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Goshia [24]
3 years ago
10

In a crash test, a 1000 kg automobile moving at 10 m/s crashes into a brick wall. How much energy goes into demolishing and warm

ing the wall and the auto?
Physics
1 answer:
marishachu [46]3 years ago
8 0

Answer:

Explanation:

Given

mass of automobile m=1000\ kg

velocity of automobile v=10\ m/s

If the automobile crashes into the brick wall then it's kinetic energy is converted into thermal and demolishing energy to wall.

Car completely stops as it crashes into the wall so Energy converted into demolition and thermal energy

E=\frac{1}{2}mv^2

E=\frac{1}{2}\times 1000\times (10)^2

E=50\ kJ  

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What is the speed of a bird of mass 8kg which has kinetic energy of 8836J?
Kay [80]

Answer:

For the bird moving in a straight line, the kinetic energy is one-half the product of the mass and the square of the speed: Ek=12mu2.

4 0
3 years ago
A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con
Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \  \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm

The magnitude of the electric field is now given as;

E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0
3 years ago
Electromagnetic waves with wavelength between about 1 cm and 20 cm are called??
kodGreya [7K]
I believe the answer is Microwaves (just to clarify, not the oven). They're widely used for communications.<span />
8 0
3 years ago
What is principal focus and focus length with examples​
Shkiper50 [21]

Explanation:

Principle focus is the point on the axis of a convex lens, where the parallel rays of light from one side of the lens. meet on other side after refraction. Distance between optical centre to principle focus point is the focal length.

7 0
3 years ago
Yashoda prepares some lime juice on a hot day. She adds 80 g of ice at a temperature of 0°C to 0.32 kg of lime juice. The temper
Vikentia [17]

Answer:

Explanation:

a )

hear energy required to melt 1 g of ice = 340 J ,

hear energy required to melt 80 g of ice = 340 x 80  J = 27220 J .

b ) energy gained by the melted ice ( water at O°C ) = m ct

where m is mass of water , s is specific heat and t is rise in temperature

= 80 x 4.2 x ( 8°C - 0°C)

= 2688 J .

c )

energy lost by lime juice = energy gained by ice and water

= 27220 J + 2688 J .

= 29908 J .

d )

Let specific heat required be S

Heat lost by lime juice = M S T

M is mass of lime juice , S is specific heat , T is decrease in temperature

= 320 g x S x ( 29 - 8 )°C

= 6720 S

For equilibrium

Heat lost = heat gained

6720 S = 29908 J

S = 4.45 J /g °C .

4 0
3 years ago
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