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3 years ago

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Roseanne heated a solution in a beaker as part of a laboratory experiment on energy transfer. After a while, she noticed the liq

uid solution began to circulate within the beaker. What process involving heat transfer did Roseanne observe inside the beaker?

1 answer:

sveta [45]3 years ago

6 0

I think it is convection hope I could help.

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I will thank and give 5 stars to whoever answers!!

uranmaximum [27]

Answer:

1.moon

2.Earth is the first planet with a satellite, or moon.

3.Mars, the fourth planet out, has two moons, but they are probably just a couple of big old rocks.

5.The incoming object was destroyed on impact, reduced to vapor, dust, and chunks. Large surviving parts were driven deep into the interior of Earth. A signicant portion of Earth was destroyed as well. The energy that resulted from the crash produced an explosion of unimaginable magnitude.

Explanation:

i did what i can

7 0

2 years ago

Read 2 more answers

A rocket engine uses fuel and oxidizer in a reaction that produces gas particles having a velocity of 1380 ms The desired thrust

bearhunter [10]

Answer:

a. 141.3 kg/s b. 5.49 m/s² c. i. 104228.9 N ii. 8.53 m/s² d. i. 97305.2 N ii. 9.84 m/s²

Explanation:

a. What must be the fuel/oxidizer consumption rate (in kg s1)?

The thrust T = Rv where R = mass consumption rate and v = velocity of rocket. Since T = 195000 N and v = 1380 m/s,

R = T/v = 195000 N/1380 m/s = 141.3 kg/s

b. If the initial weight of the rocket is 125000 N, what is its initial acceleration?

We also know that thrust T - W = ma since the rocket has to move against gravity. where M = mass of rocket = W/g = 125000 N/9.8m/s² = 12755.1 kg, W = weight of rocket = 125000 N, a = acceleration of rocket and T = thrust = 195000 N.

So, T - W = Ma

195000 N - 125000 N = (12755.1 kg)a

70000 N = ma

a = 70000 N/12755.1 kg = 5.49 m/s²

c. What are the weight and acceleration of the rocket at t 15.0 s after ignition?

We know that the loss in mass ΔM = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 15 s,

ΔM = 141.3 kg/s × 15 = 2119.5 kg

The new mass is thus M = M - ΔM = 12755.1 kg - 2119.5 kg = 10635.6 kg

i.The weight after 15 seconds is thus W' = M'g = 10635.6 kg × 9.8m/s² = 104228.9 N

ii. Since T - W' = M'a. where M' is our new mass and a our new acceleration,

a = (T - W')/M'

= (195000 N - 104228.9 N)/10635.6 kg

= 90771.1 N/10635.6 kg

= 8.53 m/s²

d. What are the weight and acceleration of the rocket at 20.0 s after ignition?

We know that the loss in mass ΔM" = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 20 s,

ΔM" = 141.3 kg/s × 20 = 2826 kg

The new mass is thus M" = M - ΔM" = 12755.1 kg - 2826 kg = 9929.1 kg

i. The weight after 20 seconds is thus W" = M"g = 9929.1 kg × 9.8m/s² = 97305.2 N

ii. Since T - W" = M"a. where M" is our new mass and a our new acceleration,

a = (T - W")/M"

= (195000 N - 97305.2 N)/9929.1 kg

= 97694.8 N/9929.1 kg

= 9.84 m/s²

4 0

2 years ago

Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s

Alenkasestr [34]

Answer: 1.289 m

Explanation:

The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:

<u>x-component: </u>

$x=V_{o}cos\theta t$ (1)

Where:

$x$ is the horizontal distance traveled by the venom

$V_{o}=3.10 m/s$ is the venom's initial speed

$\theta=47\°$ is the angle

$t$ is the time since the venom is spitted until it hits the ground

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0.44 m$ is the initial height of the venom

$y=0$ is the final height of the venom (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Let's begin with (2) to find the time it takes the complete path:

$0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}$ (3)

Rewritting (3):

$-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0$ (4)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (5)

Where:

$a=-4.9 m/s^{2$

$b=2.267 m/s$

$c=0.44 m$

Substituting the known values:

$t=\frac{-2.267 \pm \sqrt{2.267^{2}-4(-4.9)(0.44)}}{2(-4.9)}$ (6)

Solving (6) we find the positive result is:

$t=0.609 s$ (7)

Substituting (7) in (1):

$x=(3.10 m/s)cos(47\°)(0.609 s)$ (8)

We finally find the horizontal distance traveled by the venom:

$x=1.289 m$

7 0

3 years ago

Sarah, who has a mass of 55 kg, is riding in a car at 20 m/s. She sees a cat crossing the street and slams on the brakes! Her se

avanturin [10]

Answer:

-2200 N

Explanation:

The change in momentum of Sarah is equal to the impulse, which is the product between the force exerted by the seatbelt on Sarah and the time during which the force is applied:

$\Delta p=I\\m \Delta v = F \Delta t$

where

m is the mass

$\Delta v$ is the change in velocity

F is the average force

$\Delta t$ is the duration of the collision

In this problem:, we have:

m = 55 kg is Sarah's mass

$\Delta v = 0-20 = -20 m/s$ is the change in velocity

$\Delta t = 0.5 s$ is the duration of the collision

Solving for F, we find the force exerted by the seatbelt on Sarah:

$F=\frac{m\Delta v}{\Delta t}=\frac{(55)(-20)}{0.5}=-2200 N$

Where the negative sign indicates that the direction of the force is opposite to that of Sarah's initial velocity.

5 0

3 years ago

A ray of light incident in air strikes a rectangular glass block of refractive index 1.5 at angle of incidence of 45.

Gwar [14]

Refraction is said to occur when there is a change in the speed of light.

<h3>What is the angle of refraction?</h3>

We know that refraction is said to occur when there is a change in the speed of light as it travels form one medium to another.

Given that the refractive index of the rectangular glass block is 1.5. The angle of refraction can be obtained by the use of the Snell's law;

n = sin i /sinr

n = refractive index

sini = angle of incidence

sin r = angle of refraction

sinr = sini/n

sinr = sin 45/1.5

= 0.471

r = 28 degrees

b) Now;

sinr =sin 45/1.2

sinr = 0.589

r = 36 degrees

For the glass

sinr = sin 36/1.5

sin r = 0.392

r = 23 degrees

Learn more about angle of refraction:brainly.com/question/2660868

#SPJ1

4 0

1 year ago