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Ket [755]
4 years ago
11

A ball is kicked with an initial velocity of 17 m/s in the horizontal direction and 13 m/s in the vertical direction. (Assume th

e ball is kicked from the ground.) (a) At what speed (in m/s) does the ball hit the ground
Physics
1 answer:
BabaBlast [244]4 years ago
7 0

Answer:

The velocity of the ball before it hits the ground = 21.4m/s

Explanation:

The question can be solved as a projectile problem. One of the main characteristics of the parabolic motion is the symmetry it has as long as the initial height is equal to the final height with respect to a reference system, the vertical displacement of the horizontal displacement being independent. 

The vertical component of velocity is 13m/s

The horizontal component is 17m/s

The ball's pathway is defined by a right angle triangle .

Using Pythagorean

Let vertical component be a

Let horizontal component be b

Let the speed of the ball before hitting the ground be c

Pythagorean is given by:

a^2 + b^2 = c^2

13^2 + 17^2 = c^2

169 + 289 = c^2

458 = c^2

Sqrt(458) = c

21.4 m/s = c

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igomit [66]

Answer:

The net force should be of a magnitude of 64 N

Explanation:

We use Newton's second Law for this:

F_{net} = m\,*\,a

which for our case gives:

F_{net} = m\,*\,a=16\,(4)\,N= 64\,\,N

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3 years ago
4. Solid ‘X’ turns into a liquid at 80 0C and into a gas at 140 0C. Describe the changes in the
Paladinen [302]

Answer:

Explanation:

The melting point of the solid is 80°C

        Vapor point of the liquid is 140°C

What happens to particles of X when heated from 70°C to 85°C?

  • Firstly, there would be a phase change from solid to liquid.
  • Below the melting point, a substance will exist as a solid.
  • With increase in thermal energy inputted by heat, as the temperature climbs above the melting point, it changes to the liquid.
  • When the solid begins to heat up, the particles of X starts vibrating about their fixed point.
  • At they melting point, they break lose and flow to form a liquid.
  • The particles will have more kinetic energy.
5 0
3 years ago
The cephalocaudal trend can be described as the__________direction of motor development.
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It is C i think it might be C
8 0
3 years ago
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A 645-turn coil with a 20.250 m ​2 ​​ area is spun in Earth’s 5.00×10 ​−5 ​​ T magnetic field, producing a 1.25-V maximum emf. A
Dmitriy789 [7]

Answer:

\omega = 1.914\ rad/s

Explanation:

Given,

Number of turns, N = 645 N

Area, A = 20.25 m²

Earth Magnetic field, B = 5 x 10⁻⁵ T

Maximum Emf = 1.25 V.

Angular velocity, ω = ?

Using Induced Emf formula

\varepsilon = NAB\omega

\omega= \dfrac{\varepsilon}{NAB}

\omega= \dfrac{1.25}{645\times 20.25\times 5\times 10^{-5}}

\omega = 1.914\ rad/s

Angular velocity of the coil = \omega = 1.914\ rad/s

5 0
4 years ago
For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener
Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

A)

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

v=u+at

where:

v is the velocity at time t

u is the initial velocity

a=g=9.8 m/s^2 is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

u=0 (the object starts from rest)

a=9.8 m/s^2

Therefore, the velocity after

t = 2 s

is:

v=0+(9.8)(2)=19.6 m/s (downward)

B)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

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

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

KE=\frac{1}{2}(3)(19.6)^2=576 J

C)

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

a=g=9.8 m/s^2 (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

s=0+\frac{1}{2}(9.8)(2)^2=19.6 m

D)

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

KE'=\frac{1}{2}Mv^2

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

KE'=\frac{1}{2}(6)(19.6)^2=1152 J

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0
3 years ago
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