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3 years ago

Consider two less-than-desirable options.

Physics

1 answer:

Gemiola [76]3 years ago

7 0

Answer:

The force would be the same in both cases - option C.

Explanation:

The change in momentum is known as an impulse. In the two cases under consideration, the change in momentum is the same, thus impulse for both cases is the same.

Impulse is the average force multiplied by time interval.

I = F(average)*ΔT. Where F(average) is the average force and ΔT is the time interval.

The average force in both cases is the same since the collision time is the same.

Thus option C is the correct answer.

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Jerry the mouse is running along a straight desert road at a constant velocity of 18 m/s. If a certain Tom cat wants to capture

Kruka [31]

Answer:

a) t = 1.75 s

b) x = 31.5 m

Explanation:

a) The time at which Tom should drop the net can be found using the following equation:

$y_{f} = y_{0} + v_{oy}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height = 0

y₀: is the initial height = 15 m

g: is the gravity = 9.81 m/s²

$v_{0y}$ : is the initial vertical velocity of the net = 0 (it is dropped from rest)

$0 = 15m - \frac{1}{2}9.81 m/s^{2}*t^{2}$

$t = \sqrt{\frac{2*15 m}{9.81 m/s^{2}}} = 1.75 s$

Hence, Tom should drop the net at 1.75 s before Jerry is under the bridge.

b) We can find the distance at which is Jerry when Tom drops the net as follows:

$v = \frac{x}{t}$

$x = v*t = 18 m/s*1.75 m = 31.5 m$

Then, Jerry is at 31.5 meters from the bridge when Jerry drops the net.

I hope it helps you!

3 0

3 years ago

English or metric <br> What is 16in to cm

il63 [147K]

40.64 cm is the answer to your question.

6 0

3 years ago

(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.

faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let call "x₁" distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁ ( force between earth and the object )

F₁ = K * Mt * m₀/ ( x₁)² K is a gravitational constant

F₂ (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂ K*Mt*m₀ / x₁² = K*Ml*m₀ /x₂²

Or simplifying the expression

Mt/ x₁² = Ml/ x₂²

We know that x₁ + x₂ = 384000 Km then

x₁ = 384000 - x₂

Mt/( 384000 - x₂)² = Ml / x₂²

Mt * x₂² = Ml *( 384000 - x₂)²

We need to solve for x₂

Mt * x₂² = Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂² = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂² = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂² + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂ = -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 254880*10∧10 / 1160

x₂ ₁ = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁ = -56.43*10∧5 + 508* 10∧5 / 1160

x₂ ₁ = 451.27*10∧5/1160

x₂ ₁ = 4512.7*10∧4 /1160

x₂ ₁ = 3.89*10∧4 km (distance between the moon and the object)

x₂ ₁ = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁ = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0

3 years ago

If a 400-mm diameter pipe with a pipe roughness coefficient of 100 flows full of pressurized water with a head loss of 0.4 ft pe

RoseWind [281]

Answer:

Q = 913.9 gpm

Explanation:

The Hazen Williams equation can be written as follows:

$P = \frac{4.52\ Q^{1.85}}{C^{1.85}d^{4.87}}$

where,

P = Friction Loss per foot of pipe = $\frac{0.4}{1000\ ft}$ = 4 x 10⁻⁴

Q = Flow Rate in gallon/min (gpm) = ?

d = pipe diameter in inches = (400 mm)(0.0393701 in/1 mm) = 15.75 in

C = roughness coefficient = 100

Therefore,

$4\ x \ 10^{-4} = \frac{4.52\ Q^{1.85}}{(100)^{1.85}(15.75)^{4.87}}\\\\Q^{1.85} = \frac{4\ x \ 10^{-4}}{1.33\ x\ 10^{-9}} \\\\Q = (300384.75)^\frac{1}{1.85}$

5 0

3 years ago

In this problem, you will calculate the location of the center of mass for the Earth-Moon system, and then you will calculate th

Radda [10]

Answer:

a) Option D is correct.

The center of mass between the Eartg and the moon is inside the Earth.

Explanation:

Given,

Mass of the moon = (7.35×10²²) kg

Mass of the Earth = (6.00×10²⁴) kg

Mass of the Sun = (2.00×10³⁰) kg

Distance between the Earth and the moon = (3.80×10⁵) km

Distance between the Earth and the Sun = (1.50×10⁸) km

With the assumption that all.of the bodies being considered are on the same straight line on the x-axis,

Note that Centre of mass is given as

C.M = (Σmx)/(Σm)

For the Earth-moon system, let the earth be x=0, then the moon is at x = (3.80 × 10 5) km away.

C.M = (Σmx)/(Σm)

Σmx = (6.00×10²⁴)) × (0) + (7.35×10²²) × (3.80×10⁵) = (2.793 × 10²⁸) kg.km

Σm = (6.00×10²⁴) + (7.35×10²²) = (6.0735 × 10²⁴) kg

CM = (2.793 × 10²⁸) ÷ (6.0735 × 10²⁴)

CM = (4.60 × 10³) km = 4600 km

This means the centre of mass is 4600 km from the Earth.

The Earth's radius = 6378 km

Hence, the centre if mass is inside the Earth.

Hope this Helps!!!

8 0

3 years ago