertical direction,

h = vy*t + (1/2)at^2

h = 0 + (1/2)*9.8*1

h = 4.9 m

option (c) is correct.

8.

let's see how much energy needed to convert -16 oC ice to 0 deg ice,

Q = mcdT

Q = 10*0.5*16 = 80 J

remaining energy E = 160 - 80 = 80 J

this will convert 0 deg ice to 0 deg water.

80 = mL

m = 80 / 80 = 1 g

option (d) is correct.

9.

As there is no external torque, so angular momentum will be conserved.

l1w1 = l2w2

Here, l1 = lo + 2*mr1^2

l1 = 4 + 2*3*1^2 = 10 kgm^2

w1 = 0.60 rad/s^2

l2 = lo + 2*mr2^2

l2 = 4 + 2*3*0.2^2 = 4.24 kgm^2

10*0.6 = 4.24 * w2

w2 = 1.4 rad/s

option (e) is correct.

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