What is the first step for both loading and unloading a firearm?

(a) Calculate the acceleration due to gravity on the surface of the Sun.

Ira Lisetskai [31]

<h2>a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b) Factor of increase in weight is 27.95</h2>

Explanation:

a) Acceleration due to gravity

$g=\frac{GM}{r^2}$

Here we need to find acceleration due to gravity of Sun,

G = 6.67259 x 10⁻¹¹ N m²/kg²

Mass of sun, M = 1.989 × 10³⁰ kg

Radius of sun, r = 6.957 x 10⁸ m

Substituting,

$g=\frac{6.67259\times 10^{-11}\times 1.989\times 10^{30}}{(6.957\times 10^8)^2}\\\\g=274.21m/s^2$

Acceleration due to gravity on the surface of the Sun = 274.21 m/s²

b) Acceleration due to gravity in earth = 9.81 m/s²

Ratio of gravity = 274.21/9.81 = 27.95

Weight = mg

Factor of increase in weight = 27.95

8 0

3 years ago

An object accelerates in a direction that is always perpendicular to its motion.what is the effect if any of the acceleration on

mars1129 [50]

If its accelerating it will increase velocity in the direction of the acceleration which is perpendicular to the velocity.

3 0

3 years ago

If a Stone with an of original velocity of zero is falling from a ledge and takes eight seconds to hit the ground what is the fi

DaniilM [7]

Let t=time to reach the ground=8 secs, g= acceleration of gravity. The speed v on reaching the ground is gt=8g=78.4 m/s where g=9.8 m/s/s approx.

3 0

3 years ago

A ball is projected with an initial velocity of 40 meter per second and reached maximum height of 160 meters calculate tge angle

Andru [333]

There's a problem with the question as given. Even with a maximum projection angle of θ = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height y at time t would be

y = (40 m/s) t - 1/2 g t ²

Set y = 160 m, and you'll find that there is no (real) solution for t, so the ball never attains the given maximum height.

From another perspective: recall that

v ² - v₀² = 2a ∆y

where

• v₀ = initial velocity

• v = final velocity

• a = acceleration

• ∆y = displacement

At its maximum height, the ball has zero vertical velocity, and ∆y = maximum height = 160 m. The ball is in free fall once it's launched, so a = -g.

So we have

0² - (40 m/s)² = -2g (160 m)

but this reduces to

(40 m/s)² = 2 (9.8 m/s²) (160 m)

1600 m²/s² ≠ 3136 m²/s²

7 0

3 years ago

Radar uses radio waves of a wavelength of 2.9 m . The time interval for one radiation pulse is 100 times larger than the time of

Mandarinka [93]

Answer:

145 m

Explanation:

Given:

Wavelength (λ) = 2.9 m

we know,

c = f × λ

where,

c = speed of light ; 3.0 x 10⁸ m/s

f = frequency

thus,

$f=\frac{c}{\lambda}$

substituting the values in the equation we get,

$f=\frac{3.0\times 10^8 m/s}{2.9m}$

f = 1.03 x 10⁸Hz

Now,

The time period (T) = $\frac{1}{f}$

or

T = $\frac{1}{1.03\times 10^8}$ = 9.6 x 10⁻⁹ seconds

thus,

the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s

Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s

Now,

For radar to detect the object the pulse must hit the object and come back to the detector.

Hence, the shortest distance will be half the distance travelled by the pulse back and forth.

Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}

Thus, the minimum distance to target = $\frac{290}{2}$ = 145 m

6 0

3 years ago