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liq [111]
3 years ago
11

Freeeeeeeee poinnttttsssss

Physics
1 answer:
RUDIKE [14]3 years ago
5 0

Answer:

k

Explanation:

umm really

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What is the acceleration if a car slows down from 30 m/s to 15 m/s in 4 seconds?
Tamiku [17]

Answer:

11.25

Explanation:

acceleration =v-u/t

=-30-15/4

=-45/4

=11.25

hope it helps

4 0
3 years ago
Read 2 more answers
The pilot of an airplane flying at an elevation of 5000 feet sights two trees that are 300 feet apart. If the angle of depressio
OLEGan [10]

Answer:

32°

Explanation:

From the diagrammatic representation of the question,

Point A is the pilot point of view from the airplane

Point C is the foot of the first tree

Point D is the foot of the second tree

Line AB is the height of the airplane from the ground level.

Line CD is the distance between the two trees

Considering ΔABC,

tan 33° = \frac{5000}{BC}

BC = 5000 / tan 33°

BC = \frac{5000}{0.6494}

BC = 7,699.41 feet

BD = BC + CD

BD = 7,699.41 + 300

BD = 7,999.41 feet

Considering ΔABD,

tan θ = \frac{AB}{BD}

tan θ = \frac{5000}{7,999.41}

tan θ = 0.6250

θ = tan⁻¹ 0.6250

θ = 32°

7 0
3 years ago
A car or truck moving down the highway has what energy by virtue of its motion
Lana71 [14]
Kinetic energy because it is moving

6 0
3 years ago
DOUBLE POINTS!!!
spin [16.1K]
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7 0
3 years ago
A hydrogen atom contains a single electron that moves in a circular orbit about a single proton. Assume the proton is stationary
riadik2000 [5.3K]

Answer:

Radius between electron and proton= 6.804\times 10^{-10}m

Explanation:

The motion of the electron is carried out in the orbit due to the balancing of the electrostatic force between the proton and the electron and the centripetal force acting on the electron.

The electrostatic force is given as = \frac{kq_1q_2}{r^2}

Where,

k = coulomb's law constant (9×10⁹ N-m²/C²)

q₁ and q₂  = charges = 1.6 × 10⁻¹⁹ C

r = radius between the proton and the electron

Also,

Centripetal force on the moving electron is given as:

=\frac{m_eV^2}{r}

where,

m_e = mass of the electron (9.1 ×10⁻³¹ kg)

V = velocity of the moving electron (given: 6.1 ×10⁵ m/s)

Now equating both the formulas, we have

\frac{kq_1q_2}{r^2} = \frac{m_eV^2}{r}

⇒r = \frac{kq_1q_2}{m_eV^2}

substituting the values in the above equation we get,

r = \frac{9\times 10^{9}\times (1.6\times 10^{-19})^2}{9.1\times 10^{-31}\times (6.1\times 10^5)^2}

⇒r = 6.804\times 10^{-10}m

3 0
3 years ago
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