Ask question

Ask question

All categories

2 years ago

7

A 40.9 kg satellite circles planet Cruton every 5.82 h. The magnitude of the gravitational force exerted on the satellite by Cru

ton is 105 N. (a) What is the radius of the orbit

1 answer:

iragen [17]2 years ago

5 0

Answer:

r = 2.8524 x10⁷ m = 28524 km

Explanation:

First, we calculate the angular speed of the satellite:

$\omega = \frac{Arc Length}{Time}\\\\For\ One\ Complete\ Revolution\ around\ Cruton:\\\\\omega = \frac{2\pi\ rad}{5.82\ h}\frac{1\ h}{3600\ s}\\\omega = 3\ x\ 10^{-4} rad/s$

Now, we use the formula for the gravitational force. Since gravitational force will be acting as the centrifugal force due to circular motion. Therefore,

$F = \frac{mv^{2}}{r}\\$

where,

F = Gravitational Forc = 105 N

m = mass of satellite = 40.9 kg

v = linear speed of satellite = rω = r(3 x 10⁻⁴ rad/s)

r = radius of orbit = ?

Therefore,

$105\ N = \frac{(40.9\ kg)(r*3\ x\ 10^{-4}\ rad/s)^{2}}{r}\\r = \frac{105}{(40.9)(9\ x\ 10^{-8})} \\$

<u>r = 2.8524 x10⁷ m = 28524 km</u>

You might be interested in

What do the arrows in this photograph represent

babunello [35]

Answer: D. An action-reaction force pair

Explanation: When you sit in your chair, your body exerts a downward force on the chair and the chair exerts an upward force on your body. There are two forces resulting from this interaction - a force on the chair and a force on your body. Another example would be a person pushing against a wall (action force), and the wall exerts an equal and opposite force against the person.

7 0

2 years ago

Four objects are situated along the y axis as follows: a 1.99-kg object is at 2.99 m, a 2.96-kg object is at 2.57 m, a 2.43-kg o

Dominik [7]

Answer:

The center of mass for the object is $y_c = 1.063 \ m$ from the origin

Explanation:

From the question we are told that

The mass of the first object is $m_1 = 1.99 \ kg$

The position of first object with respect to origin $y_1 = 2.99 \ m$

The mass of the second object is $m_2 = 2.96 \ kg$

The position of second object with respect to origin $y_2 = 2.57 \ m$

The mass of the third object is $m_3 = 2.43 \ kg$

The position of third object with respect to origin $y_3 = 0 \ m$

The mass of the fourth object is $m_3 = 3.96 \ kg$

The position of fourth object with respect to origin $y_3 = -0.502 \ m$

Generally the center of mass of the object along the x-axis is zero because all the mass lie on the y axis

Generally the location of the center mass of the object is mathematically represented as

$y_c = \frac{m_1 * y_1 + m_2 * y_2 + m_3 * y_3 + m_4 * y_4}{m_1 + m_2 + m_3 + m_4}$

=> $y_c = \frac{1.99 * 2.99 + 2.96 * 2.57 + 2.43 * 0 + 3.96 * (-0.502)}{1.99+ 2.96 + 2.43 + 3.96}$

=> $y_c = 1.063 \ m$

3 0

3 years ago

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

3 years ago

An uncrewed mission to the nearest star, Proxima Centauri, is launched from the Earth's surface as a projectile with an initial

Anna [14]

Answer:

42.96 km/s

Explanation:

From the conservation of Energy

$(PE+KE)_i=(PE+KE)_f\\\Rightarrow -\frac{GmM}{R}+\frac{1}{2}mv_i^2=0+\frac{1}{2}mv_f^2$

Mass gets cancelled

$-\frac{GM}{R}+\frac{1}{2}v_i^2=0+\frac{1}{2}v_f^2\\\Rightarrow -2\frac{GM}{R}+v_i^2=v_f^2\\\Rightarrow -v_e^2+v_i^2=v_f^2\\\Rightarrow v_f=\sqrt{v_i^2-v_e^2}$

$v_e=\sqrt{\frac{2Gm}{R}}$ = Escape velocity of Earth = 11.2 km/s

$v_i$ = Velocity of projectile = 44.4 km/s

$v_f=\sqrt{44.4^2-11.2^2}\\\Rightarrow v_f=42.96\ km/s$

The velocity of the spacecraft when it is more than halfway to the star is 42.96 km/s

6 0

2 years ago

En la siguiente expresión matemáticas w=mg el peso w con relación a se relaciona con la masa m en una proporción

s2008m [1.1K]

Answer:

a) Directamente proporcional

Explanation:

El peso se puede definir como la fuerza que actúa sobre un cuerpo o un objeto como resultado de la gravedad.

Matemáticamente, el peso de un objeto viene dado por la fórmula;

$Peso = mg$

Donde;

m es la masa del objeto.

g es la aceleración debida a la gravedad.

De la expresión matemática, podemos deducir que el valor del peso de un objeto es directamente proporcional a la masa del objeto.

Por lo tanto, un aumento en la masa de un objeto provocaría un aumento en el peso del objeto y viceversa.

4 0

3 years ago