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3 years ago

What is the momentum of a 37-kg person riding south on an 18-kg bicycle at 1.2 m/s

Physics

1 answer:

Arlecino [84]3 years ago

3 0

The equation of momentum is:

$p=mv$

Where p is the momentum, m is the mass and v is the velocity. In this case the mass of the person and the mass of the bike so:

$m_{total}=mass_{person}+mass_{bike}=37kg+18kg=55kg$

And the velocity of the person riding the bike is 1.2 m/s and so:

$p=(55kg)(1.2m/s)=66 \frac{kgm}{s}$

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Which famous scientist is associated with the force of gravity?

jenyasd209 [6]

Answer:

Isaac Newton

Explanation:

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3 years ago

A hiker leaves her camp and walks 3.5 km in a direction of 55° south of west to the lake. after a short rest at the lake, she hi

3241004551 [841]

Let's choose east as positive x-direction and south as positive y-direction. We can resolve the two displacement along these two axes:

- Displacement 1 (3.5 km, $55^{\circ}$ south of west

$d_{1x}=-(3.5 km)( cos 55^{\circ})=-2.01 km$

$d_{1y}=(3.5 km)( sin 55^{\circ})=2.87 km$

- Displacement 2 (2.7 km, $16^{\circ}$ east of south

$d_{2x}=(2.7 km)( sin 16^{\circ})=0.74 km$

$d_{1y}=(2.7 km)( cos 16^{\circ})=2.60 km$

So, the total components on the two directions are

$d_x = -2.01 km+0.74 km=-1.27 km$

$d_y=2.87 km+2.60 km=5.47 km$

And the magnitude of the hiker's resultant displacement is

$d=\sqrt{(1.27 km)^2+(5.47 km)^2}=5.6 km$

8 0

3 years ago

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Help! Please???

Andre45 [30]

Well for the The weather on the coast of Virginia is probably warm and dry. A cold front would likely bring <span>rain or rainstorms. would be B Rainstorms</span>

3 0

3 years ago

A rocket engine uses fuel and oxidizer in a reaction that produces gas particles having a velocity of 1380 ms The desired thrust

bearhunter [10]

Answer:

a. 141.3 kg/s b. 5.49 m/s² c. i. 104228.9 N ii. 8.53 m/s² d. i. 97305.2 N ii. 9.84 m/s²

Explanation:

a. What must be the fuel/oxidizer consumption rate (in kg s1)?

The thrust T = Rv where R = mass consumption rate and v = velocity of rocket. Since T = 195000 N and v = 1380 m/s,

R = T/v = 195000 N/1380 m/s = 141.3 kg/s

b. If the initial weight of the rocket is 125000 N, what is its initial acceleration?

We also know that thrust T - W = ma since the rocket has to move against gravity. where M = mass of rocket = W/g = 125000 N/9.8m/s² = 12755.1 kg, W = weight of rocket = 125000 N, a = acceleration of rocket and T = thrust = 195000 N.

So, T - W = Ma

195000 N - 125000 N = (12755.1 kg)a

70000 N = ma

a = 70000 N/12755.1 kg = 5.49 m/s²

c. What are the weight and acceleration of the rocket at t 15.0 s after ignition?

We know that the loss in mass ΔM = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 15 s,

ΔM = 141.3 kg/s × 15 = 2119.5 kg

The new mass is thus M = M - ΔM = 12755.1 kg - 2119.5 kg = 10635.6 kg

i.The weight after 15 seconds is thus W' = M'g = 10635.6 kg × 9.8m/s² = 104228.9 N

ii. Since T - W' = M'a. where M' is our new mass and a our new acceleration,

a = (T - W')/M'

= (195000 N - 104228.9 N)/10635.6 kg

= 90771.1 N/10635.6 kg

= 8.53 m/s²

d. What are the weight and acceleration of the rocket at 20.0 s after ignition?

We know that the loss in mass ΔM" = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 20 s,

ΔM" = 141.3 kg/s × 20 = 2826 kg

The new mass is thus M" = M - ΔM" = 12755.1 kg - 2826 kg = 9929.1 kg

i. The weight after 20 seconds is thus W" = M"g = 9929.1 kg × 9.8m/s² = 97305.2 N

ii. Since T - W" = M"a. where M" is our new mass and a our new acceleration,

a = (T - W")/M"

= (195000 N - 97305.2 N)/9929.1 kg

= 97694.8 N/9929.1 kg

= 9.84 m/s²

4 0

3 years ago

Learning Goal: To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circ

andreev551 [17]

Answer:

v = 15.8 m/s

Explanation:

Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement. So the variation of mechanical energy is equal to the work of the fictional force

$W_{fr}$ = ΔEm = $Em_{f}$ -Em₀

Let's write the mechanical energy at each point

Initial

Em₀ = Ke = ½ k x²

Final

$Em_{f}$ = K + U = ½ m v² + mg y

Let's use Hooke's law to find compression

F = - k x

x = -F / k

x = 4400/1100

x = - 4 m

Let's write the energy equation

fr d = ½ m v² + mgy - ½ k x²

Let's clear the speed

v² = (fr d + ½ kx² - mg y) 2 / m

v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50) 2/60.0

v² = (160 + 8800 - 1470) / 30

v = √ (229.66)

v = 15.8 m/s

5 0

3 years ago

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