PHYSICS COLLEGE

A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.9 seconds after it is launched, the probe is at location <5600, 7200, 0> m, and at this same instant its momentum is <51000, -7000, 0> kg·m/s. At this instant, the net force on the probe due to the gravitational pull of Mars plus the air resistance acting on the probe is <-4000, -780, 0> N.Assuming that the net force on the probe is approximately constant during this time interval, what is the change of the momentum of the probe in the time interval from 22.6 seconds after the probe is launched to 22.9 seconds after the launch?

Answers

Answer 1

Answer:

Answer:

The change in momentum is $\Delta p = <-1200 , -234 , 0> kg \cdot m/s$

Explanation:

From the question we are told that

The mass of the probe is $m = 170 kg$

The location of the prob at time t = 22.9 s is $A = <5600, 7200,0>$

The momentum at time t = 22.9 s is $p = < 51000, -7000, 0> kg m/s$

The net force on the probe is $F = <-4000 , -780 , 0> N$

Generally the change in momentum is mathematically represented as

$\Delta p = F * \Delta t$

The initial time is 22.6 s

The final time is 22.9 s

Substituting values

$\Delta p = <-4000 , -780 , 0> * (22.9 - 22.6)$

$\Delta p = <-4000 , -780 , 0> * (0.3)$

$\Delta p = <-1200 , -234 , 0> kg \cdot m/s$

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What is the force required to accelerate 1 kilogram of mass at 1 meter per second per second.1 Newton

1 pound

1 kilometer

1 gram

Answers

Answer:

it's answer is 1 newton

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.60 m/s^2 . At the same instant a truck, traveling with a constant speed of 23.5 m/s , overtakes and passes the car. a. How far beyond its starting point does the car overtake the truck?b. How fast is the car traveling when it overtakes the truck?

Answers

Answer:

306.8264448 m

47.0016 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Distance traveled by car

$s_c=ut+(1)/(2)at^2\n\Rightarrow s_c=(1)/(2)at^2$

Distance traveled by truck

$s_t=ut$

In order to overtake both distances should be equal

$(1)/(2)at^2=ut\n\Rightarrow (1)/(2)at=u\n\Rightarrow t=(2u)/(a)\n\Rightarrow t=(2* 23.5)/(3.6)\n\Rightarrow t=13.056\ s$

$s_c=(1)/(2)at^2\n\Rightarrow s_c=(1)/(2)3.6* 13.056^2\n\Rightarrow s_c=306.8264448\ m$

The distance the car has to travel is 306.8264448 m

$v^2-u^2=2as\n\Rightarrow v=√(2as+u^2)\n\Rightarrow v=√(2* 3.6* 306.8264448+0^2)\n\Rightarrow v=47.0016\ m/s$

The speed of the car when it overtakes the truck is 47.0016 m/s

A^^\-> points in the -x direction with a magnitude of 21. What is the y component of A^^\->

Answers

Answer:

$A_y=-36.37^(\circ)$

Explanation:

Given that,

Vector A points in the -x direction with a magnitude of 21.

Let the x component is making an angle of 60 degrees with negative x axis. The x component of a vector is given by :

$A_x=A\ cos\theta$

$-21=A\ cos(60)$

$A=(-21)/(cos(60))$

A = -42 units

The y component of a vector is given by :

$A_y=A\ sin\theta$

$A_y=-42\ sin(60)$

$A_y=-36.37^(\circ)$

So, the y component of vector A is (-36.37) degrees. Hence, this is the required solution.

Two point charges are placed on the x axis.The firstcharge, q1= 8.00 nC, is placed a distance 16.0 mfromthe origin along the positive x axis; the second charge,q2= 6.00 nC, is placed a distance 9.00 mfrom the originalong the negative x axis.[Give the x and y components of the electric fieldas an ordered pair. Express your answer innewtons per coulomb to three significant figures.Keep in mind that an x component that points tothe right is positive and a y component thatpoints upward is positive.]

Answers

Answer:

E = (0, 0.299) N

Explanation:

Given,

Charge $q_1\ =\ 8.0\ nC$
Charge $q_2\ =\ 6.0\ nC$
Distance of the first charge from the origin = (16m, 0)
Distance of the second charge from the origin = (-9, 0)
Point where the electric field required = (0, 12m)

Let $\theta_1\ and\ theta_2$ be the angle of the electric fields by first and second charge at the point A.

$\therefore sin\theta_1\ =\ (12)/(20)\n\Rightarrow \theta_1\ =\ sin^(-1)\left ((12)/(20)\ \right )\n\Rightarrow \theta_1\ =\ 36.87^o\n\n\therefore sin\theta_1\ =\ (12)/(9)\n\Rightarrow \theta_1\ =\ sin^(-1)\left ((12)/(9)\ \right )\n\Rightarrow \theta_1\ =\ 53.13^o\n$

Electric field by charge $q_1$ at point A,

$F_1\ =\ (kq_1)/(r_1^2)\n\Rightarrow F_1\ =\ (9* 10^9* 8* 10^(-9))/(20^2)\n\Rightarrow F_1\ =\ 0.18\ N/C$

Electric field by the charge $q_2$ at point A,

$F_1\ =\ (kq_1)/(r_1^2)\n\Rightarrow F_1\ =\ (9* 10^9* 6.0* 10^(-9))/(16^2)\n\Rightarrow F_1\ =\ 0.24\ N/C$

Now,

Net electric field in horizontal direction at point A $F_x\ =\ F_(1x)\ +\ F_(2x)\n\Rightarrow F_x\ =\ F_1cos\theta_1\ +\ F_2cos\theta_2\n\Rightarrow F_x\ =\ 0.18*( -cos36.87^o)\ +\ 0.24* cos53.13^o\n\Rightarrow F_x\ =\ -0.144\ +\ 0.144\ N/C\n\Rightarrow F_x\ =\ 0\ N/C$

Net electric field in vertical direction at point A.

$F_y\ =\ F_(1y)\ +\ F_(2y)\n\Rightarrow F_y\ =\ F_1sin\theta_1\ +\ F_2sin\theta_2\n\Rightarrow F_y\ =\ 0.18* sin36.87^o\ +\ 0.24* sin53.13^o\n\Rightarrow F_y\ =\ 0.180\ +\ 0.192\n\Rightarrow F_y\ =\ 0.299\ N/C$

Hence, the net electric field at point A,

$F\ =\ ( 0, 0.299 )\ N/C.$

An infant throws 7 g of applesauce at a velocity of 0.5 m/s. All of the applesauce collides with a nearby wall and sticks to it. What is the decrease in kinetic energy of the applesauce?

Answers

Answer:

Δ KE = - 8.75 x 10⁻⁴ J

Explanation:

given,

mass of applesauce = 7 g = 0.007 Kg

initial velocity, u = 0.5 m/s

final velocity, v = 0 m/s

Decrease in kinetic energy = ?

initial kinetic energy

$KE_1=(1)/(2)mu^2$

$KE_1=(1)/(2)* 0.007 * 0.5^2$

KE₁ = 8.75 x 10⁻⁴ J

final kinetic energy

$KE_2=(1)/(2)mv^2$

$KE_2=(1)/(2)* 0.007 * 0^2$

KE₂ =0 J

Decrease in kinetic energy

Δ KE = KE₂ - KE₁

Δ KE = 0 - 8.75 x 10⁻⁴

Δ KE = - 8.75 x 10⁻⁴ J

decrease in kinetic energy of the applesauce is equal to 8.75 x 10⁻⁴ J

Final answer:

The decrease in kinetic energy of the applesauce, when it hits the wall and stops, is the initial kinetic energy of it. Using the formula of kinetic energy, the decrease is calculated to be 0.000875 Joules.

Explanation:

This question relates to the concept of kinetic energy in physics. Kinetic energy is calculated by the formula 0.5 * mass (kg) * velocity (m/s)^2. So the initial kinetic energy of the applesauce right after being thrown was 0.5 * 0.007 kg * (0.5 m/s)^2 = 0.000875 Joules.

When the applesauce hits the wall and stops, its velocity drops to 0. Thus, its kinetic energy also goes to 0 (because kinetic energy is proportional to the square of velocity).

Therefore, the decrease in kinetic energy is the same as the initial kinetic energy of the applesauce, which is 0.000875 Joules.

Learn more about Kinetic Energy here:

brainly.com/question/33783036

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A parallel-plate capacitor is constructed from two aluminum foils of 1 square centimeter area each placedon both sides of a rubber square of the same size. The rubber dielectric is 2.5 mm thick, hasr2.5, andbreakdown field strength of 25 megavolts per meter. Find the voltage rating of the capacitor using a safetyfactor of 10.

Answers

Answer:

The voltage will be 0.0125V

Explanation:

See the picture attached

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