Science, who ever gets this answer will get a brainlest

Why doesn’t a rocket in space need to use its engine to keep it moving??? Please help!

Studentka2010 [4]

Every action produces an equal and opposite reaction. When a rocket shoots fuel out one end, this propels the rocket forward

7 0

3 years ago

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You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves down a frictionless track to a height o

irina [24]

Answer: 20.765 m/s

Explanation:

This problem can be solved by the conservation of energy principle, this means the initial energy $E_{o}$ must be equal to the final energy $E_{f}$ :

$E_{o}=E_{f}$ (1)

Where each energy is the sum of kinetic energy $K$ and potential energy $U$ :

$K_{o}+U_{o}=K_{f}+U_{f}$ (2)

Where:

$K_{o}=\frac{1}{2}mV_{o}^{2}$

Being $m$ your mass and $V_{o}=0 m/s$ your initial velocity, since the roller coaster sterted from rest.

$U_{o}=mgh_{o}$

Being $g=9.8 m/s^{2}$ the acceleration due gravity and $h_{o}=25 m$ your initial height

$K_{f}=\frac{1}{2}mV_{f}^{2}$

Being $V_{f}$ your final velocity

$U_{f}=mgh_{f}$

Being $h_{f}=3 m$ your final height

Rewritting (2):

$\frac{1}{2}mV_{o}^{2}+mgh_{o}=\frac{1}{2}mV_{f}^{2}+mgh_{f}$ (3)

$mgh_{o}=m(\frac{1}{2}V_{f}^{2}+gh_{f})$ (4)

Isolating $V_{f}$ :

$V_{f}=\sqrt{2g(h_{o}-h_{f})}$ (5)

$V_{f}=\sqrt{2(9.8 m/s^{2})(25 m-3 m)}$ (6)

Finally:

$V_{f}=20.765 m/s$ This is your spedd when you arrive at 3 m height

7 0

4 years ago

1. What is the wave speed of a wave that has a frequency of 100 Hz and a wavelength of 0.30 m?

aivan3 [116]

Answer:

1. v = 30 m/s

2. v = 5 m/s

3. f = 40 Hz

4. f = 400 Hz

5. f = 300 Hz

6. λ = 0.772 m

7. λ = 0.386 m

8. λ = 0.625 m

9. v = 100 m/s

10. v = 50 m/s

Explanation:

The relationship between frequency, wavelength, and speed of a wave is given by the following formula:

$v = f\lambda$

where,

v = speed of wave

f = frequency of wave

λ = wavelength

1.

f = 100 Hz

λ = 0.3 m

Therefore,

v = (100 Hz)(0.3 m)

v = 30 m/s

2.

f = 50 Hz

λ = 0.1 m

v = (50 Hz)(0.1 m)

v = 5 m/s

3.

v = 20 m/s

λ = 0.5 m

$f = \frac{v}{\lambda} = \frac{20\ m/s}{0.5\ m}$

f = 40 Hz

4.

v = 80 m/s

λ = 0.2 m

$f = \frac{v}{\lambda}=\frac{80\ m/s}{0.2\ m}$

f = 400 Hz

5.

v = 120 m/s

λ = 0.4 m

$f = \frac{v}{\lambda}=\frac{120\ m/s}{0.4\ m}$

f = 300 Hz

6.

v = 340 m/s

f = 440 Hz

$\lambda = \frac{v}{f}=\frac{340\ m/s}{440\ Hz}\\$

λ = 0.772 m

7.

v = 340 m/s

f = 880 Hz

$\lambda = \frac{v}{f}=\frac{340\ m/s}{880\ Hz}\\$

λ = 0.386 m

8.

v = 250 m/s

f = 400 Hz

$\lambda = \frac{v}{f}=\frac{250\ m/s}{400\ Hz}\\$

λ = 0.625 m

9.

f = 50 Hz

λ = 2 m

v = (50 Hz)(2 m)

v = 100 m/s

10.

f = 100 Hz

λ = 0.5 m

v = (100 Hz)(0.5 m)

v = 50 m/s

6 0

3 years ago

The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of

adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = $ma_a (0.35)$

$1.8v^2(1.1)$ - 1200(9.8)(1.25) = 1200a(0.35)

$1.8v^2(1.1)$ - 14700 = 420 a ------- equation (1)

$F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a$ --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

${1.1 * 1200 \ a} - 14700 = 420 a$

1320 a - 14700 = 420 a

1320 a - 420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

5 0

3 years ago

The first Leyden jar was probably discovered by a German clerk named E. Georg von Kleist. Because von Kleist was not a scientist

skelet666 [1.2K]

Answer:

a) q = 4.47 10⁻⁵ C

b) ΔV = 4.47 10⁴ V

Explanation:

A Leyden bottle works as a condenser that accumulates electrical charge, so we can use the formula of the energy stored in a capacitor

U = Q² / 2C

Q = √ (2UC)

let's reduce the magnitudes to the SI system

c = 1 nF = 1 10⁻⁹ F

let's calculate

q = √ (2 1 10⁻⁹-9)

q = 0.447 10⁻⁴ C

q = 4.47 10⁻⁵ C

b) for the potential difference we use

C = Q / ΔV

ΔV = Q / C

ΔV = 4.47 10⁻⁵ / 1 10⁻⁹

ΔV = 4.47 10⁴ V

8 0

3 years ago