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Zielflug [23.3K]
4 years ago
9

Lightning is more likely to strike a metal tower than a rubber hose because:

Physics
1 answer:
Rudik [331]4 years ago
6 0

This is because metal is a conductor of electricity while rubber is not (rubber is an insulator). Therefore the metal is more amenable to 'ground' the high charge in the lightning than rubber would. This is also why, when lightning strikes metal, it is safely 'grounded' while if it strikes a non-conductor, most occasionally, it explodes into a fire.

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A cyclist is riding his bike up a mountain trail. When he starts up the trail , he is going 8 m/s. As the trail gets steeper, he
Mrac [35]

Acceleration = (change in speed) / (time for the change)

change in speed = (speed at the end) - (speed at the beginning)

Our cyclist's change in speed = (3 m/s) - (8 m/s) = -5 m/s

Acceleration = (-5 m/s) / (60 seconds)

<em>Acceleration = -1/12 m/s²</em>

7 0
4 years ago
1) A tiger leaps horizontally from a 7.5 m high rock with a speed of 4.5 m/s. How far from
Artyom0805 [142]

The distance of tiger's leap from the base of rock is 5.58 m

It is a question of two dimensional motion

The time of motion in two dimensional motion is given by:

t= \sqrt{2y/g}

where y is the height and g is  the acceleration due to gravity

y is given to be 7.5m and let us assume g to be 9.8 m/s^2

t = \sqrt{2 * 7.5/9.8}

 = 1.24s

Using time and speed,

We know that distance is the product of speed and time,

Distance= speed x time

speed is given to be 4.5 m/s

distance from the base of rock = 4.5 x 1.24

                                                    = 5.58m

Hence the distance of tiger's leap from the base of rock is 5.58 m

Disclaimer:

The acceleration due to gravity is assumed to be 9.8 m/s^2

For further reference:

brainly.com/question/11213880?referrer=searchResults

#SPJ9

8 0
1 year ago
If a current of 2.4 a is flowing in a cylindrical wire of diameter 2.0 mm, what is the average current density in this wire?
Gnom [1K]

The average current density in the wire is given by:

J=\frac{I}{A}

where I is the current intensity and A is the cross-sectional area of the wire.


The cross-sectional area of the wire is given by:

A=\pi r^2

where r is the radius of the wire. In this problem, r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m, so the cross-sectional area is

A=\pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2


and the average current density is

J=\frac{I}{A}=\frac{2.4 A}{3.14 \cdot 10^{-6} m^2}=7.64 \cdot 10^5 A/m^2

8 0
3 years ago
Read 2 more answers
A 0.05-m3 rigid tank initially contains refrigerant-134a at 0.8 MPa and 100 percent quality. The tank is connected by a valve to
lesya692 [45]

Answer:

A= 203 KJ

B= 54 Kg

Explanation:

The initial specific volumes and internal energies are obtained from A-12 for a given pressure and state. The enthalpy of the refrigerant in the supply line is determined using the saturated liquid approximation for the given temperature with data from A-11. The mass that has entered the tank is:

Δm = m₂ – m₁

= V(1/α₂ – 1/α₁)

= 0.05 (1/0.0008935 – 1/ 0.025645)Kg

= 54Kg

The heat transfer is obtained from the energy balance:

ΔU= m_i_n h_i_n+  Q_n_e_t

m₂u₂ – m₁u₂ = m_i_nh_i_n + Q_n_e_t

Q_n_e_t= m₂u₂ – m₁u₁ – m_i_n

= V/α₂u₂ - V/α₁u₁ – m_i_n

=(0.05/0.0008935 . 116.72 – 0.05/0.025645 . 246.82 – 54.108.28) Kj

= 203 KJ

8 0
3 years ago
An object of mass M oscillates on the end of a spring. To double the period, replace the object with one of mass: (a) 2M. (b)M/2
harkovskaia [24]

Answer:

(c) 4M

Explanation:

The system is a loaded spring. The period of a loaded spring is given by

T = 2\pi\sqrt{\dfrac{m}{k}}

<em>m</em> is the mass and <em>k</em> is the spring constant.

It follows that, since <em>k</em> is constant,

T\propto\sqrt{m}

\dfrac{T}{\sqrt{m}} = C where <em>C</em>  represents a constant.

\dfrac{T_1}{\sqrt{m_1}} = \dfrac{T_2}{\sqrt{m_2}}

m_2 = m_1\left(\dfrac{T_2}{T_1}\right)^2

When the period is doubled, T_2 = 2T_1.

m_2 = m_1\left(\dfrac{2T_1}{T_1}\right)^2 = 4m_1

Hence, the mass is replaced by 4M.

7 0
4 years ago
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