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2 years ago

What is the main energy source that drives global climate?

Physics

1 answer:

marysya [2.9K]2 years ago

6 0

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The correct option would be A. solar.

We also use solar energy to produce electricity.

Have A Nice Day ❤

Stay Brainly! ヅ

- Ally ✧

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A 1,700 kg car is being used to give a 1,400 kg car a push start by exerting a force of

Gwar [14]

Answer:

- 670 kg.m/s

Explanation:

Newton's third law states that to every action, there is equal and opposite reaction force. Since the force will be same but different in direction and acted in the same time then the impulses ( force multiply by time) of the two car be same in magnitude but different in direction - 670 kg.m/s

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3 years ago

A student wants to start a small business in school. Write down six items that

Fiesta28 [93]

Answer:

packets of pen

packets of pencil

copies

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bottles

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2 years ago

The surface area of a postage

ANTONII [103]

Answer:

608

Explanation:

Trust me

4 0

2 years ago

Read 2 more answers

As a capacitor is being charged in an RC circuit. the current flowing through a resistor isa) increasingb) decreasingc) constant

Montano1993 [528]

<h2>Answer: decreasing</h2>

An RC circuit is an electrical circuit composed of resistors and capacitors, where the charging time $T$ of the circuit is proportional to the magnitude of the electrical resistance $R$ and the capacity $C$ of the capacitor.

As shown below:

$T=R.C$

In this context, the electrical resistance is the opposition to the flow of electrons when moving through a conductor.

Therefore:

<h2>When a capacitor is being charged in an RC circuit, the current flowing through a resistor decreases.</h2>

And the correct option is b.

4 0

2 years ago

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

8 0

2 years ago