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2 years ago

Which One Is True? Frist to answer right will be brainlist!

Physics

1 answer:

nata0808 [166]2 years ago

8 0

Answer:

Potato is true!!! :)

Explanation:

Because I said so LOL! Also u didnt say the choices so... HHEHEHE gimme brainliest

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Please help <br>problems 2a.,2b.,3a.,and 3b.

Darina [25.2K]

Answer:

2a) x = 32 [mil/h]; 2b) t = 0.5[h]; 3a) t = 2.5 [h]; 3b) x = 185[mil]

Explanation:

2a)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

$v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 32 [\frac{mil}{h}] \\t=time = 1 [h]\\x=v*t\\x=32[\frac{mil}{h} ]*1[h]\\x=32[mil}$

2b)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

$v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{420}{840}\\ t=0.5[h]$

3a)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

$v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{35}{14}\\ t=2.5[h]$

3b)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

$v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 74 [\frac{mil}{h}] \\t=time = 2.5 [h]\\x=v*t\\x=74[\frac{mil}{h} ]*2.5[h]\\x=185[mil}$

8 0

3 years ago

PLEASE HELP!!!

PtichkaEL [24]

Answer:

Densities increase down the group

MP and BP decrease down the group

Softness increased going down the group

Speed of reacting increases going down the group

4 0

2 years ago

A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend

Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$ .

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ

According to the question, we have -

Entering Velocity (v) = 20 i m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $$\frac{2\pi r}{4}$ = $$\frac{\pi R}{2}$

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $$\frac{\pi R}{2v} = \frac{\pi R}{40}$

Hence, the time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$

To solve more questions on Force on charged particle, visit the link below-

brainly.com/question/14597200

#SPJ4

6 0

1 year ago

PLEASE HELPPPPP!!!! :)

GuDViN [60]

Answer:

F=X.F=mxq. 3. 1 N/kg=0.5kg g=9.80

4 0

2 years ago

A rowboat heads directly across a river at a speed of 3 m/s. Select the correct equations that show that if the river flows at 4

olasank [31]

Answer:

Explanation:

Check attachment for solution

6 0

3 years ago