The gasoline in a car does 40,000 J of work on a car and generates a constant force of 20 N. How far did the car go?

A couple of astronauts agree to rendezvous in space after hours. Their plan is to let gravity bring them together. She has a mas

Oduvanchick [21]

Answer:

(a) Acceleration of female astronaut = 9.33*10^-12 m/s^2; Acceleration of male astronaut = 7.56*10^-12 m/s^2

(b) 27.013 days

(c) No, their accelerations would not be constant.

Explanation:

In the given question, we have:

Mass of female astronaut $M_{1}$ = 60.0 kg

Mass of male astronaut $M_{2}$ = 74.0 kg

Distance between them (S) = 23.0 m

(a) The free-body diagram is shown in the attached figure.

Using the equations below, the initial accelerations of the two astronauts can be calculated:

Force of gravity (F) = $\frac{G*M_{1}*M_{2}}{S^{2} }$

G = 6.67*10^-11 $\frac{m^{3} }{kg*s^{2} }$

F = (6.67*10^-11 *60*74)/23^2 = 5.598*10^-10 N

For the female astronaut, her initial acceleration = F/ $M_{1}$ = 5.598*10^-10/60 = 9.33*10^-12 m/s^2

For the male astronaut, his acceleration = F/ $M_{2}$ = 5.598*10^-10/74 = 7.56*10^-12 m/s^2

(b) Since the different between their mass is not much, we can deduce that:

$a_{average} = \frac{a_{1}+a_{2}}{2}$ = (9.33*10^-12 + 7.56*10^-12)/2 = 8.445*10^-12 m/s^2

Using the equation below, we can calculate the the time:

S = ut + 1/2 (at^2) where u = 0

23 = 1/2 (8.445*10^-12)*t^2

t^2 = 5.447*10^12

t = 2333883.044 s = 27.013 days

(c) No, their accelerations will not be constant. It will increase because their radii would be decreasing.

8 0

3 years ago

Determine the velocity that a car should have while traveling around a frictionless curve of radius 100m and that is banked 20 d

alex41 [277]

Answer:

$v=18.89\frac{m}{s}$

Explanation:

From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:

$\sum F_x:Nsin(20^\circ)=ma_c(1)\\\sum F_y:Ncos(20^\circ)=mg(2)$

Solving N from (2) and replacing in (1):

$N=\frac{mg}{cos(20^\circ)}\\(\frac{mg}{cos(20^\circ)})sin(20^\circ)=ma_c\\gtan(20^\circ)=a_c$

The centripetal acceleration is given by:

$a_c=\frac{v^2}{r}$

Replacing and solving for v:

$gtan(20^\circ)=\frac{v^2}{r}\\v=\sqrt{grtan(20^\circ)}\\v=\sqrt{9.8\frac{m}{s^2}(100m)tan(20^\circ)}\\v=18.89\frac{m}{s}$

4 0

4 years ago

A thin, metallic spherical shell of radius 0.187 m has a total charge of 6.53×10−6 C placed on it. A point charge of 5.15×10−6 C

MAVERICK [17]

Answer: a) 112.88 * 10^3 N/C; b) The electric field point outward from the center of the sphere.

Explanation: In order to solve this problem we have to use the gaussian law so we use a gaussian surface at r=0.965 m and the electric flux is equal to Q inside/εo

E* 4*π*r^2= Q inside/εo

E= k*Q inside/r^2= 9*10^9*(6.53+5.15)μC/(0.965)^2=122.88 * 10 ^3 N/C

5 0

3 years ago

Which energy changes take place when a pedalling cyclist uses a generator (dynamo) to light his bicycle lamp?

Elina [12.6K]

Answer: chemical → kinetic → electrical → light

Explanation: At the same time when the generator lights up the bicycle lamp the lamp lights up using electrical energy so mechanical energy is also transformed into electrical energy.

8 0

2 years ago

Two small, positively charged spheres have a combined charge of 5.23 x 10^-5 C. If each sphere is repelled from the other by an

pshichka [43]

Answer:

The smallest charge of the dial is 2.46 10-5 C

Explanation:

The Coulomb force is responsible for the electroactive repulsion, the equation that describes it is

F = k q1 q2 / r²

Where K is the Coulomb constant that is worth 8.99109 N m² / C², q is the electric charge of each sphere and r is the distance between them.

They also give us the condition that the sum of the charge is 5.23 10-5 C

Qt = q1 + q2 = 5.23 10⁻⁵ C

Let's replace in the Coulomb equation, let's clear and calculate

F = k (Qt -q2) q2 / r²

F = k q22 / r² - k Qt q2 / r²

1.0 = 8.99 10⁹ q2² /2.04² - 8.99 10⁹ 5.23 10⁻⁵ q2 / 2.04²

1.0 = 2.16 10⁹ q2²2 - 11.30 10⁴ q2

0 = 2.16 109 q2² - 11.30 10⁴ q2 -1.0 (* 1/2.16 109)

0 = q2² - 1.05 10⁻⁵ q2 - 0.463 10⁻⁹

Let's solve the second degree equation for q2

q2 = 1.05 10⁻⁵ ±√[(1.05 10⁻⁵)² - 4 1 (-0.463 10⁻⁹)] / 2

q2 = 1.05 10⁻⁵ ±√ [1.10 10⁻¹⁰ + 18.52 10⁻¹⁰] / 2

q2 = {1.05 10⁻⁵ ± 4.43 10⁻⁵} / 2

The solutions are

q2 ’= 2.74 10-5 C

q2 ’’ = -1.69 10-5 C

As the problem tells us that the spheres are positively charged, the correct solution is 2.74 10-5 C, let's see the charge of the other sphere

Qt = q1 + q2 ’

q1 = Qt -q2 ’

q1 = 5.23 10-5 - 2.74 10-5

q1 = 2.46 10-5 C

The smallest charge of the dial is 2.46 10-5 C

5 0

4 years ago