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Zigmanuir [339]

3 years ago

15

Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed

and time are classified). What is its average acceleration in m/s2 and in multiples of g (9.80 m/s2)?

1 answer:

olasank [31]3 years ago

4 0

Answer:

Average acceleration is $(11.05)g\ m/s^2$

Explanation:

It is given that,

Initial velocity, u = 0

Final velocity, v = 6.5 km/s = 6500 m/s

Time taken, t = 60 s

Acceleration, $a=\dfrac{v-u}{t}$

$a=\dfrac{v}{t}$

$a=\dfrac{6500\ m/s}{60}$

$a=108.33\ m/s^2$

Since, $g=9.8\ m/s^2$

So, $a=(11.05)g\ m/s^2$

So, the angular acceleration of the missile is $(11.05)g\ m/s^2$ . Hence, this is the required solution.

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It is 5.0 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 km/h

Ostrovityanka [42]

Answer:

Walking burns up more energy,1740000J

Explanation:

Given that the displacement is 5.0km, and running at 10km/h and uses, walking at 3km/hr and uses 290watts:

Energy consumption for running is calculated as:

$700watts=700j/s,d=5000m,v=\frac{25}{9}m/s\\\\\therefore E_c=\frac{d}{v}\times Energy \ usage\\\\=\frac{5000}{\frac{25}{9}}\times 700j/s\\\\=1260000J$

Energy consumption for walking is calculated as:

$290watts=290j/s,d=5000m,v=\frac{5}{6}m/s\\\\\therefore E_c=\frac{d}{v}\times Energy \ usage\\\\=\frac{5000}{\frac{5}{6}}\times 290j/s\\\\=1740000J$

Walking is a slower process hence the need for more energy over longer periods raltive to running the same distance.

Hence walking burns more energy; 1,740,000J. It burns more because you walk for a greater period of time.

5 0

3 years ago

How much electromagnetic energy is contained in each cubic meter near the Earth's surface if the intensity of sunlight under cle

WITCHER [35]

Answer:

344.8 x10^-8J/m³

Explanation:

Using=> energy intensity/ speed oflight

= 1000/2.9x10^8

= 344.8 x10^-8J/m³

3 0

3 years ago

A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate (a) the average speed and (

Margarita [4]

$\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 1\ mile = 1609.34 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 1/4 \ mile = 402.33 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \ V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \ in \ m/s} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}$

4 0

2 years ago

A satellite of mass m circles a planet of mass M and radius R in an orbit at a height 2R above the surface of the planet. What m

Goryan [66]

Answer:

ΔE = GMm/24R

Explanation:

centripetal acceleration a = V^2 / R = 2T/mr

T= kinetic energy

m= mass of satellite, r= radius of earth

= gravitational acceleration = GM / r^2

Now, solving for the kinetic energy:

T = GMm / 2r = -1/2 U,

where U is the potential energy

So the total energy is:

E = T+U = -GMm / 2r

Now we want to find the energy difference as r goes from one orbital radius to another:

ΔE = GMm/2 (1/R_1 - 1/R_2)

So in this case, R_1 is 3R (planet's radius + orbital altitude) and R_2 is 4R

ΔE = GMm/2R (1/3 - 1/4)

ΔE = GMm/24R

6 0

3 years ago

Two blocks of masses 20 kg and 8.0 kg are connected togetherby

Dmitry [639]

Answer:

The tension is 14 N

Explanation:

For this problem we have to use newton's law, so:

$F=m*a$

The second string is connected to the mass of 8 kg, but the mass of 8 kg is connected to the mass of 20kg, so we can say that the second string is handling the two masses. so:

$F=28kg*0.5\frac{m}{s^2}=14N$

3 0

3 years ago