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3 years ago

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A 150-kg object takes 1.5 minutes to travel a 2,500-meter straight path. It begins the trip traveling 120 m/s and decelerates to

a velocity of 20 m/s. What was its acceleration?

2 answers:

podryga [215]3 years ago

7 0

I believe it is -1.11 m/s^2. I will let you know if its correct

Rom4ik [11]3 years ago

5 0

Your answer is -1.11m/s2. i know this cause i had it right on a test. hope this helps everyone else looking for the answer

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A 1.00-kg object is attached by a thread of negligible mass, which passes over a pulley of negligible mass, to a 2.00-kg object.

Anna [14]

Answer:

a = 3.27 m/s²

v = 2.56 m/s

Explanation:

given,

mass A = 1 kg

mass B = 2 kg

vertical distance between them = 1 m

$F_d = mg$

$F_d = 2 \times 9.8$

$F_d = 19.6\ N$

$F_u = mg$

$F_u = 1 \times 9.8$

$F_u = 9.8\ N$

$F_{net} = 19.6 - 9.8$

$F_{net}=9.8\ N$

$F = (m_1+m_2)a$

$9.8 = (2+1)a$

a = 3.27 m/s²

The speed of the system at that moment is:

v² = u² + 2×a×s

v² = 0² + 2× 3.27 × 1

v ² = 6.54

v = 2.56 m/s

3 0

3 years ago

The light coming out of a concave lens:

torisob [31]

The light coming out of a concave lens will never meet.

So, the answer is A. will never meet.

Happy Studying! ^^

4 0

3 years ago

Read 2 more answers

3. a) If the distance between two objects increases by a factor of 5 (5),

irina [24]

Answer:

Answer for A

Explanation:

F1=GmM/r1^2

If r2 becomes r2=5r

F2=GmM/(25r^2)

Multiply with 25 gives to maintain the same force

I.e.,25F2=F1

F2=G(25m)M/25r^2=F1

By the factor 25 would change to increase to same.

3 0

2 years ago

ΑΣ Sign A. keeping everything grounded on Earth. C. planes not being able to fly into space. B. making tall skyscrapers safe, D.

stepladder [879]

Answer:

i think that it is c because it not only tells us about what the moon does to the earth it also tells us what the earth does to the moon.

Explanation:

3 0

3 years ago

Read 2 more answers

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

(d) 0.225 m

The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

8 0

3 years ago