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oksian1 [2.3K]
3 years ago
7

A student throws a 130 g snowball at 6.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of

the average force on the wall if the duration of the collision is 0.18 s?
Physics
1 answer:
Alex73 [517]3 years ago
8 0

Answer:

4.7 N

Explanation:

130 g = 0.13 kg

The momentum of the snowball when it's thrown at the wall is

p = mv = 0.13*6.5 = 0.845 kgm/s

Which is also the impulse. From here we can calculate the magnitude of the average force F knowing the duration of the collision is 0.18 s

p = F\Delta t

F*0.18 = 0.845

F = 0.845 / 0.18 = 4.7 N

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Cattle ranchers and dairy farmers rarely allow all of their animals to reproduce. Instead, they practice selective breeding and
Alex Ar [27]

The answer is b becoz it meets growing demands of the country

7 0
3 years ago
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An RLC series circuit consists of a 50 Ω resistor, a 200 μF capacitor, and a 120-mH inductor whose coil has a resistance of 20 Ω
solmaris [256]

Answer:

182.9 Volts

Explanation:

R = resistance of the resistor = 50 Ω

C = capacitance of the capacitor = 200 μF = 200 x 10⁻⁶ F

L = Inductance of the inductor = 120 mH = 0.12 H

f = frequency = 60 Hz

Capacitive reactance is given as

X = (2πfC)⁻¹

X = (2(3.14) (60) (200 x 10⁻⁶))⁻¹

X = 13.3 Ω

Inductive reactance is given as

X' = 2πfL

X' = 2(3.14) (60) (0.12)

X' = 45.2 Ω

Impedance of the circuit is given as

z = √(R² + (X' - X)²)

z = √(50² + (45.2 - 13.3)²)

z = 59.31 Ω

V = rms emf of the source = 240 Volts

rms voltage across the inductor is given as

V' = V z⁻¹ X'

V' = (240) (59.31)⁻¹ (45.2)

V' = 182.9 Volts

5 0
3 years ago
Em um fio condutor uma carga de 6.000 C atravessa uma secção transversal em 5 minutos. Determinando-se a corrente no fio, encont
photoshop1234 [79]

Answer:

I = 20 A

Explanation:

The question says that, "A load of 6,000 C is conducted through a cross section in 5 minutes. Determining-if a current is not correct, we will find the value of?"

We have,

Charge, q = 6,000 C

Time, t = 5 minutes = 300 s

We need to find the current. We know that, the charge flowing per unit time is equal to current. So,

I=\dfrac{q}{t}\\\\I=\dfrac{6000}{300}\\\\I=20\ A

So, the current flowing through the circuit is 20 A.

7 0
3 years ago
A 0.25 kg ideal harmonic oscillator has a total mechanical energy of 9.8 J. If the oscillation amplitude is 20.0 cm, what is the
DanielleElmas [232]

Answer:

7.04 Hz

Explanation:

x(t) = A cos(wt)

v(t) = - wA sin (wt)

Vmax = wA

E = 1/2 * m * (Vmax)^2

E = 1/2*m*(wA)^2

w = \frac{1}{A} \sqrt{\frac{2E}{m} }

f = w/ 2*pi

f = \frac{1}{2*pi*A}*\sqrt{\frac{2E}{m} }  \\f = \frac{1}{2 * pi * 0.2}*\sqrt{\frac{2(9.8)}{0.25} } \\f = 7.04 Hz

6 0
3 years ago
What units are used to express energy?
marysya [2.9K]
Joules
Watts
Kilocalories
BTU
Electron volt

I can't remember any others. Hope that's enough : )
7 0
4 years ago
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