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alukav5142 [94]
4 years ago
6

Can you help me please?

Physics
1 answer:
egoroff_w [7]4 years ago
7 0
<h3>Answer</h3>

option B)

19N

<h3>Explanation</h3>

If the object is at equilibrium, then the net force acting upon the object should be 0 N. Thus, if all the forces are added together, horizontal and vertical forces separately, then the resultant force (the vector sum) should be 0 Newton.

As we only need to find the magnitude of x-component of force F

so find all x component/horizontal forces acting on the object.

50cos(40) - 40cos(25) + 30cos(55) + x = 0

38.30 - 36.25 + 17.21 + x + = 0

19.26 + x = 0

x = - 19.26

x ≈ 19 (magnitude only)

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Newton's laws of motion are true on earth and space.
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Answer:

False

Explanation:

It is a common misunderstanding that objects in space have no weight. If that were true, they would just float away from the Earth, the Sun and the other planets. Objects in low Earth orbit experience about 90% of the weight that they feel on the surface of the Earth.

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What type of issue does the effects manufacturer need
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Answer:

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Explanation:

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Suppose you wish to fabricate a uniform wire out of 1.10 g of copper. If the wire is to have a resistance R = 0.390 Ω, and if al
Citrus2011 [14]

To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:

Mass of copper wire(m) = 1.10g = 1.10*10^{-3} kg

Density (\rho)= 8.92*10^3kg/m^3

Resistively of copper (\gamma) = 1.7*10^{-8}\Omega \cdot m

Resistance (R) = 0.390\Omega

Volume is defined as,

V= lA \text{ and }\frac{m}{\rho}

lA= \frac{1.10*10^{-3}}{8.92*10^3}

lA = 1.233*10^{-7} m^3 (1)

We know that,

\frac{l}{A} = \frac{R}{\gamma}

\frac{l}{A}= \frac{0.390\Omega}{1.7*10^{-8}\Omega m}

\frac{l}{A} = 2.2941*10^7 m^{-1} (2)

Multiplying equation we have

l^2 = (1.233*10^{-7})( 2.2941*10^7)

l^2 = 2.8286m^2

l =\sqrt{2.8286m^2}

l = 1.68m

Therefore the length of the wire is 1.68m

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5 0
3 years ago
An ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component
Bond [772]

The y-component of the acceleration is 0.33 m/s^2

Explanation:

The y-component of the acceleration is given by:

a_y = \frac{v_y-u_y}{t}

where

v_y is the y-component of the  final velocity

u_y is the y-component of the initial velocity

t is the time elapsed

For the ice skater in this problem, we have:

u_y = u sin \theta_1 = (2.25 m/s)(sin 50.0^{\circ})=1.72 m/s

where

u = 2.25 m/s is the initial velocity

\theta_1 = 50.0^{\circ} is the initial  direction

v_y = v sin \theta_2 = (4.65)(sin 120^{\circ})=4.03 m/s, where

v = 4.65 m/s is the final velocity

\theta_2 = 120.0^{\circ} is the final direction

The time elapsed is

t = 8.33 s

Therefore, we can find the y-component of the acceleration:

a_y=\frac{4.03-1.72}{8.33}=0.33 m/s^2

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

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