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3 years ago

What is the net force acting on the object above?

Physics

1 answer:

photoshop1234 [79]3 years ago

6 0

The answer would be C (the one you picked)

Hope this helps

Have a great day/night

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Kelly sits on a rock. Her weight is an action force. Describe the reaction force.

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An equal and opposite force acting upwards

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4 years ago

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What color would the sky be if the atmosphere was 100% large molecules and particles like dust and water?

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3 years ago

Express force in terms of base units

MariettaO [177]

<h2><u>Answer:</u></h2>

<u></u>

<u>Force = kgm/s² ⟶ Newton</u>

<h2><u>Explanation:</u></h2>

<u></u>

According to the formula we've learnt,

Force = mass × acceleration.

To find force in terms of base units, we must first identify the base SI units of mass & acceleration.

Base SI unit of mass = kg (kilogram)

Base SI unit of acceleration = m/s² (meter per second squared)

So, the base unit of force is,

Force = mass × acceleration.

Force = kg × m/s²

<u>Force = kgm/s² ⟶ Newton</u>

________________

Hope it helps!

$\mathfrak{Lucazz}$

4 0

2 years ago

Do electromagnetic waves require matter to travel?

VMariaS [17]

Answer:

No.

Explanation:

Electromagnetic waves do not require a medium of matter to move through, electromagnetic waves are used in things like your cell phone and telecommunications.

6 0

2 years ago

How do I calculate the tension in the horizontal string?

matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

$\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}$

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

$\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}$

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

$T_1\cos 30\degree-m\cdot g=0$

Solve for T₁,

$T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N$

Now, we use the second equation to find the tension in the horizontal string,

$T_2-T_1\sin 30\degree=0$

Solve for T₂,

$T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N$

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

8 0

1 year ago