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3 years ago

Pushing a baby on a swing is easier than pushing an adult on the same swing.

Physics

2 answers:

Shalnov [3]3 years ago

7 0

Answer:

Law 2:

Pushing a baby on a swing is easier than pushing an adult on the same swing. Water spills from a glass carried by someone who is walking steadily and suddenly stops short.

Law 2:

When thrown with the same force, a soccer ball accelerates more than a bowling ball.

Law 1:

A magician pulls a tablecloth out from under a dish on a table without disturbing the dish.

Law 3:

A rocket launches into space, pushing fuel exhaust in one direction and the rocket in the opposite direction.

Law 1:

A book rests on top of a shelf and does not move until a student accidentally knocks it off.

Explanation:

Newton's laws of motion:

Law 1: Law of inertia

Law 2: Law of force (F = mass x acceleration)

Law 3: Law of Action & Reaction

Sergeeva-Olga [200]3 years ago

5 0

Answer:

what is the question though

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Which of the following is a component of staying positive while competing

Natalija [7]

Answer: B - complimenting others on good plays

Explanation: Reading the first words sort of give it away when staying positive you compliment, not criticize, confront angrily, or refuse.

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3 years ago

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True or false? An object at rest has an instantaneous acceleration of zero.

Tpy6a [65]

True. it’s at rest which means it’s not moving so there’s no acceleration

8 0

3 years ago

A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o

Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let $q_1$ and $q_2$ be the charges such that

$q_1+q_2=4.7$

and Force between charge particles is given by

$F=\frac{kq_1q_2}{r^2}$

$4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}$

$q_1\cdot q_2=0.522$

put the value of $q_1$

$q_2\left ( 4.7-q_2\right )=0.522$

$q_2^2-4.7q_2+0.522=0$

$q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}$

$q_2=0.114 C$

thus $q_1=4.586 C$

3 0

3 years ago

An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +44 ft/s2. After some time t1,

mash [69]

Answer:

a) t₁ = 4.76 s, t₂ = 85.2 s

b) v = 209 ft/s

Explanation:

Constant acceleration equations:

x = x₀ + v₀ t + ½ at²

v = at + v₀

where x is final position,

x₀ is initial position,

v₀ is initial velocity,

a is acceleration,

and t is time.

When the engine is on and the sled is accelerating:

x₀ = 0 ft

v₀ = 0 ft/s

a = 44 ft/s²

t = t₁

So:

x = 22 t₁²

v = 44 t₁

When the engine is off and the sled is coasting:

x = 18350 ft

x₀ = 22 t₁²

v₀ = 44 t₁

a = 0 ft/s²

t = t₂

So:

18350 = 22 t₁² + (44 t₁) t₂

Given that t₁ + t₂ = 90:

18350 = 22 t₁² + (44 t₁) (90 − t₁)

Now we can solve for t₁:

18350 = 22 t₁² + 3960 t₁ − 44 t₁²

18350 = 3960 t₁ − 22 t₁²

9175 = 1980 t₁ − 11 t₁²

11 t₁² − 1980 t₁ + 9175 = 0

Using quadratic formula:

t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22

t₁ = 4.76, 175

Since t₁ can't be greater than 90, t₁ = 4.76 s.

Therefore, t₂ = 85.2 s.

And v = 44 t₁ = 209 ft/s.

3 0

3 years ago

A charge feels a 7.89 x 10-7 N force when it moves 2090 m/s at a 29.4° angle to a 4.23 x 10-3 T magnetic field. What is the amou

tamaranim1 [39]

We have that the amount of the charge q is

$q=1.8*10^{-7}$

From the Question we are told that

Force $F=7.89 x*10^{-7}$

Velocity $V=2090m/s$

Angle $\theta=29.4$

Magnetic field $B=4.23 * 10^{-3} T$

Generally, the equation for Force F is mathematically given by

$F=qVBsin\theta\\\\q=\frac{F}{VBsin\theta}$

$q=\frac{7.89 x*10^{-7}}{4.23 * 10^{-3} T*sin29.4*2090m/s}$

$q=1.8*10^{-7}$

In conclusion

The amount of the charge q is

$q=1.8*10^{-7}$

For more information on this visit

brainly.com/question/1470439

6 0

3 years ago

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