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STatiana [176]
3 years ago
12

Which scenario did not include a chemical change?

Physics
1 answer:
Jobisdone [24]3 years ago
7 0

Answer:

what scenario i dont understand

Explanation:

step by step explenation

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Select the correct answer. Which of the following causes air pollution inside the house? A. Smoking cigarettes B. Growing housep
Troyanec [42]

Answer: Smoking cigarettes

Explanation: The other options don't cause pollution to form in the air besides smoking, from the particles it creates causing harm and damage to your lungs and fill the air with smoke particles.

3 0
2 years ago
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Find the weight of a 25 kg table. (Use g= 10 m/s?)
PtichkaEL [24]

Answer:

weight = 25*10 =250 N

Explanation:

g must be given in units of m/s^2

The weight of any type of body will be the product of his mass by the gravity

where:

m =mass [kg]

F = force [N] or [kg*m/s^2]

g = acceleration [m/s^2]

3 0
4 years ago
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Donna drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 8 hours. When Donna drove
Dennis_Churaev [7]

Answer:

d=360 miles

Donna lives 360 miles from the mountains.

Explanation:

Conceptual analysis

We apply the formula to calculate uniform moving distance[

d=v*t   Formula (1)

d: distance in miles

t: time in hours

v: speed in miles/hour

Development of problem

The distance Donna traveled to the mountains is equal to the distance back home, equal to d,then,we pose the kinematic equations for d, applying formula 1:

travel data to the mountains: t₁= 8 hours ,  v=v₁

d= v₁*t₁=8*v₁ Equation (1)

data back home : t₂=4hours ,  v=v₂=v₁+45

d=v₂*t₂=(v₁+45)*4=4v₁+180 Equation (2)

Equation (1)=Equation (2)

8*v₁=4v₁+180

8*v₁-4v₁=180

4v₁=180

v₁=180÷4=45 miles/hour

we replace v₁=45 miles/hour in equation (1)

d=8hour*45miles/hour

d=360 miles

8 0
3 years ago
F = Gm2/r2 G = ?<br> Fr2/m2<br> Fm2/r2<br> √(F)m/r<br> √(F)r/m
trapecia [35]

Answer: G=F\frac{r^{2}}{m^{2}}

Explanation:

We have the following equation:

F=G\frac{m^{2}}{r^{2}}

We have to find G this means we have to isolate it:

Fr^{2}=Gm^{2}

G=F\frac{r^{2}}{m^{2}}

8 0
3 years ago
A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa
andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

\omega=\frac{v}{r}

where

\omega is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

\omega=\frac{v}{r}

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s

(c) 5550 m

The maximum playing time of the CD is

t =74.0 min \cdot 60 s/min = 4,440 s

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

d=vt=(1.25 m/s)(4,440 s)=5,550 m

(d) -6.4\cdot 10^{-3} rad/s^2

The angular acceleration of the CD is given by

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f = 21.6 rad/s is the final angular speed (when the CD is scanned at the outermost part)

\omega_i = 50.0 rad/s is the initial angular speed (when the CD is scanned at the innermost part)

t=4440 s is the time elapsed

Substituting into the equation, we find

\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2

5 0
3 years ago
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