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WINSTONCH [101]

3 years ago

14

In punting a football, the kicker tries to maximize both the distance of the kick and its "hang time"—the time that the ball is

in the air. A kicker gets off a great punt with a hang time of 5.0 s that lands 50 yards from the kicker. What is the speed of the ball as it leaves the kicker’s foot?

1 answer:

Alex787 [66]3 years ago

3 0

Answer:

v₀ = 26.17 m/s

Explanation:

If

R = Xmax = 50 yards = (50 yards)*(0.9144 m / 1 yard) = 45.72 m

t = 5 s (Time of Flight)

we can apply the equation:

R = V₀x*t ⇒ V₀x = R / t = 45.72 m / 5 s ⇒ V₀x = 9.144 m/s

then we use the equation of Time of Flight:

t = 2*V₀y / g ⇒ V₀y = g*t / 2 = (9.81 m/s²)*(5 s) / 2 ⇒ V₀y = 24.525 m/s

Finally we apply

v₀ = √(V₀x² + V₀y²) ⇒ v₀ = √(9.144² + 24.525²) m/s

⇒ v₀ = 26.17 m/s

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Calculate the force of a particle with a net charge of 170 coulombs traveling at a speed of 135 meters/second perpendicular to t

amm1812

Answer:

F=1.14N j

Explanation:

The magnitude of the magnetic force over a charge in a constant magnetic field is given by the formula:

$|\vec{F}|=|q\vec{v} \ X\ \vec{B}|=qvsin\theta$ (|)

In this case v and B vectors are perpendicular between them. Furthermore the direction of the magnetic force is:

-i X k = +j

Finally, by replacing in (1) we obtain:

$\vec{F}=(170C)(135\frac{m}{s})(5.0*10^{-5}T)=1.14N\ \hat{j}$

hope this helps!

6 0

3 years ago

Read 2 more answers

In 1865, Jules Verne proposed sending men to the Moon by firing a space capsule from a 220-m-long cannon with final speed of 10.

Sidana [21]

Answer:

The unrealistically large acceleration experienced by the space travelers during their launch is 2.7 x 10⁵ m/s².

How many times stronger than gravity is this force? 2.79 x 10⁴ g.

Explanation:

given information:

s = 220 m

final speed, vf = 10.97 km/s = 10970 m/s

g = 9.8 m/s²

he unrealistically large acceleration experienced by the space travelers during their launch

vf² = v₀²+2as, v₀ = 0

vf² = 2as

a =vf²/2s

= (10970)²/(2x220)

= 2.7 x 10⁵ m/s²

Compare your answer with the free-fall acceleration

a/g = 2.7 x 10⁵/9.8

a/g = 2.79 x 10⁴

a = 2.79 x 10⁴ g

7 0

3 years ago

A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.

iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

$F = \frac{kq_1q_2}{d^2}$

Here

k = Coulomb's Constant

$q_{1,2} =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}$

$F_{12} = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_{21} = F_{12}$

$F_{21} = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}$

$F_{12} = F_{21} = -54 mN$

Force is negative i.e. attractive

6 0

3 years ago

Explain why the angle is important when lauching projectiles

yaroslaw [1]

Answer:

to have an accurate measure

Explanation:

6 0

4 years ago

An ideal gas is confined within a closed cylinder at atmospheric pressure (1.013 * 105 Pa) by a piston. The piston moves until t

likoan [24]

Answer:

$911700\ \text{Pa}$

Explanation:

$P_1$ = Initial pressure = $1.013\times 10^5\ \text{Pa}$

$V_1$ = Initial volume

$V_2$ = Final volume = $\dfrac{V_1}{9}\\\Rightarrow \dfrac{V_1}{V_2}=9$

Temperature is the same in the initial and final state

From the ideal gas law we have

$P_1V_1=P_2V_2\\\Rightarrow P_2=\dfrac{P_1V_1}{V_2}\\\Rightarrow P_2=P_1\times9\\\Rightarrow P_2=1.013\times 10^5\times 9\\\Rightarrow P_2=911700\ \text{Pa}$

The final pressure of the system is $911700\ \text{Pa}$ .

5 0

3 years ago