Ice core samples are used to measure…

1) A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/

mojhsa [17]

Answer: a) 49.560 and 21.13 b) i) 50 N, ii) 196 N iii) 196 N iv) 47.685 N

c) i) 594.72 ii) 0 iii) 0 iv) 0

d) 594.72

Explanation: question a)

The force is inclined at an angle of 25° to the horizontal

The horizontal component of force = 50 cos 25° = 49.560 N

The vertical component of force = 50 sin 30°= 21.130N

Question b)

i) according to the question applied force is 50 N

ii) if g = 9.8m/s², w=mg where m = mass of object = 20kg hence weight = 20* 9.8 = 196 N

iii) the normal force is the force the floor exerts on the body as a result of the weight of the object.

Normal reaction R = W = mg, we already deduced that w = mg, hence R = 196 N.

iv) according to newton's laws of motion

F - Fr = ma

F = applied force = horizontal component of force = 49.560 N.

We need to get the acceleration (a) by using Newton laws of motion before we can be able to compute the frictional force..

The body started from rest hence initial velocity u = 0

Final velocity v = 1.5m/s distance covered (s) = 12m

v ² = u² + 2as

But u = 0

v² = 2as

1.5² = 2(a) * 12

2.25 = 24a

a = 2.25/24 = 0.09735m/s²

From F - Fr = ma

49.560 - Fr = 20 * 0.09735

49.560 - Fr = 1.875

Fr = 49.560 - 1.875

Fr = 47.685 N

Question c)

i) The applied force = 49.560 N, distance covered = 12m

Work done = force * distance

Work done = 49.560 * 12

Work done = 594.72 J

ii) the weight of the object does not make the object move a distance, hence work done = 0 ( since distance covered is 0)

iii) the normal force is the same thing as the weight and they did not cover any distance hence work done is zero.

iv) the frictional force does not cover any distance, hence work done is zero.

Question d)

The total work done = work done by applied force + work done by weight + work done by normal reaction + work done by frictional force.

Total work done = 594.72 + 0 + 0 + 0 = 594.72 J

8 0

3 years ago

List small/average stars<br><br>

mario62 [17]

Answer:

Lol, you should do Nate, Bobby, Cindy, Joe, and Beth

Jk, if you want to be series and probably not fail go for these:

If it wants types of small/average stars, then go with

Small star names:

OGLE-TR-122B

Gliese 229 B

TRAPPIST-1

Teegarden's Star

Luyten 726-8 (A and B)

Proxima Centauri

Wolf 359 111400

Ross 248

Barnard's Star

CM Draconis B

Ross 154 167000

CM Draconis A

Kapteyn's Star

7 0

3 years ago

Can someone please help me? Anyone? PLEASE:(

tatuchka [14]

Answer:

There's no question-

8 0

3 years ago

A rigid tank contains 7 kg of an ideal gas at 5 atm and 30c. a valve is opened, and half of mass of the gas can escape. the fin

Readme [11.4K]

The equation of state for an ideal gas is
$pV=nRT$
where p is the gas pressure, V the volume, n the number of moles, R the gas constant and T the temperature.

The equation of state for the initial condition of the gas is
$p_1 V_1 = n_1 R T_1$ (1)
While the same equation for the final condition is
$p_2 V_2 = n_2 R T_2$ (2)

We know that in the final condition, half of the mass of the gas is escaped. This means that the final volume of the gas is half of the initial volume, and also that the final number of moles is half the initial number of moles, so we can write:
$V_1 = 2 V_1$
$n_1 = 2 n_2$
If we substitute these relationship inside (1), and we divide (1) by (2), we get
$\frac{p_1}{p_2} = \frac{T_1}{T_2}$

And since the initial temperature of the gas is $T_1 = 30 C=303 K$ , we can find the final temperature of the gas:
$T_2 = T_1 \frac{p_2}{p_1}=(303 K) \frac{1.5 atm}{5.0 atm}=90.9 K$

6 0

4 years ago

Read 2 more answers

two forces equal in magnitude and opposite in direction act at the same point on an object. is it possible for there to be a net

elena55 [62]

If the forces are equal, at a distance equidistant it is not possible to act a pair on the body since both torques cancel each other. Being of the same magnitude and in the opposite direction, the sum of the torques will be zero.

5 0

3 years ago