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ycow [4]
3 years ago
15

Jason applies a force of 4.00 newtons to a sled at an angle 62.0 degrees from the ground. What is the component of force effecti

ve in pulling the sled horizontally along the ground?

Physics
2 answers:
Naddika [18.5K]3 years ago
6 0
When a force, F, is applied to an object at an angle,θ, to compute for the horizontal and vertical components, we have the following

F_{vertical} = Fsin(\theta)
F_{horizontal} = Fcos(\theta)

So, for this particular problem, given that the force applied to the sled is 4.00 N at 62⁰. Thus, the horizontal component of the force is equal to

F = 4.00(cos62⁰) = 1.88 N

Therefore, the horizontal force is 1.88 N.
Answer: 1.88 N

kaheart [24]3 years ago
5 0
You want to draw a free body diagram of the forces on the sled in the horizontal x-direction.

If you visualize the system in an x-y coordinate plane, the force along the x-direction is the angle it makes with the x-axis multiples by the force.

The angle made with the x-axis is cosine of the angle theta.

Please see picture attached.

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Brilliant_brown [7]

\large \sf{What \:  is \:  the  \: shape  \: of  \: a  \: solid}

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Solids have a definite shape and volume. Liquids have a definite volume, but take the shape of the container. Gases have no definite shape or volume.

Solids have a fixed shape and a fixed size. The particles are very close together and held in place by strong forces (bonds). Their particles cannot move around, but they do vibrate. Because the particles cannot move around, a solid has a fixed shape.

Solids can be hard like a rock, soft like fur, a big rock like an asteroid, or small rocks like grains of sand. The key is that solids hold their shape and they don't flow like a liquid.

\sf{ \huge}{ \colorbox{pink}{  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:}}

Hope It Helps!

7 0
3 years ago
n isolated charged soap bubble of radius R0=7.45 cmR0=7.45 cm is at a potential of V0=307.0 volts.V0=307.0 volts. If the bubble
Gnesinka [82]

Complete Question

An isolated charged soap bubble of radius R0 = 7.45 cm  is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r

Answer:

The difference is    U_f -U_i = 16 *10^{-7} J

Explanation:

From the question we are told that

     The radius of the soap bubble  is  R_o =  7.45 \ cm =  \frac{7.45}{100} =  0.0745 \ m

      The potential of the soap bubble is  V_1  =307.0 V

      The new radius of the soap bubble  is R_1 =  0.19 * 7.45=1.4155\ cm = 0.014155 \ m

The initial electric potential is mathematically represented as

     U_i  = \frac{V_1^2 R_o }{2k }

The final  electric potential is mathematically represented as

    U_f  = \frac{V_2^2 R_1 }{2k }

The initial potential is mathematically represented as

     V_1 =  \frac{kQ}{R_o}

The final  potential is mathematically represented as

        V_2 =  \frac{kQ}{R_1}

Now  

         \frac{V_2}{V_1}  =  \frac{R_o}{R_1}

substituting values

        \frac{V_2}{V_1}  =  \frac{7.45}{1.4155} =   \frac{1}{0.19}

=>      V_2 =  \frac{V_1}{0.19}

    So

         U_f  = \frac{V_1^2 R_2 }{2k * 0.19^2}

Therefore

        U_f -U_i = \frac{V_1^2 R_2 }{2k * 0.19^2} - \frac{V_1^2 R_o }{2k }

       U_f -U_i =     \frac{V_1^2}{2k} [\frac{ R_1 }{ * 0.19^2} - R_o]

where k is the coulomb's constant with value 9*10^{9} \  kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

       U_f -U_i =     \frac{307^2}{9 * 10^{9}} [\frac{ 0.014155 }{ 0.19^2} - 0.0745]

       U_f -U_i = 16 *10^{-7} J

           

     

8 0
3 years ago
Suppose you add two vectors A and B . What relative direction between them produces the resultant with the greatest magnitude? W
Akimi4 [234]

Answer:

Same direction to produce maximum magnitude and opposite direction to produce minimum magnitude

Explanation:

Let a be the angle between vectors A and B. Generally when we add A to B, we can split A into 2 sub vectors, 1 parallel to B and the other perpendicular to B.

Also let A and B be the magnitude of vector A and B, respectively.

We have the parallel component after addition be

Acos(a) + B

And the perpendicular component after addition be

Asin(a)

The magnitude of the resulting vector would be

\sqrt{(Acos(a) + B)^2 + (Asin(a))^2}

= \sqrt{A^2cos^2a + B^2 + 2ABcos(a) + A^2sin^2a}

= \sqrt{A^2(cos^2a + sin^2a) + B^2 + 2ABcos(a)}

= \sqrt{A^2 + B^2 + 2ABcos(a)}

As A and B are fixed, the equation above is maximum when cos(a) = 1, meaning a = 0 degree and vector A and B are in the same direction, and minimum with cos(a) = -1, meaning a = 180 degree and vector A and B are in opposite direction.

8 0
3 years ago
If mass is conserved during a nuclear reaction, then _____.
aleksley [76]
Then, the reaction does not have any kinetic energy
7 0
4 years ago
g Drop the object again and carefully observe its motion after it hits the ground (it should bounce). (Consider only the first b
Anestetic [448]

Answer:

a) quantity to be measured is the height to which the body rises

b) weighing the body , rule or fixed tape measure

c)   Em₁ = m g h

d) deformation of the body or it is transformed into heat during the crash

Explanation:

In this exercise of falling and rebounding a body, we must know the speed of the body when it reaches the ground, which can be calculated using the conservation of energy, since the height where it was released is known.

a) What quantities must you know to calculate the energy after the bounce?

The quantity to be measured is the height to which the body rises, we assume negligible air resistance.

So let's use the conservation of energy

starting point. Soil

          Em₀ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₀ = Em_f

         Em₀ = m g h₀

b) to have the measurements, we begin by weighing the body and calculating its mass, the height was measured with a rule or fixed tape measure and seeing how far the body rises.

c) We use conservation of energy

starting point. Soil

          Em₁ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₁ = Em_f

         Em₁ = m g h

d) to determine if the energy is conserved, the arrival energy and the output energy must be compared.

There are two possibilities.

* that have been equal therefore energy is conserved

* that have been different (most likely) therefore the energy of the rebound is less than the initial energy, it cannot be stored in the possible deformation of the body or it is transformed into heat during the crash

7 0
3 years ago
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