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Katen [24]
2 years ago
13

Who has the most power? athlete A (bench presses 100 kg over 0.6 m in 0.5 s) athlete B (bench presses 150 kg over 0.6 m in 1.0 s

) athlete C (bench presses 200 kg over 0.6 m in 2.0 s) athlete D (bench presses 250 kg over 0.6 m in 2.5 s)
Physics
1 answer:
MrRissso [65]2 years ago
4 0

Answer:

Athlete A

Explanation:

Power is the rate of doing work and it is calculated as follows:

Power = work done/time taken = mgh/t

(for work being done against gravity)

So for athlete A

P = (100 kg * 9.8 N/kg* 0.6m)/0.5 s = 1176 W

For athlete B

P = (150 kg * 9.8 N/kg* 0.6m)/1 s = 882 W

For athlete C

P = (200 kg * 9.8 N/kg* 0.6m)/2 s = 588 W

For athlete D

P = (250 kg * 9.8 N/kg* 0.6m)/2.5 s = 588

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F = G mM / r^2, where 
<span>F = gravitational force between the earth and the moon, </span>
<span>G = Universal gravitational constant = 6.67 x 10^(-11) Nm^2/(kg)^2, </span>
<span>m = mass of the moon = 7.36 × 10^(22) kg </span>
<span>M = mass of the earth = 5.9742 × 10^(24) and </span>
<span>r = distance between the earth and the moon = 384,402 km </span>

<span>F </span>
<span>= 6.67 x 10^(-11) * (7.36 × 10^(22) * 5.9742 × 10^(24) / (384,402 )^2 </span>
<span>= 1.985 x 10^(26) N</span>
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3 years ago
Velocity v (m/s)
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Explanation:

given solution

h=45m v^2=u^2+2gh

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  • v=30
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2 years ago
A whale comes to the surface to breathe and then dives at an angle of 20.0 ∘ below the horizontal (see the figure (Figure 1)). I
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Answer:

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3 years ago
A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force o
umka2103 [35]

Answer:

a) The proton's speed is 5.75x10⁵ m/s.

b) The kinetic energy of the proton is 1723 eV.  

Explanation:

a) The proton's speed can be calculated with the Lorentz force equation:

F = qv \times B = qvBsin(\theta)     (1)          

Where:

F: is the force = 9.14x10⁻¹⁷ N

q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C

v: is the proton's speed =?

B: is the magnetic field = 3.28 mT

θ: is the angle between the proton's speed and the magnetic field = 17.6°

By solving equation (1) for v we have:

v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s

Hence, the proton's speed is 5.75x10⁵ m/s.

b) Its kinetic energy (K) is given by:

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

Where:

m: is the mass of the proton = 1.67x10⁻²⁷ kg

K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV  

Therefore, the kinetic energy of the proton is 1723 eV.

I hope it helps you!        

3 0
3 years ago
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