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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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Let’s allow ourselves to be a little more optimistic with our bus passengers. Earlier in the movie Sandy has to make the bus jum

Roman55 [17]

Answer:

25.71486 seconds

Explanation:

t = Time taken

u = Initial velocity = 60 mph

v = Final velocity = 80 mph

s = Displacement = 0.5 mile

a = Acceleration

Converting to m/s

$60\times \dfrac{1609.34}{3600}=26.8223\ m/s$

$80\times \dfrac{1609.34}{3600}=35.7631\ m/s$

From equation of motion

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{35.7631^2-26.8223^2}{2\times \dfrac{1609.34}{2}}\\\Rightarrow a=0.34769\ m/s^2$

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{35.7631-26.8223}{0.34769}\\\Rightarrow t=25.71486\ s$

Time taken is 25.71486 seconds

4 0

3 years ago

Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the ball's final speed, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

3 years ago

How does`ezzzzz321erfq23c2f

Nataliya [291]

Lol what???? i don’t understand

7 0

2 years ago

Suppose you had two magnets, who were attracted to each other, and you brought them together. Then took them apart by separating

IgorLugansk [536]

Answer:4. Two charged objects have a repulsive force of 0.080 N. If the distance separating the objects is tripled, then what is the new force? Explanation: The electrostatic force is inversely related to the square of the separation distance.

Explanation:

8 0

2 years ago

Need help anybody thanks

mars1129 [50]

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only on the room and what's in it, not on anything outside.
The time of day has no effect. The other things all do.

4 0

3 years ago