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2 years ago

13

In this activity, you will analyze the motion of a cart being pulled up an inclined plane at a constant speed. The angle of the

incline can be modified by 10 degrees increments between the values of 30 degrees and 90 degrees. Three different masses can be selected - 2.0kg, 3.0kg, and 4.0kg in each simulation the cart is pull d to the same height - 1.0 meter above the original starting position. For each simulation, the force that must be applied is reported on the screen. The displacement of the cart can be measured using the cm-ruler that is displayed for each trial. Can some help me

1 answer:

Verizon [17]2 years ago

8 0

Answer:

darn this too much lololpop

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select all that apply the force of gravity a. changes slightly with the location on the earth b. decreases with height above sea

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The answer to your question is C

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2 years ago

Consider an insulated tank with a volume V = 2 L is separated into two equal-volume parts by a thin wall. On the left is an idea

steposvetlana [31]

Answer

given,

V = 2 L

the left is an ideal gas at P = 100 k Pa and T = 500 K

mass is constant

$m_1 = m_2$

$\dfrac{P_1V_1}{RT_1} = \dfrac{P_2V_2}{RT_2}$

Pressure is same because it's not changing due to process

$\dfrac{V}{500} = \dfrac{2 V}{T_2}$

$T_2 = 1000\ K$

$\Delta S_{univ} = \Delta S_{sys} + (\Delta S)_{surr}$

$\Delta S_{univ} =m(C_v ln (\dfrac{T_2}{T_1}))+ R ln (\dfrac{V_2}{V_1})$

$m = \dfrac{P_1V_1}{RT_1}$

$m = \dfrac{100 \times 10^3 \times 2 \times 10^{-3}}{287\times 500}$

m = 1.39 x 10⁻³ Kg

$\Delta S_{univ} =1.39\times 10^{-3}(0.718 ln\ 2+ 0.287 ln (2)$

$\Delta S_{univ} =0.968\times 10^{-3}\ kJ/K$

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2 years ago

Tropical winds that blow east to west are?

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Its called the "westerly"

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3 years ago

A 295-kg object and a 595-kg object are separated by 4.10 m.

kodGreya [7K]

Answer:

a)F=3 x 10⁻⁷ N

b)x=2.405 m

Explanation:

Given that

m₁=295 kg

m₂=595 kg

d= 4.1 m

a)

m₃=63 kg

r=d/2 = 2.05 m

The force between the mass m₁ and m₃

$F_{13}=\dfrac{Gm_1m_3}{r^2}$

by putting the values

$F_{13}=\dfrac{Gm_1m_3}{r^2}$

$F_{13}=\dfrac{6.67\times 10^{-11}\times 295\times 63 }{2.05^2}$

F₁₃=2.94 x 10⁻⁷ N

The force between the mass m₂ and m₃

by putting the values

$F_{23}=\dfrac{Gm_2m_3}{r^2}$

$F_{23}=\dfrac{6.67\times 10^{-11}\times 595\times 63 }{2.05^2}$

F₂₃=5.94 x 10⁻⁷ N

The net force F

F= F₂₃- F₁₃

F=5.94 x 10⁻⁷ N-2.94 x 10⁻⁷ N

F=3 x 10⁻⁷ N

b)

Lest take at distance x from mass m₂ net force is zero.

$F_{23}=\dfrac{Gm_2m_3}{x^2}$

$F_{13}=\dfrac{Gm_1m_3}{(4.1-x)^2}$

Form above two equation

$\dfrac{Gm_1m_3}{(4.1-x)^2}=\dfrac{Gm_2m_3}{x^2}$

$\dfrac{m_1}{(4.1-x)^2}=\dfrac{m_2}{x^2}$

$\dfrac{295}{(4.1-x)^2}=\dfrac{595}{x^2}$

x²=2.01(4.1-x)²

x=1.42 (4.1-x)

x=5.82 - 1.42x

x=2.405 m

4 0

2 years ago

A ball is hit so that it accelerates at 10 m/s2. The ball hits a glove with a force of 5 N. What is the mass of the ball?

jenyasd209 [6]

Answer:

0.5 kg

Explanation:

» <u>Concepts</u>

Newton's second law, the Law of Acceleration, states that F = ma, where F = Force in Newtons, m = mass in kg, and a = acceleration in m/s^2.

» <u>Application</u>

We are asked to find the mass of the ball using the equation F = ma. We're also given the force and acceleration, so the equation looks like 5 = 10(m).

» <u>Solution</u>

Step 1: Divide both sides by 10.

$5/10=10m/10$
$m=0.5$

Thus, the mass of the ball is 0.5 kg.

8 0

2 years ago