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GrogVix [38]
3 years ago
5

The term relative intensity is used to denote the amount of energy expended per minute.

Physics
1 answer:
sergeinik [125]3 years ago
7 0
I think the answer is True
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kinetic

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PLZ ANSWER FAST WILL GIVE BRAINILEST TO FIRST PERSON WHO ANSWERS:P
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A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for
stealth61 [152]

Answer:

(a) a_s=0.62\frac{m}{s^2}

(b) a_s=0.13\frac{m}{s^2}

(c) x_f=2.6m

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}

(c) Using the kinematics equation:

x_f=x_0+v_0t \pm  \frac{at^2}{2}

For the girl, we have x_0=0 and v_0=0. So:

x_f_g=\frac{a_gt^2}{2}(1)

For the sled, we have v_0=0. So:

x_f_s=x_0_s-\frac{a_st^2}{2}(2)

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s

Now, we solve (1) for x_f_g

x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m

5 0
3 years ago
A ball rolling down a hill was displaced 21.9 m while uniformly accelerating from rest. If the final velocity was 7.14 m/s, what
belka [17]

Answer:

a = 1.16 m/s²

Explanation:

In order to find the acceleration of the ball we will use 3rd equation of motion.

2as = Vf² - Vi²

where,

a = acceleration = ?

s = displacement = 21.9 m

Vf = Final Velocity = 7.14 m/s

Vi = Initial Velocity = 0 m/s (Since, ball starts from rest)

Therefore, using the values, we get:

2a(21.9 m) = (7.14 m/s)² - (0 m/s)²

a = (50.97 m²/s²)/(43.8 m)

<u>a = 1.16 m/s²</u>

4 0
3 years ago
A 0.500-V potential difference is maintained across a 1.50-m length of tungsten wire that has a cross-sectional area of 0.800 mm
Genrish500 [490]

Answer:

5.95 A

Explanation:

From the question

R = ρL/A..................... Equation 1

Where R = resistance of the tungsten wire, ρ = Resistivity of the tungsten wire, L = length, A = cross sectional area.

Given: L = 1.5 m, A = 0.8 mm² = 0.8×10⁻⁶ m, ρ = 5.60×10⁻⁸ Ω.m

Substitute these values into equation 1

R = 1.5(5.60×10⁻⁸)/0.8×10⁻⁶

R = 0.084 Ω.

Finally, using Ohm law,

V = IR

Where V = Voltage, I = current

Make I the subject of the equation

I = V/R............... Equation 2

I = 0.5/0.084

I = 5.95 A

4 0
3 years ago
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