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3 years ago

What is the magnitude of the linear momentum of a 7.30 kg bowling ball going down the

Physics

1 answer:

padilas [110]3 years ago

3 0

Momentum = mass x velocity
m=7.3kg
v=20 m/s

momentum= 7.3(20) = 146 kg m/s

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A sound source is located somewhere along the x-axis. Experiments show that the same wave front simultaneously reaches listeners

guapka [62]

Answer:

a) x₀ = - 2 m , b) y = 4.47 m

Explanation:

A wave travels in the middle with constant speed, let's use the equation of uniform motion

v = d / t

t = d / v

The distance to the first listeners, see attached

d₁ = x₀-x

t = (x₀ +7) / v

The distance to the second listener

d₂ = x - x₀

t = (+ 3- x₀) / v

As the wave arrives at the same time, we can equal the two equations

(x₀ +7) / v = (3 -x₀) / v

x₀ + 7 = 3 - x₀

2 x₀ = 3 - 7

x₀ = -4/2

x₀ = - 2 m

b) The time it takes for the wave to reach the listeners of the x-axis, where the speed of sound is 340 m / s

t = 5/340

t = 0.0147 s

Let's look for the distance the wave travels for the listener axis and

v = d₃ / t

d₃ = v.t

d₃ = 340 * 0.0147

d₃ = 5 m

For the distance component we use the Pythagorean triangle

d₃² = x₀² + y²

y² = d₃² - x₀²

y = √ (d₃² -4)

y = √ (5² -4)

y = 4.47 m

6 0

3 years ago

Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed

olasank [31]

Answer:

Average acceleration is $(11.05)g\ m/s^2$

Explanation:

It is given that,

Initial velocity, u = 0

Final velocity, v = 6.5 km/s = 6500 m/s

Time taken, t = 60 s

Acceleration, $a=\dfrac{v-u}{t}$

$a=\dfrac{v}{t}$

$a=\dfrac{6500\ m/s}{60}$

$a=108.33\ m/s^2$

Since, $g=9.8\ m/s^2$

So, $a=(11.05)g\ m/s^2$

So, the angular acceleration of the missile is $(11.05)g\ m/s^2$ . Hence, this is the required solution.

4 0

3 years ago

An archer shoots an arrow with a velocity of 45.0m/s at an angle of 50.0degrees with the horizontal.An assistant standing on the

garri49 [273]

Answer:

a) u = 30.29 m/s

b) t = 2.09 s

Explanation:

given,

velocity = 45 m/s

angle (θ) = 50°

horizontal velocity = 45 cos 50°

time taken to reach 150 m.

times = $\dfrac{150}{45 cos 50^0}$

t = 5.19 s

a) height of arrow

$s = u t +\dfrac{1}{2}gt^2$

$s = v sin \theta \times t+\dfrac{1}{2}gt^2$

$s = 45 sin 50^0 \times 5.19 -\dfrac{1}{2}\times 9.81\times 5.19^2$

s = 46.78 m

v² - u² = 2 g s

u² = 2 × 9.81 × 46.78

u = 30.29 m/s

b) time taken by the apple = $\dfrac{u}{g}=\dfrac{30.29}{9.81}$

= 3.09 s

time after which it has to be thrown = 5.19-3.09 = 2.1 s

5 0

3 years ago

Using the formula 1/2 (m x v2), what is the kinetic energy of a 4 kg rock falling through the air at 5 m/s

aivan3 [116]

Answer:

KE = 50J

Explanation:

$KE = \frac{1}{2}mv^{2} \\\\ KE = \frac{1}{2}(4)(5)^{2} \\\\ KE = 2(25) \\\\ KE = 50J$

5 0

3 years ago

The kinetic energy of a moving object is E=12mv2. A 61 kg runner is moving at 10kmh. However, her speedometer is only accurate t

jek_recluse [69]

Answer:

$e=3367.2J$

% $e=1.43$ %

Explanation:

From the exercise we know two information. The real speed and the experimental measured by the speedometer

$v_{r}=10km/h=2.77m/s$

Since the speedometer is only accurate to within 0.1km/h the experimental speed is

$v_{e}=10km/h-0.1km/h=9.9km/h=2.75m/s$

Knowing that we can calculate Kinetic energy for the real and experimental speed

$E_{r}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.77m/s)^2=234023J$

$E_{e}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.75m/s)^2=230656J$

Now, the potential error in her calculated kinetic energy is:

$e=E_{r}-E_{e}=(234023-230656)J=3367.2J$

% $e=\frac{E_{r}-E_{e}}{E_{r}}x100=\frac{(234023-230656)J}{234023J}x100=1.43$ %

4 0

3 years ago