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3 years ago

7

A man is walking while riding a train. He says he is moving at 2 mph. A woman standing on a platform at a train station says the

man on the train is moving at 72 mph. Which person is correct?

1 answer:

-BARSIC- [3]3 years ago

5 0

The woman is correct because it is the pace he is moving not walking.

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A pulley has an initial angular speed of 12.5 rad/s and a constant angular acceleration of 3.41 rad/s2. Through what angle does

Novosadov [1.4K]

Answer:

Angle turn pulley is 113 rad

Explanation:

given data

angular speed w = 12.5 rad/s

angular acceleration a 3.41 rad/s²

time t = 5.26 s

to find out

what angle does the pulley turn

solution

we get here angle by this equation that is here

angle turn = w × t + 0.5 × a × t² ......................1

put here value we get

angle turn = w × t + 0.5 × a × t²

angle turn = 12.5 × 5.26 + 0.5 × 3.41 × 5.26²

Angle turn = 65.75 + 47.17

Angle turn = 112.92 rad = 113 rad

4 0

3 years ago

How to convert 8 feet to metres

ohaa [14]

Answer:

2.4384m

Explanation:

1 ft in m is 0.3048

8ft in m is x

therefore 8ft in m is

8 * 0.3048 = 2.4384

7 0

2 years ago

Read 2 more answers

What happens when the thermal energy of a substance increases?

Sonja [21]

When thermal energy of a substance increases, it's entropy(randomness) & Kinetic energy increases.

For more appropriate answer, you should put the options 'cause there could be more than one answer for this question.

4 0

3 years ago

A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp

babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0

3 years ago

e force acting between two charged particles A and B is 5.2 × 10-5 newtons. Charges A and B are 2.4 × 10-2 meters apart. If the

anastassius [24]

The force acting between the particles is

$F=k \frac{Q_{1}Q_{2}}{r^2}$
Then
$Q_{2}= \frac{5.2 \times 10^-^5 \times 0.024^2}{ 9.0 \times 10^9 7.2 \times 10^-^8} =4.622 \times 10^-^1^1C$

7 0

3 years ago

Read 2 more answers