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Natalka [10]
4 years ago
5

What would be the advantages and disadvantages of using a jet pack for transportation

Physics
1 answer:
scoundrel [369]4 years ago
3 0
PRO;
<span>-Huge impact on today's technology </span>
<span>-Easier transportation </span>
<span>-Light weight </span>

<span>CON; </span>
<span>-Jet pack may burn your back </span>
<span>-Only travels short distances </span>
<span>-Bad ending if your high in the air and you run out of fuel.</span>
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A solid sphere rolls along a horizontal, smooth surface at a constant linear speed without slipping. What is the ratio between t
dsp73

Answer:

option E

Explanation:

given,

Rotational Kinetic Energy, KE_r = \dfrac{1}{2}I\omega^2

Moment of inertia of the solid,

I = \dfrac{2}{5}MR^2

\omega = \dfrac{V}{R}

now,

KE_r = \dfrac{1}{2}I\omega^2

KE_r = \dfrac{1}{2}\times \dfrac{2}{5}MR^2 (\dfrac{V}{R})^2

KE_r =\dfrac{1}{5}MV^2......(1)

transnational kinetic energy

KE_t =\dfrac{1}{2}MV^2

Total kinetic energy

KE = \dfrac{1}{2}MV^2 + \dfrac{1}{5} MV^2

KE = \dfrac{7}{10}MV^2

ratio of rotational kinetic energy to the total kinetic energy

\dfrac{KE_r}{KE_t}=\dfrac{\dfrac{1}{5}MV^2}{\dfrac{7}{10}MV^2}

\dfrac{KE_r}{KE_t}=\dfrac{2}{7}

hence, the correct answer is option E

4 0
3 years ago
A race car traveling at 10. meters per second accelerates at the rate of 1.5 meters per seconds while traveling a distance of 7,
Arturiano [62]

The final speed of the car is 2) 150 m/s

Explanation:

Since the motion of the car is a uniformly accelerated motion, we can solve the problem by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the car in this problem, we have

u = 10 m/s

a=1.5 m/s^2

s = 7,467 m

Solving for v, we find the final velocity (and speed) of the car:

v=\sqrt{u^2+2as}=\sqrt{10^2+2(1.5)(7,467)}=150 m/s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

4 0
3 years ago
Read 2 more answers
A car moving at constant speed in a straight line with the engine providing a driving force equal to the resistive force F.
Lunna [17]

125 b

simultaneous kinematic equations two variables are F and stopping distance

6 0
3 years ago
Read 2 more answers
Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on
Romashka [77]

Answer:

q_1=\pm0.03 \mu C and q_2=\pm0.02 \mu C.

Explanation:

According to Coulomb's law, the magnitude of  force between two point object having change q_1 and q_2 and by a dicstanced is

F_c=\frac{1}{4\pi\spsilon_0}\frac{q_1q_2}{d^2}-\;\cdots(i)

Where, \epsilon_0 is the permitivity of free space and

\frac{1}{4\pi\spsilon_0}=9\times10^9 in SI unit.

Before  dcollision:

Charges on both the sphere are q_1 and q_2, d=20cm=0.2m, and F_c=1.35\times10^{-4} N

So, from equation (i)

1.35\times10^{-4}=9\times10^9\frac{q_1q_2}{(0.2)^2}

\Rightarrow q_1q_2=6\times10^{-16}\;\cdots(ii)

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

\frac{q_1+q_2}{2}

d=20cm=0.2m, and F_c=1.406\times10^{-4} N

So, from equation (i)

1.406\times10^{-4}=9\times10^9\frac{\left(\frac{q_1+q_2}{2}\right)^2}{(0.2)^2}

\Rightarrow (q_1+q_2)^2=2.50\times10^{-15}

\Rightarrow q_1+q_2=\pm5\times 10^{-8}

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

\Rightarrow q_1+q_2=5\times 10^{-8}\;\cdots(iii)

\Rightarrow q_1=5\times 10^{-8}-q_2

The equation (ii) become:

(5\times 10^{-8}-q_2)q_2=6\times10^{-16}

\Rightarrow -(q_2)^2+5\times 10^{-8}q_2-6\times10^{-16}=0

\Rightarrow q_2=3\times10^{-8}, 2\times10^{-8}

From equation (iii)

q_1=2\times10^{-8}, 3\times10^{-8}

So, the magnitude of initial charges on both the sphere are 3\times10^{-8} Coulombs=0.03 \mu C and 2\times10^{-8} Colombs or 0.02 \mu C.

Considerion the nature of charges too,

q_1=\pm0.03 \mu C and q_2=\pm0.02 \mu C.

4 0
3 years ago
An ice cube measures 1.38 in. on each edge and weighs 1.39 oz. What is the density of the ice cube in g/cm^3
brilliants [131]

Answer:

0.91 g/cm³

Explanation:

Density: This can be defined as the ratio of mass of an object to its volume.

The mathematical expression for density is given as,

D = m/v .............................. Equation 1

Where D = Density of the ice cube, m = mass of the ice cube, v = volume of the ice cube.

v = l³

Where l = length of each side of the cube.

l = 1.38 in = (1.38×2.54) cm = 3.51 cm

v = 3.51³

v = 43.24 cm³

Given: m = 1.39 oz = (1.39×28.35) g = 39.41 g

Substitute into equation 1

D = 39.41/43.24

D = 0.91 g/cm³

Hence the density of the ice cube = 0.91 g/cm³

4 0
4 years ago
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