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salantis [7]
3 years ago
13

A straight line is drawn on the surface of a 14-cm-radius turntable from the center to the perimeter. A bug crawls along this li

ne from the center outward as the turntable spins counterclockwise at a constant 45 rpm. Its walking speed relative to the turntable is a steady 3.8 cm/s. Let its initial heading be in the positive x-direction, where the x- and y-directions are fixed relative to the laboratory. As the bug reaches the edge of the turntable (still traveling at 3.8 cm/s radially, relative to the turntable) what are the x and y components of the velocity of the bug?
Physics
1 answer:
sleet_krkn [62]3 years ago
8 0

Answer:

v = \left[\begin{array}{c}0.66&0\end{array}\right]m/s

Explanation:

The position vector r of the bug with linear velocity v and angular velocity ω in the laboratory frame is given by:

\overrightarrow{r}=vtcos(\omega t)\hat{x}+vtsin(\omega t)\hat{y}

The velocity vector v is the first derivative of the position vector r with respect to time:

\overrightarrow{v}=[vcos(\omega t)-\omega vtsin(\omega t)]\hat{x}+[vsin(\omega t)+\omega vtcos(\omega t)]\hat{y}

The given values are:

t=\frac{x}{v}=\frac{14}{3.8}=3.7 s

\omega=\frac{45\times2\pi}{60s}=4.7\frac{1}{s}

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Lori wants to send a box of oranges to a friend by mail. The box of oranges cannot exceed a mass of 10.222 Kg. If each orange ha
Sergeu [11.5K]

Explanation:

Given that,

The box of oranges cannot exceed a mass of 10.222 Kg if we are sending to a friend by mail.

The mass of each orange is 198 g

We know that,

1 kg = 1000 g

10.222 kg = 10.222×1000 g

Let there are n number of oranges. So,

n=\dfrac{10.222\times 1000\ g}{198\ g}\\\\n=51.92\approx 52\ \text{oranges}

It means she can send 52 oranges and it is maximum quantity.

4 0
3 years ago
Which of the following is NOT one of the four elements of emotion?
Nezavi [6.7K]

Answer:

Attention

Explanation:

Hello there, fellow peer! The answer to question is attention. Let's say someone is the control. The behavioral expression is an element of expression, so the control will feel emotions. Subjective Experience is when someone felt the way you feel and they are trying to help you. That is a type of emotion which can lead to empathy for you. This is also not the answer. Physiological Arousal is also not the answer because this is when you can feel what someone else is feeling and you try to give them therapy.

Using the process of elimination, our answer is therefore attention.

4 0
3 years ago
A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target
stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

\theta=35^{\circ}

a.Let v_0 be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

v_x=v_0cos\theta=v_0cos35

v_y=v_0sin\theta=v_0sin35

x=v_0cos\theta\times t=v_0cos35\times t

t=\frac{30}{v_0cos35}

h=v_yt-\frac{1}{2}gt^2

Substitute the values

1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2

1.8=30tan35-\frac{6574.6}{v^2_0}

\frac{6574.6}{v^2_0}=21-1.8=19.2

v^2_0=\frac{6574.6}{19.2}

v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

t=\frac{30}{18.5cos35}

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0
3 years ago
You have a bowling ball with a mass of 4kg. You throw it with an acceleration of 10 m/s/s. With how much force will it hit the p
Eddi Din [679]

Answer:

<h3>The answer is 40 N</h3>

Explanation:

The force acting on an object can be found by using the formula

<h3>force = mass × acceleration</h3>

From the question

mass = 4 kg

acceleration = 10 m/s²

So we have

force = 4 × 10

We have the final answer as

<h3>40 N</h3>

Hope this helps you

6 0
3 years ago
A sample of monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A). It is warmed at constant volume to
leonid [27]

Answer:

(a) 0.203 moles

(b) 900 K

(c) 900 K

(d) 15 L

(e) A → B, W = 0, Q = Eint = 1,518.91596 J

B → C, W = Q ≈ 1668.69974 J Eint = 0 J

C → A, Q = -2,531.5266 J, W = -1,013.25 J, Eint = -1,518.91596 J

(g) ∑Q = 656.089 J, ∑W =  655.449 J, ∑Eint = 0 J

Explanation:

At point A

The volume of the gas, V₁ = 5.00 L

The pressure of the gas, P₁ = 1 atm

The temperature of the gas, T₁ = 300 K

At point B

The volume of the gas, V₂ = V₁ = 5.00 L

The pressure of the gas, P₂ = 3.00 atm

The temperature of the gas, T₂ = Not given

At point C

The volume of the gas, V₃ = Not given

The pressure of the gas, P₃ = 1 atm

The temperature of the gas, T₂ = T₃ = 300 K

(a) The ideal gas equation is given as follows;

P·V = n·R·T

Where;

P = The pressure of the gas

V = The volume of the gas

n = The number of moles present

R = The universal gas constant = 0.08205 L·atm·mol⁻¹·K⁻¹

n = PV/(R·T)

∴ The number of moles, n = 1 × 5/(0.08205 × 300) ≈ 0.203 moles

The number of moles in the sample, n ≈ 0.203 moles

(b) The process from points A to B is a constant volume process, therefore, we have, by Gay-Lussac's law;

P₁/T₁ = P₂/T₂

∴ T₂ = P₂·T₁/P₁

From which we get;

T₂ = 3.0 atm. × 300 K/(1.00 atm.) = 900 K

The temperature at point B, T₂ = 900 K

(c) The process from points B to C is a constant temperature process, therefore, T₃ = T₂ = 900 K

(d) For a constant temperature process, according to Boyle's law, we have;

P₂·V₂ = P₃·V₃

V₃ = P₂·V₂/P₃

∴ V₃ = 3.00 atm. × 5.00 L/(1.00 atm.) = 15 L

The volume at point C, V₃ = 15 L

(e) The process A → B, which is a constant volume process, can be carried out in a vessel with a fixed volume

The process B → C, which is a constant temperature process, can be carried out in an insulated adjustable vessel

The process C → A, which is a constant pressure process, can be carried out in an adjustable vessel with a fixed amount of force applied to the piston

(f) For A → B, W = 0,

Q = Eint = n·cv·(T₂ - T₁)

Cv for monoatomic gas = 3/2·R

∴ Q = 0.203 moles × 3/2×0.08205 L·atm·mol⁻¹·K⁻¹×(900 K - 300 K) = 1,518.91596 J

Q = Eint = 1,518.91596 J

For B → C, we have a constant temperature process

Q = n·R·T₂·㏑(V₃/V₂)

∴ Q = 0.203 moles × 0.08205 L·atm/(mol·K) × 900 K × ln(15 L/5.00 L) ≈ 1668.69974 J

Eint = 0

Q = W ≈ 1668.69974 J

For C → A, we have a constant pressure process

Q = n·Cp·(T₁ - T₃)

∴ Q = 0.203 moles × (5/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -2,531.5266 J

Q = -2,531.5266 J

W = P·(V₂ - V₁)

∴ W = 1.00 atm × (5.00 L - 15.00 L) = -1,013.25 J

W = -1,013.25 J

Eint = n·Cv·(T₁ - T₃)

Eint = 0.203 moles × (3/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -1,518.91596 J

Eint = -1,518.91596 J

(g) ∑Q = 1,518.91596 J + 1668.69974 J - 2,531.5266 J = 656.089 J

∑W = 0 + 1668.69974 J -1,013.25 J = 655.449 J

∑Eint = 1,518.91596 J + 0 -1,518.91596 J = 0 J

5 0
3 years ago
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