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Elis [28]
3 years ago
8

Is this a scientific model? Use complete sentences to explain why or why not.

Physics
1 answer:
notsponge [240]3 years ago
4 0

Answer:

it is a scientific model if you want to research where in the world is vegetation located.

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Ron has not used heroin for several days. He has a
katen-ka-za [31]
I wanna say withdrawal? Ron needs to go to rehab
6 0
3 years ago
How much force does it take to bring a 1,375 N car from rest to a velocity of 26 m/s in 6 seconds?
V125BC [204]

Explanation:

<u>Mass of car</u> = 137.5 kg

<u>Acceleration</u> = v - u / t = 26 - 0 / 6 = 4.33 m/sec^2

Force = m * a = 137.5 * 4.33 = 595.3 N

6 0
3 years ago
What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 22 km/h and the coef
Aleks [24]
First, let's put 22 km/h in m/s:

22 \frac{km}{h} \times  \frac{1000m}{1km}  \times  \frac{1h}{3600s}=6.11 \frac{m}{s}

Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by

F_{rad}=m \frac{v^2}{R}

The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:

F_{fric}= mg \mu_{s}

Notice we don't use the kinetic coefficient even though the bike is moving.  This is because when the tires meet the road they are momentarily stationary with the road surface.  Otherwise the bike is skidding.

Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve:

m\frac{v^2}{R} = mg \mu_{s} \\ \\ \frac{v^2}{R} = g \mu_{s} \\ \\R= \frac{v^2}{g \mu_{s}} \\ \\ R= \frac{6.11}{9.8 \times 0.37}=1.685m

3 0
3 years ago
If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle
ad-work [718]

This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C  ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E =  5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

7 0
3 years ago
Jason walks 20 m East, turns around and 20 m West, Finally, he walks 10 rn North. This takes 20 s. what is Jason's velocity​
serious [3.7K]

Answer:

0.5 m/s north

Explanation:

Take east to be +x, west to be -x, north to be +y, and south to be -y.

His displacement in the x direction is:

x = 20 m − 20 m = 0 m

His displacement in the y direction is:

y = 10 m

His total displacement is therefore 10 m north.

His velocity is equal to displacement divided by time.

v = 10 m north / 20 s

v = 0.5 m/s north

3 0
3 years ago
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