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Yuliya22 [10]
3 years ago
8

A football player kicks the ball at a 45 degree angle. Without an effect from the wind, the ball would travel 60.0m horizontally

. A) what is the initial speed of the ball? B) when the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.50 m/s. What distance does the ball travel horizontally from the initial kick point? (In m)
Physics
1 answer:
Ronch [10]3 years ago
5 0
B when the ball is at its maxium height 
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A wire must be replaced in a circuit by a new wire of the same material but nine times longer. If the new wire's resistance is t
Blizzard [7]

Answer:

three times the original diameter

Explanation:

From the wire's resistance formula, we can calculate the relation between the diameter of the wire and its length:

R=\rho\frac{l}{\pi \frac{d^2}{4}}\\d=\sqrt{\rho \frac{4 l}{\pi R}}\\

Here, d is the wire's diameter, \rho is the electrical resistivity of the material and R is the resistance of the wire. We have l'=9l

d'=\sqrt{\rho \frac{4 l'}{\pi R}}\\d'=\sqrt{\rho \frac{4 (9l)}{\pi R}}\\d'=3\sqrt{\rho \frac{4 l}{\pi R}}\\d'=3d

7 0
3 years ago
A light spring has a force constant of 70 N/m and is used to pull a 10 kg box on a horizontal frictionless surface. If the box h
Sergio039 [100]

Answer:

82.4 cm

Explanation:

∑F = ma

kx cos θ = ma

x = ma / (k cos θ)

x = (10 kg) (5 m/s²) / (70 N/m cos 30.0°)

x = 0.824 m

x = 82.4 cm

8 0
3 years ago
PLEASE HELP ASAPPPPPPPPP
Svetlanka [38]
60kg would be the answer
8 0
2 years ago
If an organism starts from only one cell,how do we get different body parts and organs
Vladimir [108]

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Every cell has the complete set of instructions in it.  In the past
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8 0
3 years ago
Consider two identical objects of mass m = 0.250 kg and charge q = 4.00 μC. The first charge is held in place at the origin of a
Gnom [1K]

Answer:

a = 640 m/s²

Explanation:

From work-kinetic energy principles,

The net force acting on the second object is the gravitational force and the electric force due to the first object.

So, the gravitational force on the mass is F₁ = Gm₁m₂/r² since m₁ = m₂ = m, U = -Gm²/r²

Also, the electric force on the charge is F₂ = kq₁q₂/r² since q₁ = q₂ = q, U = kq²/r²

The net Force F = ma

So, -F₁ + F₂ = F     (F₁ is negative since it is an attractive force in the negative x -direction and F₂ is positive since it is a repulsive force in the positive x- direction)

-Gm²/r² + kq²/r² = ma

ma = -Gm²/r² + kq²/r²

a = (-Gm²/r² + kq²/r²)/m

a = (-G + kq²/m²)m/r²

Since m = 0.250 kg, q = 4.00 μC = 4.00 × 10⁻⁶ C, r = 3.00 cm = 3.00 × 10⁻² m, G = 6.67 × 10⁻¹¹ Nm²/kg², k = 9 × 10⁹ Nm²/C² and a = acceleration of second mass.

Substituting the variables into the equation, we have

a = (m/r²)(-G + k(q/m)²)]

a = (0.250 kg/{3.00 × 10⁻² m}²)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(4.00 × 10⁻⁶ C/0.250 kg)²)

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(16 × 10⁻⁶ C/kg)²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(256 × 10⁻¹² C²/kg²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 2304 × 10⁻³ Nm²/kg²  ]

a = (0.250 kg/9.00 × 10⁻⁴ m)(2.304 Nm²/kg²)

a = 0.576 Nm²/kg /9.00 × 10⁻⁴ m²

a = 0.064 × 10⁴N/kg

a = 64 × 10 N/kg)

a = 640 m/s²

8 0
2 years ago
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