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1 year ago

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A 15.0 g bullet traveling horizontally at 865 m>s passes through a tank containing 13.5 kg of water and emerges with a speed

of 534 m>s. What is the maximum temperature increase that the water could have as a result of this event

1 answer:

Murrr4er [49]1 year ago

3 0

Answer:

The rise in temperature is 0.06 K.

Explanation:

mass of bullet, m = 15 g

initial speed, u = 865 m/s

final speed, v = 534 m/s

mass of water, M = 13.5 kg

specific heat of water, c = 4200 J/kg K

The change in kinetic energy

$K = 0.5 m(u^2 - v^2)\\\\K = 0.5\times 0.015\times (865^2-534^2)\\\\K = 3473 J$

According to the conservation of energy, the change in kinetic energy is used to heat the water.

K = m c T

where, T is the rise in temperature.

3473 = 13.5 x 4200 x T

T = 0.06 K

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Answer:

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Explanation:

Given that

Length= 2L

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$sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}$

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So

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IgorLugansk [536]

Answer:

E. downward and constant

Explanation:

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An ambulance moving at 42 m/s sounds its siren whose frequency is 450 hz. a car is moving in the same direction as the ambulance

Korvikt [17]

(a) Since the ambulance and the car are moving one relative to each other, we have to use the general formula of the Doppler effect, which gives us the shift of the frequency of the siren as heard by an observer in the car:
$f'=( \frac{v+v_o}{v+v_s} )f$
where
f' is the apparent frequency as heard by the observer in the car
v is the velocity of the wave
$v_o$ is the velocity of the observer (positive if it is moving towards the source, negative if it is moving away)
$v_s$ is the velocity of the source (positive if the source is moving away from the observer, negative if is is moving towards it)
f is the real frequency of the sound

In the first part of the problem:
$v=343 m/s$ (speed of the sound wave)
$v_o =-25 m/s$ (the car is moving away from the ambulance)
$v_s = -42 m/s$ (the ambulance is moving towards the car)
$f=450 Hz$ (original frequency of the sound)

If we plug the numbers into the formula, we find
$f'=( \frac{343 m/s-25 m/s}{343 m/s-42 m/s} )(450 Hz)=475 Hz$

b) This time, the ambulance passes the car, so the ambulance is now moving away from the car; this means that $v_s$ must be positive:
$v_s=+42 m/s$
Moreover, the car is now moving towards the ambulance, so we should reverse also the sign of $v_o$ :
$v_o=+25 m/s$
All the other data do not change, so if we use the same formula as before, we find
$f'=( \frac{343 m/s+25 m/s}{343 m/s+42 m/s} )(450 Hz)=430 Hz$

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1 year ago

An object moving on the x axis with a constant acceleration increases its x coordinate by 82.9 m in a time of 2.51 s and has a v

Aneli [31]

We are given: Final velocity ( $v_f$ )=20 m/s .

Time t= 2.51 s and

distance s = 82.9 m.

We know, equation of motion

$v_f = v_i + at.$

Let us plug values of final velocity, and time in above equation.

$20=v_i+a(2.51)$

$20=v_i+2.51a$

Subtracting 2.51a from both sides, we get

$20-2.51a=v_i$ -----------equation(1)

Using another equation of motion

$v_f-v_i=2as$

Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.

We get,

$20-(20-2.51a)=2*a(82.90)$

Now, we need to solve it for a.

20-20+2.51a=165.8a.

-163.29a=0

a=0.

So, the acceleration would be 0 m/s^2.

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A person wants to lose weight by "pumping iron". The person lifts an 80 kg weight 1 meter. How many times must this weight be li

statuscvo [17]

Answer:

37357 sec

or 622 min

or 10.4 hrs

Explanation:

GIVEN DATA:

Lifting weight 80 kg

1 cal = 4184 J

from information given in question we have

one lb fat consist of 3500 calories = 3500 x 4184 J

= 14.644 x 10^6 J

Energy burns in 1 lift = m g h

= 80 x 9.8 x 1 = 784 J

lifts required $= \frac{(14.644 x 10^6)}{784}$

= 18679

from the question,

1 lift in 2 sec.

so, total time = 18679 x 2 = 37357 sec

or 622 min

or 10.4 hrs

3 0

1 year ago