Ask question

Ask question

All categories

katen-ka-za [31]

2 years ago

12

An astronaut is standing on the surface of a planet that has a mass of 6.42×1023 kg and a radius of 3397 km. The astronaut fires

a 2.6-g bullet straight up into the air with an initial velocity of 406 m/s. What is the greatest height the bullet will reach? The planet has no atmosphere.

1 answer:

mr_godi [17]2 years ago

7 0

Answer:

22.2 km

Explanation:

3397 km = 3397000m

Let gravitational constant $G = 6.674\times10^{-11}m^3/kgs^2$ . We can calculate the (constant) gravitational acceleration on this planet using Newton's gravitational law

$a = G\frac{M}{R^2}$

where M and R are the mass and radius of the plannet, respectively

$a = 6.674\times10^{-11}\frac{6.42\times10^{23}}{3397000^2} = 3.71 m/s^2$

When the bullet is travelling to its highest point, its kinetic energy is converted to potential energy:

$E_p = E_k$

$mah = mv^2/2$

where m is the bullet mass and h is the vertical distance traveled, v = 406 m/s is the bullet velocity at the firing point

We can divide both sides by m

$ah = v^2/2$

$h = \frac{v^2}{2a} = \frac{406^2}{2*3.71} = 22196.88m$ or 22.2 km

You might be interested in

A mercury thermometer has a glass bulb of interior volume 0.100 cm3 at 10°c. the glass capillary 10) tube above the bulb has an

Nadya [2.5K]

Initial volume of mercury is
V = 0.1 cm³

The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.

Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³

The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm

Answer: 4.5 cm

7 0

2 years ago

Read 2 more answers

A system delivers 1275 j of heat while the surroundings perform 855 j of work on it. calculate ∆esys in j.

kakasveta [241]

The first law of thermodynamics says that the variation of internal energy of a system is given by:
$\Delta U = Q + W$
where Q is the heat delivered by the system, while W is the work done on the system.

We must be careful with the signs here. The sign convention generally used is:
Q positive = Q absorbed by the system
Q negative = Q delivered by the system
W positive = W done on the system
W negative = W done by the system

So, in our problem, the heat is negative because it is releaed by the system:
Q=-1275 J
while the work is positive because it is performed by the surrounding on the system:
W=+855 J

So, the variation of internal energy of the system is
$\Delta U = -1275 J+855 J=-420 J$

6 0

2 years ago

A 17 g audio compact disk has a diameter of 12 cm. The disk spins under a laser that reads encoded data. The first track to be r

sladkih [1.3K]

Answer:

0.00066518 Nm

Explanation:

v = Velocity = 1.2 m/s

r = Distance to head = 2.3 cm

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

t = Time taken = 2.4 s

Angular speed is given by

$\omega=\dfrac{v}{r}\\\Rightarrow \omega=\dfrac{1.2}{0.023}\\\Rightarrow \omega=52.17391\ rad/s$

From equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{52.17391-0}{2.4}\\\Rightarrow \alpha=21.73912\ rad/s^2$

Torque

$\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mR^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}0.017\times 0.06^2\times 21.73912\\\Rightarrow \tau=0.00066518\ Nm$

The torque of the motor is 0.00066518 Nm

6 0

2 years ago

A golfer hits a golf ball at an angle of 25.0Â° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is

kvasek [131]

<u>Answer:</u>

Maximum height reached = 35.15 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Considering upward vertical motion of projectile.

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g $m/s^2$ and final velocity = 0 m/s.

0 = u sin θ - gt

t = u sin θ/g

Total time for vertical motion is two times time taken for upward vertical motion of projectile.

So total travel time of projectile = 2u sin θ/g

Horizontal motion:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

Vertical motion (Maximum height reached, H) :

We have equation of motion, $v^2=u^2+2as$ , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

$0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}$

In the give problem we have R = 301.5 m, θ = 25° we need to find H.

So $\frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g$

Now we have $H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m$

So maximum height reached = 35.15 meter.

7 0

2 years ago

Imagine that the above hoop is a tire. the coefficient of static friction between rubber and concrete is typically at least 0.9.

Stels [109]

The hoop is attached.

Consider that the friction force is given by:
F = μ·N
   = μ·m·g·cosθ

We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ

Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
</span>μ = (1/2)·m·g·sinθ / <span>m·g·cosθ
   = </span>(1/2)·tanθ

Now, solve for θ:
θ = tan⁻¹(2·μ)
   = tan⁻¹(2·0.9)
   = 61°

Therefore, the maximum angle <span>you could ride down without worrying about skidding is 61°.</span>

5 0

2 years ago