Answer:
5.01 J
Explanation:
Info given:
mass (m) = 0.0780kg
height (h) = 5.36m
velocity (v) = 4.84 m/s
gravity (g) = 9.81m/s^2
1. First, solve for Kinetic energy (KE)
KE = 1/2mv^2
1/2(0.0780kg)(4.84m/s)^2 = 0.91 J
so KE = 0.91 J
2. Next, solve for Potential energy (PE)
PE = mgh
(0.0780kg)(9.81m/s^2)(5.36m) = 4.10 J
so PE = 4.10 J
3. Mechanical Energy , E = KE + PE
Plug in values for KE and PE
KE + PE = 0.91J + 4.10 J = 5.01 J
A nitrogen laser generates a pulse containing 10.0 mj of energy at a wavelength of 340.0 nm and has 1785 x 10¹⁹ photons in the pulse.
<h3>How many photons are in the pulse?</h3>
Energy of a single photon is
E=hcλ
E=6.626×10⁻³⁴ J s×3×108 m/s /340×10⁻⁹ m
E=6.31×10⁻¹⁹ J
Number of photons in the laser is
n=Total Energy/Energy per photon
n=10⁷×10⁻³J /5.90×10⁻¹⁹J/photon
n= 1785 x 10¹⁹ photons
To learn about photons, refer: brainly.com/question/20912241?referrer=searchResults
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Answer:
3525.19 kg
Explanation:
The computation of the mass of the car is shown below:
As we know that
Fc = m × V^2 ÷ R
m = Fc × R ÷ V^2
Provided that:
Fc = 34.652 kN = 34652 N
R = Radius = 24.98 m
V = speed = 15.67 m/s
So,
m = 34652 × 24.98 ÷ 15.67^2
= 3525.19 kg
Answer:
a) 0.138J
b) 3.58m/S
c) (1.52J)(I)
Explanation:
a) to find the increase in the translational kinetic energy you can use the relation

where Wp is the work done by the person and Wg is the work done by the gravitational force
By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

the change in the translational kinetic energy is 0.138J
b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

the new speed is 3.58m/s
c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

hence, the change in Er is about 1.52J times the initial rotational energy