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miss Akunina [59]

2 years ago

6

A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward a

cceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 64 m and acquired a velocity of 60 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The time interval (in s), during which the rocket engine provides upward acceleration, is closest to: ? .

1 answer:

Dvinal [7]2 years ago

4 0

Answer:

2.13 s

Explanation:

Hi!

At t = 0s the rocket is at rest in its platform, so the intial speed is zero. I f the acceleration is A, then the height Y, and the speed V are:

$Y=\frac{A}{2}t^2$

$V=At$

We nedd to find time T during which the rocket engine provides upward acceleration. We know that:

$64\;m=\frac{A}{2}T^2\\ 60\frac{m}{s} =AT\\$

With these 2 equations we can find A and T (dropping units for simplicity):

$A=\frac{60}{T} \\64 =\frac{30}{T} T^2=30T\\T=\frac{64}{30}\approx 2.13\;s$

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A professional boxer hits his opponent with a 1025 N horizontal blow that lasts 0.150 s. The opponent's total body mass is 116 k

mylen [45]

Answer:

The impulse is $I = 153.8 \ N \cdot s$

The opponents velocity is $v= 1.33 m/s$

The opponents head recoils velocity $v_r = 30.8 \ m/s$

Explanation:

From the question we are told that

The force of the blow is $F= 1025 \ N$

The duration of the blow is $t = 0.150$

The mass of the opponent is $m_o = 116 \ kg$

The mass of the opponents head is $m_h = 5 \ kg$

The impulse the boxer imparts is mathematically represented as

$I = F * t$

substituting values

$I = 1025 * 0.150$

$I = 153.8 \ N \cdot s$

The impulse can also be mathematically evaluated as

$I = m_o * v$

substituting values

$153.8 = 116 * v$

$v= \frac{153.8}{116}$

$v= 1.33 m/s$

The recoil velocity is mathematically represented as

$v_r = \frac{I}{m_h}$

substituting values

$v_r = \frac{153.8}{5}$

$v_r = 30.8 \ m/s$

5 0

2 years ago

Calculate the distance d from the center of the sun at which a particle experiences equal attractions from the earth and the sun

fredd [130]

Answer:

149.34 Giga meter is the distance d from the center of the sun at which a particle experiences equal attractions from the earth and the sun.

Explanation:

Mass of earth = m = $5.976\times 10^{24} kg$

Mass of Sun = M = 333,000 m

Distance between Earth and Sun = r = 149.6 gm = 1.496\times 10^{11} m[/tex]

1 giga meter = $10^{9} meter$

Let the mass of the particle be m' which x distance from Sun.

Distance of the particle from Earth = (r-x)

Force between Sun and particle:

$F=G\frac{M\times m'}{x^2}=G\frac{333,000 m\times m'}{x^2}$

Force between Sun and particle:

$F'=G\frac{mm'}{(r-x)^2}$

Force on particle is equal:

F = F'

$G\frac{333,000 m\times m'}{x^2}=G\frac{mm'}{(r-x)^2}$

$\frac{x}{r-x}=\sqrt{333,000}$ = ±577.06

Case 1:

$\frac{x}{r-x}=577.06$

x = $1.49\times 10^{11} m=149.34 Gm$

Acceptable as the particle will lie in between the straight line joining Earth and Sun.

Case 2:

$\frac{x}{r-x}=-577.06$

x = $1.49\times 10^{11} m=149.86 Gm$

Not acceptable as the particle will lie beyond on line extending straight from the Earth and Sun.

3 0

2 years ago

A car drives off a cliff next to a river at a speed of 30 m/s and lands on the bank on theother side. The road above the cliff i

dezoksy [38]

Answer:1.301 s

Explanation:

Given

Initial Velocity(u)=30 m/s

Height of cliff=8.3 m

Time taken to cover 8.3 m

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

here Initial vertical velocity is 0

$8.3=\frac{gt^2}{2}$

$t^2=1.69$

$t=1.301 s$

Horizontal distance

$R=u\times t$

$R=30\times 1.301=39.04 m$

7 0

2 years ago

A 0.110 kg cube of ice (frozen water) is floating in glycerine. The glycerine is in a tall cylinder that has inside radius 3.70

Sonbull [250]

Answer:

the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm

Explanation:

The change in volume of glycerin when the ice cube is placed on the surface of the glycerin can be represented as:

$V = \frac{m}{ \rho}$

Given that ;

the mass of the ice cube (m) = 0.11 kg = 0.110 × 10³ g

density of the glycerine ( $\rho$ ) = 1.260 kg/L = 1.260 g/cm³

Then:

V = $\frac{0.110*10^3 \ g}{1.260 \ g/cm^3}$

V = $0.0873*10^3 \ cm^3 (\frac{1L}{10^3 cm^3})$

V = 0.0873 L

Now;Initially the volume of the glycerin before the ice cube starts to melt is:

$V_1 = V_i + V\\\\V_1 = V_i+ 0.0873 \ L$

However; the volume of the water produced by the 0.11 kg ice cube = $0.11*10^3 \ cm^3$

The expression for change in the volume of glycerin after the ice cube starts to melt is as follows:

$V_2 = V_i + V"$

replacing $V"$ with $0.11*10^3 \ cm^3$ ; we have:

$V_2 = V_i (0.11*10^3 \ cm^3 )(\frac{1 \ L }{10^3 \ cm^3})$

$V_2 = V_i + 0.11 \ L$

The overall total change in the volume of the glycerin is illustrated as:

$V_f = V_2 - V_1$

Now; from the foregoing ; lets replace the respective value of $V_2$ and $V_1$ in the above equation ; we have;

$V_f = (V_i + 0.11 \ L) - (V_i + \ 00873 \ L)\\ \\V_f = 0.11 L - 0.0873 \ L\\\\V_f = 0.0227 \ L$

The formula usually known to be the volume of a cylinder is :

$V = \pi r ^2 h$

For the question ; we will have:

$V_f = \pi r ^2 h$

making h the subject of the formula ; we have:

$h = \frac{V_f}{\pi r^2}$

replacing 0.0227 L for $V_f$ and the given value of radius which is = 3.70 cm; we have:

$h = \frac{0.0227 \ L ( \frac{10^3 \ cm^3}{1\ L})}{\pi * (3.70 cm)^2}$

$h = \frac{22.7 \ cm^3}{\pi * (3.70 cm)^2}\\\\h = 0.528 \ \ cm$

Thus ; the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm

8 0

2 years ago

A 0.20 kg mass on a horizontal spring is pulled back 2.0 cm and released. If, instead, a 0.40 kg mass were used in this same exp

KIM [24]

Answer:

The total mechanical energy does not change if the value of the mass is changed. That is, remain the same

Explanation:

The total mechanical energy of a spring-mass system is equal to the elastic potential energy where the object is at the amplitude of the motion. That is:

$E=U=\frac{1}{2}kA^2$ (1)

k: spring constant

A: amplitude of the motion = 2.0cm

As you can notice in the equation (1), the total mechanical energy of the system does not depend of the mass of the object. It only depends of the amplitude A and the spring constant.

Hence, if you use a mass of 0.40kg the total mechanical energy is the same as the obtained with a mas 0.20kg

Remain the same

8 0

2 years ago