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Mrac [35]
3 years ago
11

An observer stands 24.7 m behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of

the bullets is 814.8 m/s, and the air temperature is 20°C. How far does each bullet travel before the observer hears the report of the rifle? Assume that the bullets encounter no obstacles during this interval, and ignore both air resistance and the vertical component of the bullets' motion.
Physics
1 answer:
GaryK [48]3 years ago
8 0

Explanation:

The given data is as follows.

     Velocity of bullet, c_{p} = 814.8 m/s

    Observer distance from marksman, d = 24.7 m

Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.

               t = \frac{24.7}{343}      (velocity in air = 343 m/s)

                 = 0.072 sec

Now, before the observer hears the report the distance traveled by the bullet is as follows.

               d_{b} = c_{b} \times t

                          = 814.8 \times 0.072

                          = 58.66

                          = 59 (approx)

Thus, we can conclude that each bullet will travel a distance of 59 m.

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A uniform, 0.0300-kg rod of length 0.400 m rotates in a horizontal plane about a fixed axis through its center and perpendicular
AURORKA [14]

Answer:

A) ω₂ = 28 rev/min

B) ω_3 = 28 rev/min

Explanation:

A) Initial moment of inertia (I₁) is the rod's own ((1/12)ML²) plus that of the two rings (two lots of mr² if we treat each ring as a point mass).

Thus;

I₁ = ((1/12)ML²) + 2(mr²)

Where;

M is mass of rod = 0.03 kg

L is length of rod = 0.4m

m is mass of small rings = 0.02 kg

r is radius of small rings = 0.05 m

Thus;

I₁ = ((1/12) x 0.03 x 0.4²) + 2(0.02 x 0.05²)

I₁ = 0.0021 kg.m²

Now, when the rings slide and reach the ends of the rod i.e at r = L/2 = 0.4/2 = 0.2m from axis, the new Moment of inertia is;

I₂ = ((1/12) x 0.03 x 0.4²) + 2(0.02 x 0.2²)

I₂ = 0.0036 kg.m²

From conservation of angular momentum, we know that:

I₁ω₁ = I₂ω₂

We are given ω₁ = 48 rev/min.

Thus; plugging in the relevant values;

0.0021 x 48 = 0.0036ω₂

0.1008 = 0.0036ω₂

ω₂ = 0.1008/0.0036

ω₂ = 28 rev/min

b) For this, we have the same scenario as the case above where the ring just reaches the ends of the rods.

Thus,

I₂ω₂ = I_3•ω_3

So,0.0021 x 48 = 0.0036ω_3

ω_3 = 28 rev/min

7 0
3 years ago
How can the momentum of an object be changed?
pickupchik [31]
C. You’re welcome





Leave a like
3 0
4 years ago
Calculate Neptune's mass given the acceleration due to gravity at the north pole is 11.529 m/s2 and the radius of Neptune at the
dolphi86 [110]

Answer:

The mass of Neptune is 1.023\times 10^{26} kilograms.

Explanation:

From Newton's Law of Gravitation, the gravitational acceleration of Neptune is determined by the following formula:

g = \frac{G\cdot M}{R^{2}} (1)

Where:

G - Gravitational constant, measured in cubic meters per kilogram-square second.

M - Mass of the planet, measured in kilograms.

R - Radius of the planet, measured in meters.

g - Gravitational acceleration, measured in meters per square second.

If we know that G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, g = 11.529\,\frac{m}{s^{2}} and R = 24.340\times 10^{6}\,m, then the mass of Neptune is:

M = \frac{g\cdot R^{2}}{G}

M = \frac{\left(11.529\,\frac{m}{s^{2}}\right)\cdot (24.340\times 10^{6}\,m)^{2} }{6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} }

M = 1.023\times 10^{26}\,kg

The mass of Neptune is 1.023\times 10^{26} kilograms.

5 0
3 years ago
HEY CAN YALL HELP ME IN DIS!!
Natasha_Volkova [10]

It might me same volume

5 0
4 years ago
A 50kg block slides down a slope that forms an angle of 54 degrees if it is known that when descending it has a force of 40N and
Harrizon [31]

Answer:

The acceleration in the block is 2.1 m/s²

Explanation:

Given that,

Mass = 50 kg

Angle = 54°

Force = 40 N

Coefficient of friction = 0.33

We need to calculate the acceleration in the block

Using balance equation

F_{net}=F_{f}-F\cos\theta

ma=\mu mg\sin\theta-F\cos\theta

a=\dfrac{\mu mg\sin\theta-F\cos\theta}{m}

Put the value into the formula

a=\dfrac{0.33\times50\times9.8\sin54-40\cos54}{50}

a=2.1\ m/s^2

Hence, The acceleration in the block is 2.1 m/s²

3 0
3 years ago
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