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8_murik_8 [283]
3 years ago
8

ASAP pls answer right if can’t see picture don’t answer

Physics
1 answer:
Anna35 [415]3 years ago
6 0

Answer:

Not sure but

F = m* a

32= 5 * a

a= 6.4 m/s^2

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On Earth, a brick has a mass of 10 kg and a weight of 5 lbs. What predictions could we make about the mass and weight of the bri
belka [17]

Answer:

Mass remains constant but weight reduces

Explanation:

Mass is the amount of matter in an object so whether on moon or any other planet, it does not change despite the changes in acceleration.

Weight is a product of mass and acceleration due to gravity, expressed as W=mg where m is the mass, W is weight and g is acceleration. From the above formula, it is evident that when you decrease g, then W also decreases while m is constant. Similarly, when m is constant and g is increased then W also increases.

Therefore, for this case, since g decreases, the weight decreases but mass remains constant.

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With certain exceptions, Class E airspace extends upward from either 700 feet or 1,200 feet AGL to, but does not include,A) 14,5
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B) 18,000 feet MSL

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There are three-dimensional parts in the navigation airspace in the world. The class E airspace is mostly used in the regions with coastal areas that are relatively populated. If we consider certain forms of exceptions, the class E airspace can move in the upward direction to few feet (i.e. 1200 ft). However, this doesn't include 18,000 feet MSL.

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Atoms of which two elements could combine with atoms of chromium (Cr) to
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D and E

Ionic bonds are formed between metals and non-metals and both of those are metals

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Jake is helping Fin push a box at a constant velocity up an incline that makes an angle of 30.0° above the horizontal by applyin
andre [41]

Given data

The angle of inclination of the plane is theta = 30 degree

The applied force in the inclined plane is F = 94 N

The distance moved in the inclined plane is d = 2.30 m

The coefficient of kinetic friction is u_k = 0.280

The free-body diagram of the above configuration is shown below:

Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.

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The expression for the work done by the pushing force is given as:

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Thus, the work done by the pushing force is 216.2 J.

(b)

The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.

\begin{gathered} F=F_f+mg\sin \theta \\ F=\mu_kN+mg\sin \theta \end{gathered}

The expression for the normal reaction force is given as:

N=mg\cos \theta

The expression for the mass of the box is given as:

\begin{gathered} F=\mu_k\times mg\cos \theta+mg\sin \theta \\ m=\frac{F}{\mu_kg\cos \theta+g\sin \theta} \end{gathered}

Substitute the value in the above equation.

\begin{gathered} m=\frac{94\text{ N}}{0.28\times9.8m/s^2\times\cos 30^o+9.8m/s^2\times\sin 30^0} \\ m=12.9\text{ kg} \end{gathered}

Thus, the mass of the box is 12.9 kg.

7 0
1 year ago
Four forces are exerted on a disk of radius R that is free to spin about its center, as shown above. The magnitudes are proporti
Dmitry_Shevchenko [17]

The given magnitude of forces of F₁ = F₄, F₂ = F₃, F₁ = 2·F₂, give the

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The given parameters are;

F₁ = F₄

F₂ = F₃

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Therefore;

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\vec{F_2} = \mathbf{ -F_2 \,  \hat j}

Clockwise moment due to F₄, M₁ = -0.5 \times F_4 \,  \hat j  \times \dfrac{R}{2}

Therefore;

M_1  =- 0.5 \times 2 \times  F_2 \,  \hat j  \times \dfrac{R}{2} =   \mathbf{ -F_2 \,  \hat j  \times \dfrac{R}{2}}

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Learn more about the resolution of vectors here:

brainly.com/question/1858958

4 0
2 years ago
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