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Damm [24]
3 years ago
5

A ball is thrown off the top of a building and lands on the ground below.

Physics
1 answer:
natita [175]3 years ago
4 0

Answer:

Mass and velocity.

Explanation:

Kinetic energy <u>is the energy that an object has due to its movement</u>, mathematically it is represented as follows:

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

where m is the mass of the object, and v is its velocity at a given point in time.

So we can see that to find the kinetic energy just before the ball hits the gound, we need the quantities:

  • mass of the ball
  • velocity of the ball before it hits the ground

With the knowledge of these two quantities the kinetic energy of the ball  before touching the gound can be determined.

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How much centripetal force is needed to make a body of mass 0.5kg to move in a circle of radius 50cm with speed 3ms^-1 ?​
scoundrel [369]

Centripetal force is given by F= mv²/r.

Given: m = 0.5 kg, v = 3 m/s, r = 0.5 m

Putting values,

F= mv²/r = 0.5× 3²/0.5 = 9 N

7 0
2 years ago
Read 2 more answers
A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc
adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

f_{c_1 = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

f_{c_1 = 3 × 10¹⁰ / 2( 0.0711  )

f_{c_1 = 3 × 10¹⁰ cm/s / 0.1422  cm

f_{c_1 =  21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

so

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( (f_{c_1 / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀e^{-\alpha _1l_{min = -100dB

we substitute

20log₁₀e^{-(310.7np/m)l_{min = -100dB

l_{min = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0
3 years ago
A 70.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. Suppos
brilliants [131]

Answer:

v₂=- 34 .85 m/s

v₁=0.14 m/s

Explanation:

Given that

m₁=70 kg ,u₁=0 m/s

m₂=0.15 kg ,u₂=35 m/s

Given that collision is elastic .We know that for elastic  collision

Lets take their final speed is v₁ and v₂

From momentum conservation

m₁u₁+m₂u₂=m₁v₁+m₂v₂

70 x 0+ 0.15 x 35 = 70 x v₁ + 0.15 x v₂

70 x v₁ + 0.15 x v₂=5.25                   --------1

v₂-v₁=u₁-u₂            ( e= 1)

v₂-v₁ = -35        --------2

By solving above equations

v₂=- 34 .85 m/s

v₁=0.14 m/s

4 0
3 years ago
A body weighs 10newton in air and 9.5 in water. How will it weigh in alchohol if density 0.8gcm-³​
Katen [24]

Answer:

The weight of object in alcohol is, W₀ = 9.6 N

Explanation:

Given,

The weight of body in air, Wₐ = 10 N

The weight of the body in water, Wₓ = 9.5 N

The density of the alcohol, ρ = 0.8 g/cm³

                                                = 800 kg/m³

The weight of body in water = weight of object in air - weight of water displaced

                        9.5 N = 10 N - weight of water displaced

                       Weight of water displaced = 0.5 N

The mass of the displaced water, m = 0.5 / 9.8

                                                            = 0.051 kg

Therefore the volume of the water displaced,

                                              V = m / ρ

                                                  = 0.051 / 1000

                                                   = 5.1 x 10⁻⁵ m³

<em>The volume of water displaced is equal to the volume of alcohol displaced</em>

Therefore, the mass of the alcohol displaced, = V x ρ

                                                                             = 5.1 x 10⁻⁵ x 800

                                                                             = 0.0405 kg

The weight of the alcohol displaced, w = 0.0405 x 9.8

                                                                    = 0.4 N

Therefore,

The weight of body in alcohol = weight of object in air - weight of alcohol displaced

                             W₀ = W - w

                                    = 10 N - 0.4 N

                                     = 9.6 N

Hence, the weight of object in alcohol is, W₀ = 9.6 N

4 0
3 years ago
Your teacher gives you a piece of cardboard with two pinholes on it, saying that they are separated by 205 μm ± 3%. In order to
wlad13 [49]

Answer:

m,lkj,mkn,njn

Explanation:because she is telling you to do a project

7 0
3 years ago
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