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3 years ago

All of the following are true about scientific evidence except: *

Physics

1 answer:

goldenfox [79]3 years ago

4 0

Answer:

Your answer is D It does not need to be repeatable.

the reason for this is because C is correct.

You need to be able to experiment on something multiple times so that you can gather further data to imbedded your evidence in facts.

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How much power is required to light a lightbulb at 100V of voltage when the lightbulb has a resistance of 500 Ohms?

salantis [7]

Answer:

Power = 20 Watts

Explanation:

Given the following data;

Voltage = 100 V

Resistance = 500 Ohms

To find the power that is required to light a lightbulb;

Mathematically, power can be calculated using the formula;

$Power = \frac {Voltage^{2}}{resistance}$

Substituting into the formula, we have;

$Power = \frac {100^{2}}{500}$

$Power = \frac {10000}{500}$

Power = 20 Watts

3 0

3 years ago

Warning triangles, flares, a vehicle's hazard lights, or emergency vehicles ahead, are all clues that you might be approaching _

daser333 [38]

Answer:

B. A collision scene

Explanation:

It could have been a parade ceremony, but, if you notice the vehicle's hazard lights or an emergency vehicle ahead, it is common sense to figure that they is a collision scene nearby.

8 0

3 years ago

Describe your acceleration as you skate down your neighborhood sidewalk.

Paraphin [41]

Answer: when you increase or decrease your speed.

Explanation:

Moving a skate at rest, you need to apply force in order to cause acceleration.

F = ma Where

F = force applied

m = mass of the skate

a = acceleration

The initial velocity u will be equal to zero and the skate will acceleration to a certain velocity.

as you skate down your neighborhood sidewalk, you will accelerate when you increase your speed. Because

Acceleration is the rate of change of velocity. That is,

Acceleration = change in velocity/ time.

And also, you will decelerate when you reduce the speed or velocity down your neighborhood sidewalk.

6 0

3 years ago

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

6 0

2 years ago

Read 2 more answers

Complete the equation???

bogdanovich [222]

which problem theres 3

4 0

3 years ago