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Travka [436]
4 years ago
14

Which situation is contrary to Newton’s first law of motion? An object at rest begins to move when an external force is applied.

A moving object has a constant velocity under the influence of balanced forces. An object at rest stays at rest as long as unbalanced forces act on it. A moving object slows down when an external force is applied.
Physics
2 answers:
vaieri [72.5K]3 years ago
7 0

Newton's first law establishes the following:

An object will remain at rest or with uniform rectilinear motion unless an external force acts on it.

Therefore, we have that a case contrary to Newton's first law is:

Unbalanced forces acting on an object at rest, cause the object to remain at rest.

It is false, because if the forces are unbalanced, then the object will not remain at rest because external forces act on the object.

Answer:

A situation that is contrary to Newton's first law of motion is:

An object at rest stays at rest as long as unbalanced forces act on it.

Dmitrij [34]3 years ago
6 0

C. An object at rest stays at rest as long as unbalanced forces act on it.

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Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

F=\dfrac{kQ^2}{r^2} ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}.....(2)

Dividing equation (1) and (2), we get :

\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}

Hence, the correct option is (d) i.e. " 4F/9"

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Microscopes are often equipped with blue filters. give a reason for using blue light in a microscope
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The blue light removes red parts of the color spectrum giving it a cooler color

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An archer puts a 0.30 kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m.a. Assuming
Vlad [161]

Answer:

Explanation:

Given

mass of archer m=0.3\ kg

Average force F_{avg}=201\ N

extension in arrow x=1.3\ m

Work done to stretch the bow with arrow

W=F\cdot x

W=201\times 1.3=261.3\ m

This work done is converted into kinetic Energy of arrow

W=\frac{1}{2}mv^2

where v= velocity of arrow

261.3=\frac{1}{2}\times 0.3\times v^2

v=\sqrt{1742}

v=41.73\ m/s

(b)if arrow is thrown vertically upward then this energy is converted to Potential energy

W=mgh

261.3=0.3\times 9.8\times h

h=\frac{261.3}{0.3\times 9.8}

h=88.87\ m

   

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3 years ago
The action force is the balloon pushing the air out. What is the magnitude of the reaction force of the air pushing on the ballo
aleksley [76]
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An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and stic
Vika [28.1K]

Answer:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

Given that,

Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

R =u\cos\theta t

Put the value into the formula

\dfrac{230}{6} = u\cos\theta

u\cos\theta=38.33.....(I)

We need to calculate the height

Using vertical component

H=u\sin\theta t-\dfrac{1}{2}gt^2

Put the value in the equation

16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2

u\sin\theta=\dfrac{16+9.8\times18}{6}

u\sin\theta=32.06.....(II)

Dividing equation (II) and (I)

\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}

\tan\theta=0.8364

\theta=\tan^{-1}0.8364

\theta=39.90^{\circ}

(a). We need to calculate the initial speed

Using equation (I)

u\cos\theta\times t=38.33

Put the value into the formula

u =\dfrac{230}{6\times\cos39.90}

u=49.96\ m/s

(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

5 0
3 years ago
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