Ask question

Ask question

All categories

2 years ago

6

During a test a rocket travels upward at 90 m/s , and when it is 50 m from the ground its engine fails. Determine the maximum he

ight sb reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of 9.81 m / s 2 due to gravity. Neglect the effect of air resistance.

2 answers:

Maslowich2 years ago

8 0

Answer:

Maximum height sb = 462.84m

Speed before it hits the ground v = 95.29 m/s

Explanation:

Given;

Initial height before engine fails h1 = 50 m

Rocket speed before engine fail u =90m/s

Acceleration due to gravity g = 9.81m/s^2

Maximum height from point of engine fail h2;

h2 = u^2/2g = 90^2/(2×9.81)

h2 = 412.84 m

Maximum height sb = h1 + h2 = 50m + 412.84m

sb = 462.84m

The speed before touching the ground v;

v = √2gsb = √(2×9.81×462.84)

v = 95.29 m/s

katrin [286]2 years ago

6 0

Answer:

Maximum height above ground = 462.8m

Speed before hitting ground = 95.3m/s

Explanation:

Detailed explanation and calculation is shown in the image below

You might be interested in

Item 19 Question 1 A Ferris wheel has a radius of 75 feet. You board a car at the bottom of the Ferris wheel, which is 10 feet a

sergey [27]

Answer:

$y = 104.4 ft$

Explanation:

As we know that we board in the car of ferris wheel at the bottom position

So we will have

final height of the car at angular displacement given as

$y = y_o + R + R sin(270 - 255)$

$y = y_o + R + R sin 15$

here we know that

$y_o = 10 ft$

$R = 75 ft$

so we have

$y = 10 + 75 + 75 sin15$

$y = 104.4 ft$

4 0

2 years ago

A 61.0-kg person jumps from rest off a 10.0-m-high tower straight down into the water. Neglect air resistance. She comes to rest

9966 [12]

Answer:

Explanation:

In this case, law of conservation of energy will be implemented. It states that "the energy of the system remains conserved until or unless some external force act on it. Energy of the system may went through the conversion process like kinetic energy into potential and potential into kinetic energy.But their total always remain the same in conserved systems."

Given data:

Height of tower = 10.0 m

Depth of the pool = 3.00 cm

Mass of person = 61.0 kg

Solution:

Initial energy = Final energy

$U_{i} = (K.E) + U_{f}$

As the person was at height initially so it has the potential energy only.

$mg(h_{1} +h_{2}) = K.E + mgh_{2}$

$K.E = mgh_{1}$

$K.E = (61.0)(9.8)(10)\\K.E = 5978 J$

Lets find out the magnitude of the force that the water is exerting on the diver.

W =ΔK.E

$F.h_{2} = 5978\\$

$F = \frac{5978}{3}$

F = 1992.67 N

7 0

2 years ago

Calculate the wave length of a water wave with a speed of 20 m/s and a frequency of 2.5 Hz

12345 [234]

Wavelength of the water wave is 8 m

Explanation:

Wavelength measures the distance between two successive crests or troughs of the wave. It is given by the following equation

λ = v/f, where f is the frequency, v is the velocity of the wave

Here, v = 20 m/s and f = 2.5 Hz

⇒ λ = 20/2.5

= 8 m

5 0

2 years ago

A charged cloud system produces an electric field in the air near earth's surface. a particle of charge -2.0 × 10-9 c is acted o

defon

Part a)

Magnitude of electric field is given by force per unit charge

$E = \frac{F}{q}$

$E = \frac{4.3 * 10^{-6}}{2 * 10^{-9}}$

$E = 2150 N/C$

Part b)

Electrostatic force on the proton is given as

F = qE

$F = 1.6 * 10^{-19} * 2150$

$F = 3.44 * 10^{-16} N$

PART C)

Gravitational force is given by

$F_g = mg$

$F_g = 1.6 * 10^{-27}*9.8$

$F_g = 1.57 * 10^{-26} N$

PART d)

Ratio of electric force to weight

$\frac{F_e}{F_g} = \frac{3.44 * 10^{-16}}{1.57*10^{-26}}$

$\frac{F_e}{F_g} = 2.2 * 10^{10}$

7 0

3 years ago

The acceleration due to gravity on the surface of Mars is about one-third the acceleration due to gravity on Earthâ€™s surface.

aksik [14]

Answer:

one-third of its weight on Earth's surface

Explanation:

Weight of an object is = W = m*g

Gravity on Earth = g₁ = 9.8 m/s

Gravity on Mars = g₂ = $\frac{1}{3}$ g₁

Weight of probe on earth = w₁ = m * g₁

Weight of probe on Mars = w₂ = m * g₂ -------- ( 1 )

As g₂ = g₁/3 --------- ( 2 )

Put equation (2) in equation (1)

so

Weight of probe on Mars = w₂ = m * g₁ /3

Weight of probe on Mars = $\frac{1}{3}$ m * g₁ = $\frac{1}{3}$ w₁

⇒Weight of probe on Mars = $\frac{1}{3}$ Weight of probe on earth

6 0

3 years ago