question 25 of 29\nan astronaut floating in space is trying to use her jetpack to get back to her space…

question 25 of 29\nan astronaut floating in space is trying to use her jetpack to get back to her space station, but she is being pulled away by a nearby planet, as shown in the image below. her jetpack provides a constant thrust of 165 n. if she angles her jetpack in such a way that it cancels out the vertical force due to the planets gravity, what is her net horizontal force?\njetpack thrust 165 n\n45°\n20°\nplanets gravitational pull 80 n\na. 98.5 n toward the space station\nb. 132.6 n toward the space station\nc. 53.3 n away from the space station\nd. 112.8 n away from the space station

question 25 of 29\nan astronaut floating in space is trying to use her jetpack to get back to her space station, but she is being pulled away by a nearby planet, as shown in the image below. her jetpack provides a constant thrust of 165 n. if she angles her jetpack in such a way that it cancels out the vertical force due to the planets gravity, what is her net horizontal force?\njetpack thrust 165 n\n45°\n20°\nplanets gravitational pull 80 n\na. 98.5 n toward the space station\nb. 132.6 n toward the space station\nc. 53.3 n away from the space station\nd. 112.8 n away from the space station

Answer

Explanation:

Step1: Resolve jet - pack thrust into horizontal component

The jet - pack thrust is $F = 165$ N and the angle with the horizontal is $\theta=20^{\circ}$. The horizontal component of the jet - pack thrust is given by $F_x=F\cos\theta$. So, $F_x = 165\times\cos(20^{\circ})$. Using a calculator, $\cos(20^{\circ})\approx0.9397$, then $F_x=165\times0.9397 = 155.0505$ N.

Step2: Resolve gravitational pull into horizontal component

The gravitational pull is $F_g = 80$ N and the angle with the horizontal is $45^{\circ}$. The horizontal component of the gravitational pull is $F_{gx}=F_g\cos(45^{\circ})$. Since $\cos(45^{\circ})=\frac{\sqrt{2}}{2}\approx0.7071$, then $F_{gx}=80\times0.7071 = 56.568$ N.

Step3: Calculate net horizontal force

The net horizontal force $F_{net,x}=F_x - F_{gx}$. Substitute the values: $F_{net,x}=155.0505-56.568 = 98.4825\approx98.5$ N. The positive sign indicates the direction is toward the space station.

Answer:

A. 98.5 N toward the space station