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2 years ago

5

a lunar probe is moving 1670 m/s at a 73.0 angle. it needs to land on the moon 3.71 x 10^8 m away in a 46.0 direction in 8.64 x

10^4 s. what is the magnitude of the acceleration that the engine must produce?

1 answer:

Oduvanchick [21]2 years ago

8 0

The magnitude of the acceleration that the engine must produce is 0.065 m/s².

<h3>What is acceleration?</h3>

Acceleration is given by the ratio of Resultant or total force acting on any object and the its mass.

It can also be defined as the rate change of velocity with time.

acceleration a = (Δv) / (Δt)

Given is a lunar probe is moving 1670 m/s at a 73° angle. it needs to land on the moon 3.71 x 10⁸ m away in a 46° direction in 8.64 x 10⁴ s.

The initial velocity will be

1670 x cos (73°-46°) = 1487.981 m/s

Using the second equation of motion, we have

s = ut + 1/2 at²

The final velocity will be zero after landing on Moon.

Substituting the values, we have

3.71 x 10⁸= 1487.981 x 8.64 x 10⁴ + 1/2 a ( 8.64 x 10⁴)²

Solving the equation, we get

acceleration, a = 0.065 m/s²

Thus, the magnitude of the acceleration that the engine must produce is 0.065 m/s²

Learn more about acceleration.

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The overall energy involved in the formation of CsCl from Cs(s) and Cl2(g) is −443 kJ/mol. Given the following information: heat

german

Answer : The magnitude of the lattice energy for CsCl is, 667 KJ/mole

Explanation :

The steps involved in the born-Haber cycle for the formation of $CsCl$ :

(1) Conversion of solid calcium into gaseous cesium atoms.

$Cs(s)\overset{\Delta H_s}\rightarrow Cs(g)$

$\Delta H_s$ = sublimation energy of calcium

(2) Conversion of gaseous cesium atoms into gaseous cesium ions.

$Ca(g)\overset{\Delta H_I}\rightarrow Ca^{+1}(g)$

$\Delta H_I$ = ionization energy of calcium

(3) Conversion of molecular gaseous chlorine into gaseous chlorine atoms.

$Cl_2(g)\overset{\frac{1}{2}\Delta H_D}\rightarrow Cl(g)$

$\Delta H_D$ = dissociation energy of chlorine

(4) Conversion of gaseous chlorine atoms into gaseous chlorine ions.

$Cl(g)\overset{\Delta H_E}\rightarrow Cl^-(g)$

$\Delta H_E$ = electron affinity energy of chlorine

(5) Conversion of gaseous cations and gaseous anion into solid cesium chloride.

$Cs^{1+}(g)+Cl^-(g)\overset{\Delta H_L}\rightarrow CsCl(s)$

$\Delta H_L$ = lattice energy of calcium chloride

To calculate the overall energy from the born-Haber cycle, the equation used will be:

$\Delta H_f^o=\Delta H_s+\Delta H_I+\Delta H_D+\Delta H_E+\Delta H_L$

Now put all the given values in this equation, we get:

$-443KJ/mole=76KJ/mole+376KJ/mole+121KJ/mole+(-349KJ/mole)+\Delta H_L$

$\Delta H_L=-667KJ/mole$

The negative sign indicates that for exothermic reaction, the lattice energy will be negative.

Therefore, the magnitude of the lattice energy for CsCl is, 667 KJ/mole

5 0

3 years ago

The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the

tigry1 [53]

Answer:

A) The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) The maximum speed of the plastic sphere will be 2.2 m/s

Explanation:

Hi there!

I´ve found the complete problem on the web:

<em>A toy launcher that is used to launch small plastic spheres horizontally contains a spring with a spring constant of 50. newtons per meter. The spring is compressed a distance of 0.10 meter when the launcher is ready to launch a plastic sphere.</em>

<em>A) Determine the elastic potential energy stored in the spring when the launcher is ready to launch a plastic sphere.</em>

<em>B) The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched. [Neglect friction.] [Show all work, including the equation and substitution with units.]</em>

<em />

A) The elastic potential energy (EPE) is calculated as follows:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compressing distance

EPE = 1/2 · 50 N/m · (0.10 m)²

EPE = 0.25 J

The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) Since there is no friction, all the stored potential energy will be converted into kinetic energy when the spring is released. The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the sphere.

v = velocity

The kinetic energy of the sphere will be equal to the initial elastic potential energy:

KE = EPE = 1/2 · m · v²

0.25 J = 1/2 · 0.10 kg · v²

2 · 0.25 J / 0.10 kg = v²

v = 2.2 m/s

The maximum speed of the plastic sphere will be 2.2 m/s

6 0

3 years ago

A Brayton cycle has air into the compressor at 95 kPa, 290 K, and has an efficiency of 50%. The exhaust temperature is 675 K. Fi

motikmotik

Answer:

The specific heat addition is 773.1 kJ/kg

Explanation:

from table A.5 we get the properties of air:

k=specific heat ratio=1.4

cp=specific heat at constant pressure=1.004 kJ/kg*K

We calculate the pressure range of the Brayton cycle, as follows

n=1-(1/(P2/P1)^(k-1)/k))

where n=thermal efficiency=0.5. Clearing P2/P1 and replacing values:

P2/P1=(1/0.5)^(1.4/0.4)=11.31

the temperature of the air at state 2 is equal to:

P2/P1=(T2/T1)^(k/k-1)

where T1 is the temperature of the air enters the compressor. Clearing T2

11.31=(T2/290)^(1.4/(1.4-1))

T2=580K

The temperature of the air at state 3 is equal to:

P2/P1=(T3/T4)^(k/(k-1))

11.31=(T3/675)^(1.4/(1.4-1))

T3=1350K

The specific heat addition is equal to:

q=Cp*(T3-T2)=1.004*(1350-580)=773.1 kJ/kg

3 0

3 years ago

What is the difference between contact and action-at-a-distance forces?

Amiraneli [1.4K]

Answer:

A

Contact forces are forces that require the actual contact (touching) of two pieces of matter. ... A field force is a force that works at a distance. No touching is required. Gravity is a good example of a field force, because it works whether or not an object is touching something or touching nothing at all.

4 0

3 years ago

Two spherical asteroids have the same radius R. Asteroid 1 hasmass M and asteroid 2 has mas 2M. The two asteroids are releasedfr

gtnhenbr [62]

Answer:

$v_1 =\sqrt{\dfrac{16GM}{15R}}$

$v_2 =\sqrt{\dfrac{4GM}{15R}}$

Explanation:

given,

mass of asteroid 1 = M

mass of asteroid 2 = 2M

radius of two asteroid = R

Distance between the asteroid = 10 R

Speed of the asteroid before collision = ?

using conservation of momentum

M u + 2M u' = M v₁ + 2 M v₂

initial speed of asteroid is equal to zero

0 = v₁ + 2 v₂

v₁ = -2 v₂

using conservation of momentum

initial potential energy is converted into potential energy and the kinetic energy of both the asteroids.

$\dfrac{GM(2M)}{10R}=\dfrac{GM(2M)}{2R}+\dfrac{1}{2}Mv_1^2 + \dfrac{1}{2}(2M)v_2^2$

$\dfrac{GM(2M)}{10R}-\dfrac{GM(2M)}{2R}=\dfrac{1}{2}M(-2v_2)^2 + \dfrac{1}{2}(2M)v_2^2$

$6v_2^2 = \dfrac{8GM}{5R}$

$v_2 =\sqrt{\dfrac{4GM}{15R}}$

now,

$v_1 =-2\sqrt{\dfrac{4GM}{15R}}$

$v_1 =\sqrt{\dfrac{16GM}{15R}}$

hence, the velocity of asteroid are

$v_1 =\sqrt{\dfrac{16GM}{15R}}$

$v_2 =\sqrt{\dfrac{4GM}{15R}}$

6 0

3 years ago