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1 year ago

If your mass is 63.7 kg, and you are standing 7.5 m away from a boulder with

Physics

1 answer:

murzikaleks [220]1 year ago

4 0

Answer:

Explanation:

F= $\frac{Gm1m2}{r^2}$

F=(6.67 x 10^-11) $\frac{63.7X9,750.6}{7.5^2}$

F=7.36502254e-7

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A garden hose having with an internal diameter of 1.1 cm is connected to a (stationary) lawn sprinkler that consists merely of a

trapecia [35]

Answer:

Water leaves the sprinkler at a speed of 2.322 m/sec

Explanation:

We have given internal diameter of the garden $d_1=1.1cm$

Speed of water in the hose is $v_1=0.95m/sec$

Number of holes n = 22

Diameter of each holes $d_2=15cm$

According to continuity equation $A_1v_1=A_2v_2$

$d_1^2\times v_1=22\times d_2^2v_2$

$1.1^2\times 0.95=22\times 0.15^2\times v_2$

$v_2=2.322m/sec$

So water leaves the sprinkler at a speed of 2.322 m/sec

6 0

2 years ago

Which property do gas particles at the same temperature share?

atroni [7]

It’s potential energy

8 0

3 years ago

Read 2 more answers

Dos cargas puntuales están fijas en el eje x: q1 = 6.0µC está en el origen, O, con x1 = 0.0 cm, y q2 = –3.0 µC está situada en e

erik [133]

Answer:

E_total = 1.30 10¹⁰ C / m²

Explanation:

The intensity of the electric field is

E = k q / r²

on a positive charge proof

The total electric field at the midpoint is

as q₁= 6 10⁻⁶ C the field is outgoing to the right

for charge q₂ = -3 10⁻⁶ C, the field is directed to the right, therefore

E_total = E₁ + E₂

E_total = k q₁ / r₁² + k q₂ / r₂²

r₁ = r₂ = r = 4 10⁻² m

E_total = k/r² (q₁ + q₂)

we calculate

E_total = 9 10⁹ / (4 10⁻²)² (6.0 10⁻⁶ +3.0 10⁻⁶)

E_total = 1.30 10¹⁰ C / m²

8 0

3 years ago

Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an

ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

$Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega$

Power factor is given by

$F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802$

The power factor is 0.28802

The average power to the circuit is given by

$P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W$

The average power to the circuit is 2.57162 W

Power to resistor

$P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W$

Power to resistor is 14.28 W

Power to inductor

$P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W$

Power to the inductor is 53.55 W

Power to the capacitor

$P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W$

The power to the capacitor is 6.07142 W

8 0

2 years ago

What is the formula of finding displacement

xenn [34]

Displacement is the final position of the object minus the initial position of the object.
Xf - Xi. Displacement is not the distance of the object. If you go to the right 10m and to the left another 10m, your displacement is 0m. But your distance is 20m

8 0

3 years ago