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svlad2 [7]
3 years ago
5

Please help. Brainliest will be given! 25 points. Show all work.

Physics
1 answer:
Drupady [299]3 years ago
4 0

Explanation:

Given:

v₀ₓ = 15 m/s cos 20° = 14.10 m/s

aₓ = 0 m/s²

v₀ᵧ = 15 m/s sin 20° = 5.13 m/s

aᵧ = -9.8 m/s²

t = 1.5 s

Find: Δx and Δy

Δx = v₀ₓ t + ½ aₓ t²

Δx = (14.10 m/s) (1.5 s) + ½ (0 m/s²) (1.5 s)²

Δx = 21.1 m

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (5.13 m/s) (1.5 s) + ½ (-9.8 m/s²) (1.5 s)²

Δy = -3.33 m

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There are NO true statements on that list of choices.
3 0
3 years ago
A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.
densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

KE = \frac{1}{2} mv^2

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (5*10^{-3})(300)^2

KE =  225J

On the other hand the required Energy to heat up t melting point is

Q_1 = mC_p \Delta T

Q_2 = L_f m

Where,

m = Mass

C_p =Specific Heat

\Delta T =Change at temperature

L_f = Latent heat of fussion

Heat required to heat up to melting point,

Q = Q_1+Q_2

Q = mC_p \Delta T+L_f m

Q = 5*0.128*(327-20) + 5*24.7

Q = 310J

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

4 0
3 years ago
What do you want to learn about simple machines? (could someone come up with a question about simple machines for me)​
bezimeni [28]

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6 0
3 years ago
What is one behavior that a plant has that a human would be suprised by
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3 0
3 years ago
A resin tube and a PVC rod are both rubbed against styrofoam. Do
Ivan

The concepts of electrostatics allow us to find the result for the question about the electric forces in the two tubes is:

  • The tubes are attracted by having charges of different signs.

<h3>Electrostatics</h3>

Electrostatics studies the transfer of charge between bodies when they rub together, in generating the charges of the insulating bodies they are not mobile, so to transfer them from one body to another they must be in contact.

The force electric between charges is of two types:

  • Attractive. If the charges are of different signs.
  • Repulsive.  If  the charges are of the same sign.

The transferred charge depends on the bodies involved in the tables showns that the resin is positively charged when rubbed, the PVC acquires very little charge and polystyrene acquires a negative charge due to rubbing.

They indicate that the resin is rubbed with polystyrene, therefore the resin must acquire a positive charge and the polystyrene a negative charge.

Then the PVC is rubbed with the polystyrene, which initially we will assume neutral, the PVC does not acquire a charge and the polystyrenes acquires a negative charge, if the polystyrene is isolated from the ground, it shares this negative charge with the PVC, therefore the two materials remain with half of the negative charge.

Finally, we bring the resin, which is positively charged, closer to the PVC, which has a slight negative charge, therefore the two bodies must attract each other.

In conclusion using the concepts of electrostatics we can find the result for the question about the electric forces in the two tubes is:

  • The tubes are attracted by having charges of different signs.

Learn more about electrostatics here: brainly.com/question/17692887

6 0
2 years ago
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