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9966 [12]
4 years ago
10

Suppose that a passenger intent on lunch during his first ride in a hot-air balloon accidently drops an apple over the side duri

ng the balloon’s liftoff. At the moment of the apple’s release, the balloon is accelerating upward with a magnitude of 4.0 m/s2 and has an upward velocity of magnitude 2 m/s. What is the magnitude of the acceleration of the apple just after it is released?
Physics
1 answer:
mel-nik [20]4 years ago
5 0
The magnitude of <span>the acceleration of the apple just after it is released would be 9.8 m/sec^2. The reason being that gravity would be the only force acting on the apple and the acceleration of the balloon would be totally unimportant. I hope that this is the answer that has actually come to your help.</span>
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Does a light switch have electrical potential energy or mechanical energy
nordsb [41]
Turning on a light switch converts mechanical energy to electrical and radiant energy. So Mechanical energy
7 0
4 years ago
Read 2 more answers
A 50.0 kg object is moving at 18.2 m/s when a 200 N force is applied opposite the direction of the objects motion, causing it to
Gekata [30.6K]

Answer:

t = 1.4[s]

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.

P=m*v\\or\\P=F*t

where:

P = impulse or lineal momentum [kg*m/s]

m = mass = 50 [kg]

v = velocity [m/s]

F = force = 200[N]

t = time = [s]

Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.

(m_{1}*v_{1})-F*t=(m_{1}*v_{2})

where:

m₁ = mass of the object = 50 [kg]

v₁ = velocity of the object before the impulse = 18.2 [m/s]

v₂ = velocity of the object after the impulse = 12.6 [m/s]

(50*18.2)-200*t=50*12.6\\910-200*t=630\\200*t=910-630\\200*t=280\\t=1.4[s]

3 0
3 years ago
two identical springs of spring constant 7580 N/m are attached to a block of mass 0.245 kg. What is the frequency of oscillation
ad-work [718]

The frequency of oscillation on the frictionless floor is 28 Hz.

<h3>Frequency of the simple harmonic motion</h3>

The frequency of the oscillation is calculated as follows;

f = (1/2π)(√k/m)

where;

  • k is the spring constant
  • m is mass of the block

f = (1/2π)(√7580/0.245)

f = 28 Hz

Thus, the frequency of oscillation on the frictionless floor is 28 Hz.

Learn more about frequency here: brainly.com/question/10728818

#SPJ1

3 0
2 years ago
These UV photons can break chemical bonds in your skin to cause sunburn—a form of radiation damage. If the 305 nm radiation prov
astraxan [27]

Answer:

energy = 391.902 kJ /mol

Explanation:

given data

wavelength = 305 nm = 305 ×10^{-9} m

to find out

average energy

solution

we know speed of light is 3 ×10^{8} m/s

so we find frequency here first by speed of light formyla

speed = wavelength × frequency

3 ×10^{8}  = 305 ×10^{-9} × frequency

frequency = 9.8360 ×10^{14} s^{-1}

so energy is

energy = hf

here h = 6.62 ×10^{-34} J-s

so

energy =  6.62 ×10^{-34}  × 9.8360 ×10^{14}

energy = 6.51 ×10^{-14} J

so

energy = 6.51 ×10^{-14}  × \frac{6.02*10^23}{1000} kJ/mol

energy = 391.902 kJ /mol

7 0
4 years ago
A proton moves with a speed of 1.00 x 106 m/s perpendicular to a magnetic field, B. As a result, the proton moves in circle of r
Softa [21]

Answer:

0.0109 m ≈ 10.9 mm

Explanation:

proton speed = 1 * 10^6 m/s

radius in which the proton moves = 20 m

<u>determine the radius of the circle in which an electron would move </u>

we will apply the formula for calculating the centripetal force for both proton and electron ( Lorentz force formula)

For proton :

Mp*V^2 / rp  = qp *VB   ∴  rp = Mp*V / qP*B    ---------- ( 1 )

For electron:

re = Me*V/ qE * B -------- ( 2 )

Next: take the ratio of equations 1 and 2

re / rp = Me / Mp                                 ( note: qE = qP = 1.6 * 10^-19 C )

∴ re ( radius of the electron orbit )

= ( Me / Mp ) rp

= ( 9.1 * 10^-31 / 1.67 * 10^-27 ) 20

= ( 5.45 * 10^-4 ) * 20

= 0.0109 m ≈ 10.9 mm

5 0
3 years ago
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