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3 years ago

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A model rocket is fired vertically from rest. It has a constant acceleration of 16.4 m/s2 for the first 1.50 s. Then its fuel is

exhausted, and it is in free fall. The rocket has a mass of 78.2 g; the mass of the fuel is much less than 78.2 g. Ignore air resistance. What was the downward net force on the rocket after its fuel was spent?

1 answer:

il63 [147K]3 years ago

3 0

Answer:

The force downward acting o the rocket is F=m*g=0.0782kg*9.8m/s^2=0.76636 N

Explanation:

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A pendulum of length L = 1 m is released from an initial angle of 15° . After 1000 s, its amplitude is reduced to 2.5º. What is

jonny [76]

Answer:

The value of the time constant is 558.11 sec.

Explanation:

Given that,

Pendulum length = 1 m

Initial angle = 15°

Time = 1000 s

Reduced amplitude = 2.5°

We need to calculate the value of the time constant

Using formula of damping oscillation

$\theta=\theta_{0}e^{\dfrac{t}{\tau}}$

Where, $\theta$ =amplitude

$\theta_{0}$ =amplitude at t = 0

Put the value into the formula

$2.5=15e^{\dfrac{-1000}{\tau}}$

$\dfrac{1}{6}=e^{\dfrac{-1000}{\tau}}$

$ln\dfrac{1}{6}=\dfrac{-1000}{\tau}$

$\tau=\dfrac{1000}{-ln\dfrac{1}{6}}$

$\tau=558.11\ sec$

Hence, The value of the time constant is 558.11 sec.

6 0

3 years ago

A 0.140 kg stone rests on a frictionless, horizontal surface. A bullet of mass 8.50 g , traveling horizontally at 320 m/s , stri

Novay_Z [31]

Answer:

a) $v = 34.607\,\frac{m}{s}$ (Positive), b) $e = 0.803$ . The collision is not perfectly elastic.

Explanation:

a) The collision can be described by the Principle of Momentum Conservation and Principle of Energy Conservation:

$(0.140\,kg)\cdot (0\,\frac{m}{s} ) + (0.0085\,kg)\cdot (320\,\frac{m}{s} ) = (0.0085\,kg)\cdot (-250\,\frac{m}{s} ) + (0.140\,kg)\cdot v$

The final velocity of the rock is:

$v = 34.607\,\frac{m}{s}$

b) The coefficient of restitution is the best criterion to distinguish elastic collsions from inelastic collisions, such criterion is the ratio of final energy of the system to initial energy of the system:

$e = \frac{\frac{1}{2}\cdot [(0.140\,kg)\cdot (34.607\,\frac{m}{s} )^{2}+(0.0085\,kg)\cdot (-250\,\frac{m}{s} )^{2}] }{\frac{1}{2}\cdot [(0.140\,kg)\cdot (0\,\frac{m}{s} )^{2}+(0.0085\,kg)\cdot (320\,\frac{m}{s} )^{2}] }$

$e = 0.803$

The collision is not perfectly elastic.

6 0

4 years ago

Read 2 more answers

Why are gases poor conductors of sound?

vesna_86 [32]

Answer:

Their atoms are spread farther apart

5 0

2 years ago

Read 2 more answers

An object with mass of 4kg is thrown with initial velocity of 20m/s from point A and follows the track of ABCD.

soldier1979 [14.2K]

<u>The distance of the length YD is 3.57m, the velocity of the object at point D is 27m/s and the time required for the object to reach the ground is 0.17s</u>

Data given;

Mass = 4kg

Initial velocity = 20m/s

length CD = 5m (from the image given)

a)

<h3>Determine The Length of YD</h3>

$YD=CDsin$ θ = $5sin45=\frac{5}{\sqrt{2} } = 3.57m$

b)

<h3>The velocity of the object at point D</h3>

The change in kinetic energy is given as

Δ in kinetic energy = Δ in potential energy + work done by friction

K.E - 1/2m $v^2$ = mgh $_1$ - mgh $_2$ + (-μmg.x)

K.E = mg(50 - 3.57) + (-mg(0.3*100) + 1/2 m $v^2$

$\frac{1}{2}mv^2_f=mg(46.43)-mg(30)+\frac{1}{2} m(400)\\\\v^2_f=20(16.43)+400\\v^2_f=728.6\\v=\sqrt{728.6} \\v_f=26.99 = 27m/s$

The velocity of the object at D with a distance of 5m.

c)

<h3>The the required for the object to reach ground</h3>

The velocity of the object in the y-axis is

$v_y=vsin45=19.09$

Acceleration in y-axis = 9.8

Height = 3.57m

h = $ut+\frac{1}{2}at^2$

$3.57=19.28(t)+\frac{1}{2}(9.8)t^2\\3.57=19.28t+4.9t^2\\4.9t^2+19.28t-3.57=0\\a=4.9, b=19.28, c= -3.57\\t=\frac{-19.28+\sqrt{(19.28)^2-4(4.9)(-3.57)} }{(2*4.9)} \\t=\frac{-19.28+21.02}{9.8}$

Taking the positive value

$\frac{-19.28+21.02}{9.8}=0.17s$

The time required for the object to reach ground is 0.17s

From the calculations above,<u> the distance YD is 3.57m, the velocity of the object at point D is 27m/s and the time required for the object to reach ground is 0.17s</u>

Learn more on projectile motion here;

brainly.com/question/1130127

<h3 />

5 0

3 years ago

In Part 3.2.2 of the experiment, a pair of students measure θmirror = 19.5 degrees. To find θbench, they use the theoretical rel

KATRIN_1 [288]

Answer:

discrepancy between theory and experiment 19.23 %

Explanation:

given data

Experimental theta (E) = 46.5 degrees

and

Theoretical theta (T) = 2 × 19.5 = 39 degree

so we can say % discrepancy will be

% discrepancy = $\frac{E-T}{T} * 100$ .....................1

so

% discrepancy = $\frac{46.5-39}{39} * 100$

% discrepancy = $\frac{7.5}{39} * 100$

% discrepancy = 19.23 %

so discrepancy between theory and experiment 19.23 %

5 0

3 years ago