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sergij07 [2.7K]

2 years ago

4

The IKAROS spacecraft, launched in 2010, was designed to test the feasibility of solar sails for spacecraft propulsion. These la

rge, ultralight sails are pushed on by the force of light from the sun, so the spacecraft doesn't need to carry any fuel. The force on IKAROS's sails was measured to be 1.12 mN. If this were the only force acting on the 290 kg spacecraft, by how much would its speed increase after 3.0 months of flight? Assume there are 30 days in each month.

1 answer:

jeka942 years ago

7 0

Answer:

increase in speed, Δv = 30.0m/s

Explanation:

Given:

Force on the spacecraft, F = 1.12 x 10⁻³ N,

Mass on spacecraft, m=290 kg

time, Δt = 3 month= 3 x 30 days =90 x 24 x 60x 60 seconds = 7776000 s

Now,

the acceleration (a) is given as:

a = F/m = (1.12 x 10⁻³)/290 = 3.86 x 10⁻⁶ m/s²

Applying Newton's equation of motion

v = u +aΔt

where,

v is the final velocity of the spacecraft after 3 months

u is the velocity of the spacecraft initially

v - u = (3.86 x 10⁻⁶) x 7776000

⇒ Δv = 30.0 m/s

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A bobsled is pushed with a force of 190.08 N. The sled has a mass of 28 kg. What is the acceleration of the bobsled? Report to t

Usimov [2.4K]

By definition it is known that force equals mass by acceleration. In other words F = m * a. To find the acceleration, you must clear the formula mentioned. Therefore, for a force of 190.08N and a mass of 28 Kg, we have that the acceleration is a = F / m = (190.08) / (28) = 6.79 m / s ^ 2

6 0

2 years ago

Which is NOT a "commonsense" psychological myth?

disa [49]

Answer:A. infants who listen to Mozart are smarter than those that don't.

Explanation: Common Sense is a sound practical judgement of the scheme of things, it can be easily understood as one thinks about life situation.

All other options such as MOST OLDER PEOPLE LIVE SAD AND SOLITARY LIVES ARE CORRECT AS THEY ARE KNOWN TO BE MORE LIKELY TO FALL SICK EASILY.

EATING "COMFORT FOOD" MAKES YOU FEEL HAPPIER IS KNOWN TO BE TRUE AND "COMMONSENSE"

UNLEASHING ANGER CAN MAKE YOU MORE AGGRESSIVE IS A "COMMONSENSE".

only the first option which says infants who listen to Mozart are smarter than those that don't listen to Mozart is not a "commonsense" psychological myth.

6 0

2 years ago

A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th

zheka24 [161]

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(c) E = 17.1416 J

(d) K = 11.8835 J

U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40 ⇒ f = 8.40 / (2π)

⇒ f = 1.3368 Hz

(c) the total energy,

we use the formula

E = m*ω²*A² / 2

⇒ E = (1.15)(8.40)²(0.650)² / 2

⇒ E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²) (the kinetic energy)

and

U = (1/2)*m*ω²*x² (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒ K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒ U = 5.2581 J

4 0

2 years ago

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

2 years ago

A river flows with a uniform velocity vr. A person in a motorboat travels 1.22 km upstream, at which time she passes a log float

storchak [24]

Answer:

t ’= $\frac{1450}{0.6499 + 2 v_r}$ , v_r = 1 m/s t ’= 547.19 s

Explanation:

This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.

By the time the boat goes up the river

v_b - v_r = d / t

By the time the boat goes down the river

v_b + v_r = d '/ t'

let's subtract the equations

2 v_r = d ’/ t’ - d / t

d ’/ t’ = 2v_r + d / t

$t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }$

In the exercise they tell us

d = 1.22 +1.45 = 2.67 km= 2.67 10³ m

d ’= 1.45 km= 1.45 1.³ m

at time t = 69.1 min (60 s / 1min) = 4146 s

the speed of river is v_r

t ’= $\frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \ + 2 \ v_r}$

t ’= $\frac{1450}{0.6499 + 2 v_r}$

In order to complete the calculation, we must assume a river speed

v_r = 1 m / s

let's calculate

t ’= $\frac{ 1450}{ 0.6499 + 2 \ 1}$

t ’= 547.19 s

8 0

2 years ago