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-Dominant- [34]

2 years ago

13

Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a

nd Jackson’s is 54.0 kg. Jackson is running at a velocity of 3.0 m/s. What must Tammy’s minimum velocity be in order to stop forward momentum?

1 answer:

Naily [24]2 years ago

8 0

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

$m_1v_1 + m_2v_2 = 0$

final they both comes to rest so here we can say that final momentum must be zero

now we have

$34 v + 54 (3 m/s) = 0$

$v = -4.76 m/s$

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A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s.

marishachu [46]

Answer:

a) the magnitude of the force is

F= Q( $\frac{kqs}{r^3}$ ) and where k = 1/4πε₀

F = Qqs/4πε₀r³

b) the magnitude of the torque on the dipole

τ = Qqs/4πε₀r²

Explanation:

from coulomb's law

E = $\frac{kq}{r^{2} }$

where k = 1/4πε₀

the expression of the electric field due to dipole at a distance r is

E(r) = $\frac{kp}{r^{3} }$ , where p = q × s

E(r) = $\frac{kqs}{r^{3} }$ where r>>s

a) find the magnitude of force due to the dipole

F=QE

F= Q( $\frac{kqs}{r^3}$ )

where k = 1/4πε₀

F = Qqs/4πε₀r³

b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces

τ = F sinθ × s

θ = 90°

note: sin90° = 1

τ = F × r

recall F = Qqs/4πε₀r³

∴ τ = (Qqs/4πε₀r³) × r

τ = Qqs/4πε₀r²

8 0

2 years ago

A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle

liq [111]

Answer:

0.0031792338 rad/s

Explanation:

$\theta$ = Angle of elevation

y = Height of balloon

Using trigonometry

$tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta$

Differentiating with respect to t we get

$\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta$

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

$h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft$

Now y = 2500 ft

$cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974$

$\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s$

The angle is changing at 0.0031792338 rad/s

6 0

3 years ago

A 74 kg firefighter slides, from rest, 4.9 m down a vertical pole. (a) If the firefighter holds onto the pole lightly, so that t

In-s [12.5K]

Answer:

Her speed is 9.8 meter per second

Explanation:

Newton's second law states that acceleration (a) is related with force (F) by:

$\sum\overrightarrow{F}=m\overrightarrow{a}$ (1)

Here the only force acting on the firefighter is the weight F=mg so (1) is:

$mg=ma$

Solving for a:

$a=g$

Now with the acceleration we can use the Galileo's kinematic equation:

$Vf^{2}=Vo^{2}+2a\varDelta x$ (2)

With Vf the final velocity, Vo the initial velocity and Δx the displacement, because the firefighter stars from rest Vo=0 so (2) is:

$Vf^{2}=2a\varDelta x$

Solving for Vf

$Vf=\sqrt{2g\varDelta x}=\sqrt{2(9.81)(4.9)}$

$Vf=9.8\frac{m}{s}$

6 0

2 years ago

A 0.125 kg hockey puck moving at 20.0 m/s is caught and held by an 85.0 kg goalkeeper at rest. After catching the puck, with wha

jenyasd209 [6]

Answer:

a

Explanation:cuz i saiad so

4 0

3 years ago

An underground gasoline tank can hold 1.07 103 gallons of gasoline at 52.0°F. If the tank is being filled on a day when the outd

Damm [24]

Answer:

1069.38 gallons

Explanation:

Let V₀ = 1.07 × 10³ be the initial volume of the gasoline at temperature θ₁ = 52 °F. Let V₁ be the volume at θ₂ = 97 °F.

V₁ = V₀(1 + βΔθ) β = coefficient of volume expansion for gasoline = 9.6 × 10⁻⁴ °C⁻¹

Δθ = (5/9)(97°F -52°F) °C = 25 °C.

Let V₂ be its final volume when it cools to 52°F in the tank is

V₂ = V₁(1 - βΔθ) = V₀(1 + βΔθ)(1 - βΔθ) = V₀(1 - [βΔθ]²)

= 1.07 × 10³(1 - [9.6 × 10⁻⁴ °C⁻¹ × 25 °C]²)

= 1.07 × 10³(1 - [0.024]²)

= 1.07 × 10³(1 - 0.000576)

= 1.07 × 10³(0.999424)

= 1069.38 gallons

7 0

3 years ago