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2 years ago

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Two test tubes are filled with a solution of bromthymol blue. A student exhales through a straw into each tube, and the bromthym

ol blue turns yellow. An aquatic green plant is placed in each tube, and the tubes are corked. One tube is placed in the dark, and the other tube is placed in direct sunlight. The yellow solution in the tube in sunlight turns blue, while the one in the dark remains yellow. Which statement best explains why the solution in the tube placed in sunlight returns to a blue color

1 answer:

saw5 [17]2 years ago

8 0

Answer:

oxygen was produced by phosynthesis

Explanation:

kkkkk

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A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 18.5 kg, and the

meriva

Answer:

a. $2.86 m/s^2$

b.1058 N

Explanation:

We are given that

Mass of each dog,M=18.5 kg

Mass of sled with rider,m=250 kg

a.Average force,F=185 N

$\mu_s=0.14$

$g=9.8 m/s^2$

By Newton's second law

$8F-f=(8M+m)a$

$a=\frac{8F-f}{8M+m}=\frac{8(185)-(0.14)(9.8)(250)}{8(18.5)+250}$

$a=2.86 m/s^2$

b.By Newton's second law

$T=ma+\mu_s mg$

Substitute the values

$T=250\times 2.86+0.14(250)(9.8)=1058 N$

Hence, the force in the coupling between the dogs and the sled=1058 N

8 0

3 years ago

A baseball is batted from a height of 1.09 m with a speed of

kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

$H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m$

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

$V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s$

<h3>Resultant speed of the ball and crosswind</h3>

$V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s$

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0

1 year ago

The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t

dem82 [27]

Answer:

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

now for the time period of moon around the earth we can say

$T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}$

here we know that

$T_1 = 0.08 year$

$r_1 = 0.0027 AU$

$M_e$ = mass of earth

Now if the same formula is used for revolution of Earth around the sun

$T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}$

here we know that

$r_2 = 1 AU$

$T_2 = 1 year$

$M_s$ = mass of Sun

now we have

$\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}$

$\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}$

$12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}$

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

4 0

3 years ago

A ping pong ball with a dent in it can be put into a pan of boiling water. After a short amount of time, the dent will pop out.

kifflom [539]

heat from water goes into air in ball

air expands

ping goes the dent

6 0

3 years ago

Read 2 more answers

Two particles are moving with positions given by x= 4t^2−2t and x= 6t^3+8t respectively.

Evgen [1.6K]

Explanation:

<em>a)Which of the two has uniform acceleration?</em>

Acceleration is the second derivative of position. The acceleration of the first particle is:

x = 4t² − 2t

v = 8t − 2

a = 8

The acceleration of the second particle is:

x = 6t³ + 8t

v = 18t² + 8

a = 36t

The first particle has uniform acceleration.

<em>b)Which one is likely to come to rest at some time during its motion?</em>

The particles come to rest when v = 0. The first particle's velocity has a real zero at t = 4. The second particle's velocity has only imaginary zeros, meaning v is never 0.

6 0

2 years ago