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QveST [7]
3 years ago
14

A small boat coasts at constant speed under a bridge. A heavy sack of sand is dropped from the bridge onto the boat. The speed o

f the boat
a. does not change.
b. increases.
c. Without knowing the mass of the boat and the sand, we can't tell.
d. decreases
Physics
1 answer:
Tju [1.3M]3 years ago
7 0

Answer:

d. decreases

Explanation:

The law of conservation of momentum tells us that the sum of momenta before the collision is equal to the sum of momenta after the collision. The bag has no momentum as it falls onto the boat because its velocity is zero in the horizontal direction. But after it hits the boat, it's momentum increases while the momentum of the system remains the same. That means a component of the system must decrease somewhere else. And that component is the velocity, not the mass, of the boat.

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A string of length L, fixed at both ends, is capable of vibrating at 309 Hz in its first harmonic. However, when a finger is pla
Scorpion4ik [409]

Answer:

 i = 0.3326 L

Explanation:

A fixed string at both ends presents a phenomenon of standing waves, two waves with the same frequency that are added together. The expression to describe these waves is

    2 L = n λ           n = 1, 2, 3…

The first harmonic or leather for n = 1

Wave speed is related to wavelength and frequency

     v = λ f

     λ = v / f

Let's replace in the first equation

    2 L = 1 (v / f₁)

For the shortest length L = L-l

   2 (L- l) = 1 (v / f₂)

These two equations form our equation system, let's eliminate v

    v = 2L f₁

    v = 2 (L-l) f₂

    2L f₁ = 2 (L-l) f₂

    L- l = L f₁ / f₂

    l = L - L f₁ / f₂

    l = L (1- f₁ / f₂)

.

Let's calculate

    l / L = (1- 309/463)

    i / L = 0.3326

4 0
3 years ago
A car in motion has kinetic energy. What happens to the kinetic energy when the car brakes to a stop? The kinetic energy is tran
Pie

Answer:

<em>When a moving car brakes to a stop the </em><em>kinetic energy of the car is converted to heat energy. </em>

Explanation:

A moving car has kinetic energy.

It is given by the equation k=\frac{1}{2} mv^2

Where m denotes mass of the car and v denote sits velocity. When the brakes are applied the velocity becomes zero and the car doesn’t possess kinetic energy anymore.

According to law of conservation of energy can neither be created nor be destroyed but can only be transformed from one form to another. On coming to a stop,  the kinetic energy of the car  gets converted to heat. The friction between the tyre and the road heats up the tyre.

7 0
3 years ago
Choose the law each sentence describes. This law relates a planet's orbital period and its average distance to the Sun. The orbi
hram777 [196]

These are the Kepler's laws of planetary motion.

This law relates a planet's orbital period and its average distance to the Sun. - Third law of Kepler.

The orbits of planets are ellipses with the Sun at one focus. - First law of Kepler.

The speed of a planet varies, such that a planet sweeps out an equal area in equal time frames. - Second law of Kepler.

7 0
3 years ago
Read 2 more answers
018 10.0 points
-BARSIC- [3]

Answer:

jfjcgufnfhfufm TV fifnricnrhkddufnfif km fgkfkvntfmrugrhfifnh r

6 0
2 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
2 years ago
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