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3 years ago

Anton created a chart listing different types of materials.

Physics

2 answers:

kari74 [83]3 years ago

4 0

Answer:

Explanation:

DochEvi [55]3 years ago

3 0

Answer:The answer is C

Explanation:

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A train brakes from 25 m/s to rest in 30 sec. What is its deceleration?

givi [52]

Answer:

a= -0.83m\s^2

Explanation:

a = v \ t

a = -25 \ 30 = -0.833 m\s^2

the object is slowing down 0.83 meter every second

8 0

3 years ago

Explain why the top of the loop cannot be the same height as (or higher than) the top of the first hill. Assume the roller coast

Ivahew [28]

Answer:

By conservation of energy, it can climb up to a height equal to that it went down before. However, due to the friction in the machines, the total mechanical energy of the roller coaster will decrease. As a result, the first "hill" of many roller coasters are the highest, but the followings will have decreasing heights.

Explanation:

7 0

3 years ago

Read 2 more answers

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

$f = f_{0}\frac{v + v_{o}}{v - v_{s}}$

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer

v: is the speed of the sound = 343 m/s

$v_{o}$ : is the speed of the observer = 0 (it is heard in the town)

$v_{s}$ : is the speed of the source =?

The frequency of the train before slowing down is given by:

$f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}$ (1)

Now, the frequency of the train after slowing down is:

$f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}$ (2)

Dividing equation (1) by (2) we have:

$\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}$

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}$ (3)

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

$v_{s_{b}} = 2v_{s_{a}}$ (4)

Now, by entering equation (4) into (3) we have:

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}$

$\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}$

By solving the above equation for $v_{s_{a}}$ we can find the speed of the train after slowing down:

$v_{s_{a}} = 11.06 m/s$

Finally, the speed of the train before slowing down is:

$v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s$

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

I hope it helps you!

7 0

3 years ago

A circuit contains two 2,000 resistors connected in parallel. What is the total resistance of the circuit?

GenaCL600 [577]

RT = R1 R2/ R1 + R2

R1 = R2 = 2k ohm

RT = R/2 = 2k/2 = 1k ohm

5 0

3 years ago

The voltage across the terminals of a generator is 5.7 v when it supplies a current of 0.3 A. It becomes 5.1 V when I=0.9A. Find

snow_tiger [21]

Answer:

The emf of the generator is 6V

The internal resistance of the generator is 1 Ω

Explanation:

Given;

terminal voltage, V = 5.7 V, when the current, I = 0.3 A

terminal voltage, V = 5.1 V, when the current, I = 0.9 A

The emf of the generator is calculated as;

E = V + Ir

where;

E is the emf of the generator

r is the internal resistance

First case:

E = 5.7 + 0.3r -------- (1)

Second case:

E = 5.1 + 0.9r -------- (2)

Since the emf E, is constant in both equations, we will have the following;

5.1 + 0.9r = 5.7 + 0.3r

collect similar terms together;

0.9r - 0.3r = 5.7 - 5.1

0.6r = 0.6

r = 0.6/0.6

r = 1 Ω

Now, determine the emf of the generator;

E = V + Ir

E = 5.1 + 0.9x1

E = 5.1 + 0.9

E = 6 V

6 0

3 years ago