HELP ME WITH THIS ONE PLEASE!!!!

3.00 textbook rests on a frictionless, horizontal tabletop surface. A cord attached to the book passes over a pulley whose diame

sammy [17]

Answer:

a1 = 3.56 m/s²

Explanation:

We are given;

Mass of book on horizontal surface; m1 = 3 kg

Mass of hanging book; m2 = 4 kg

Diameter of pulley; D = 0.15 m

Radius of pulley; r = D/2 = 0.15/2 = 0.075 m

Change in displacement; Δx = Δy = 1 m

Time; t = 0.75

I've drawn a free body diagram to depict this question.

Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;

ΣF_x = T1 = m1 × a1

a1 is acceleration and can be calculated from Newton's 2nd equation of motion.

s = ut + ½at²

our s is now Δx and a1 is a.

Thus;

Δx = ut + ½a1(t²)

u is initial velocity and equal to zero because the 3 kg book was at rest initially.

Thus, plugging in the relevant values;

1 = 0 + ½a1(0.75²)

Multiply through by 2;

2 = 0.75²a1

a1 = 2/0.75²

a1 = 3.56 m/s²

6 0

2 years ago

What’s the velocity of a sound wave traveling through air at a temperature of 18°C (64.4°F)?

Anna11 [10]

Answer:

342 m/s

Explanation:

The velocity of sound in air is approximated as:

v ≈ 331.4 + 0.6 T

where v is the velocity in m/s and T is the temperature in Celsius.

At T = 18:

v ≈ 331.4 + 0.6 (18)

v ≈ 342.2

The velocity is approximately 342 m/s.

4 0

3 years ago

Read 2 more answers

A weather balloon is filled to a volume of 6000 L while it is on the ground, at a pressure of 1 atm and a temperature of 273 K.

avanturin [10]

Answer : The final volume of the balloon at this temperature and pressure is, 17582.4 L

Solution :

Using combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 1 atm

$P_2$ = final pressure of gas = 0.3 atm

$V_1$ = initial volume of gas = 6000 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = 273 K

$T_2$ = final temperature of gas = 240 K

Now put all the given values in the above equation, we get the final pressure of gas.

$\frac{1atm\times 6000L}{273K}=\frac{0.3atm\times V_2}{240K}$

$V_2=17582.4L$

Therefore, the final volume of the balloon at this temperature and pressure is, 17582.4 L

4 0

3 years ago

Read 2 more answers

Problem 1: Spherical mirrorConsider a spherical mirror of radius 2 m, and rays which go parallel to the optic axis. What is thep

SIZIF [17.4K]

Answer:

1) iii i= 1m, 2) iii and iv, 3) i = f₂ (L-f₁) / (L - (f₁ + f₂))

Explanation:

Problem 1

For this problem we use two equations the equations of the focal distance in mirrors

f = r / 2

f = 2/2

f = 1 m

The builder's equation

1 / f = 1 / o + 1 / i

Where f is the focal length, "o and i" are the distance to the object and the image respectively.

For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0

1 / f = 0 + 1 / i

i = f

i = 1 m

The image is formed at the focal point

The correct answer is iii

Problem 2

For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat

Case 1 Flat lens - convex (convergent)

R₂ = infinity

R₁ > 0

Cas2 Flat-concave (divergent) lens

R₂ = infinity

R₁ <0

Why the correct answers are iii and iv

Problem 3

For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface

1 / f₂ = 1 / (L-f₁) + 1 / i

1 / i = 1 / f₂ - 1 / (L-f₁)

1 / i = (L-f₁-f₂) / f₂ (L-f₁)

i = f₂ (L-f₁) / (L - (f₁ + f₂))

This is the image of the rays that enter parallel to the first surface

6 0

3 years ago

What is the kinetic energy of a 1130 kg truck that is moving with a velocity of 40 m/s?

Makovka662 [10]

Answer:

The answer is 904,000.

Kinetic energy=1/2mv^2.

1/2×1130×40^2.

1/2×1808000=904,000Joules.

8 0

3 years ago