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3 years ago

What type of evidence is needed for a hypothesis to be supported or not supported?

Physics

1 answer:

NISA [10]3 years ago

8 0

Answer:

Hypotheses must be testable, and once tested, they can be supported by evidence. If a statement is made that cannot be tested and disproved, then it is not a hypothesis.

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In Example 2.7, we investigated a jet landing on an aircraft carrier. In a later maneuver, the jet comes in for a landing on sol

igor_vitrenko [27]

Answer:

a) 20 seconds

b) No.

Explanation:

t = Time taken for jet to stop

u = Initial velocity = 100 m/s (given in the question)

v = Final velocity = 0 (because the jet will stop at the end)

s = Displacement of the jet (Distance between the moment the jet touches the ground to the point the point it stops)

a = Acceleration = -5.00 m/s² (slowing down, so it is negative)

a) Equation of motion

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-100}{-5}\\\Rightarrow t=20\ s$

The time required for the plane to slow down from the moment it touches the ground is 20 seconds.

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=100\times 20+\frac{1}{2}\times -5\times 20^2\\\Rightarrow s=1000\ m$

The distance it requires for the jet to stop is 1000 m so in a small tropical island airport where the runway is 0.800 km long the plane would not be able to land. The runway needs to be atleast 1000 m long here the runway on the island is 1000-800 = 200 m short.

5 0

3 years ago

An electron in a television tube is accelerated uniformly from rest to a speed of 8.4\times 10^7~\text{m/s}8.4×10 7 m/s over

stich3 [128]

Answer:

P=3.42×10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that :

vf^2=vi^2+2*a(xf-xi)

Where :

• vf , vi, are the the final and the initial velocity of the electron

• a is the acceleration of the electron

• xf , xi are the final and the initial position of the electron .

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.

Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m

vf^2 =vi^2+2*a(xf-xi)

vf^2-vi^2=2*a(xf-xi)

2*a(xf-xi)= vf^2-vi^2

a = (vf^2-vi^2)/2(xf-xi)

Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

= 1.411 × 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m

so,

vf^2 =vi^2+2*a(xf-xi)

vf^2 =5.312× 10^7

From the following Eq. we can calculate the time elapsed in this motion .

xf =xi+vi*t+1/2*a*t

t=√2(xf-xi)/a

t=3.765×10^-10 s

now we can use the power P Eq.

P=W/Δt => ΔK/Δt

Where: the work done W change the kinetic energy K of the electron ,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42×10^-6 J/s

6 0

3 years ago

Helpp pls

Kipish [7]

Answer:

The intensity of the electric field is

$|E|=10654.37 \:N/C$

Explanation:

The electric field equation is given by:

$|E|=k\frac{q}{d^{2}}$

Where:

k is the Coulomb constant
q is the charge at 0.4100 m from the balloon
d is the distance from the charge to the balloon

As we need to find the electric field at the location of the balloon, we just need the charge equal to 1.99*10⁻⁷ C.

Then, let's use the equation written above.

$|E|=(9*10^{9})\frac{1.99*10^{-7}}{0.41^{2}}$

$|E|=10654.37 \:N/C$

I hope it helps you!

5 0

3 years ago

A material that does not conduct heat very well

MrRissso [65]

Answer:

Explanation:

Insulator is like rubber, it cant physically hold heat well

7 0

3 years ago

Read 2 more answers

An unstrained horizontal spring has a length of 0.31 m and a spring constant of 220 N/m. Two small charged objects are attached

Kruka [31]

The solution you should use is Hooke's law: F=-kx

It should have the same signs because they repel due to the stretch of the spring.

a. Since there is a constant energy within the spring, then Hooke's law will determine the possible algebraic signs. The solution should be
<span>F = kx
270 N/m x 0.38 m = 102.6 N
</span>
b. Then use Coulomb's law; F=kq1q2/r^2 to find the charges produced in the force.

8 0

3 years ago