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3 years ago

SOS HELP ME

Physics

1 answer:

Dafna1 [17]3 years ago

5 0

Answer:

Angle Vy Vy^2/(2*g)

25 6.3 2.02

30 7.5 2.87

45 10.6 5.73

50 11.5 6.75

65 13.6 9.44

70 14.1 10.1

Vy = vertical velocity = V sin theta

2 g H = Vy^2 since the final velocity is zero

H = Vy^2 / (2 * g)

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A ball is moving with a velocity of 0.5 m/s its velocity is decreasing at the rate of 0.05 m/s² what is its velocity in 5 second

Law Incorporation [45]

Answer:

v after 5s = 0.25 m/s, it took 10s to stop, it has traveled 2.5m before stopping

Explanation

We can use the equation of motion with constant acceleration

Given: v0= 0.5 m/s a= -0.05 m/s²

v(5s) = v0 + a×t = 0.25 m/s

Stop => v=0 => v0 + a×t = 0 => t=10s

Distance at t=10s ⇒ x(10) = 0.5×10 + 0.5x(-0.05)x10² = 2.5m

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3 years ago

A manufacturer claims that a carpet will not generate more than 5.8 kV of static electricity What magnitude of charge would have

DiKsa [7]

Answer:

4.4×10⁻⁷ Coulomb

Explanation:

V = Voltage = 5.8 kV

d = Potential distance = 2.8 mm = 0.0028 m

A = Area = 0.3×0.08 = 0.024 m²

ε₀ = permittivity constant in a Vacuum= 8.85×10⁻¹² F/m

$\frac{Q}{V}=\frac{A\epsilon_0}{d}\\\Rightarrow \Q=V\frac{A\epsilon_0}{d}\\\Rightarrow Q=5.8\times 10^3\frac{0.024\times 8.85\times 10^{-12}}{0.0028}\\\Rightarrow Q=4.4\times 10^{-7}\ C$

Magnitude of charge transferred between a carpet and a shoe is 4.4×10⁻⁷ Coulomb.

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3 years ago

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

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3 years ago

Describe why you weigh less on moon than on earth. Give examples

dmitriy555 [2]

Weight doesn't really mean much as it just means gravity the bigger a space object is the more force it has to pull on something since the moon is smaller than the earth then it has less gravity and then less weight on scales.

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3 years ago

Two cars are heading towards each other but are 12 km apart. one car is going 70 km/hr, and the other is going 50 km/hr. how muc

vova2212 [387]

<span>12-50t=70t, t= 0.1h = 6 minutes.</span>

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3 years ago