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3 years ago

According to newtos first law of motion. what is required to make a object slow down.

Physics

1 answer:

ladessa [460]3 years ago

5 0

Answer:

Friction

Explanation:

Friction is a force that slows down moving objects. If you roll a ball across a shaggy rug, you can see that there are lumps and bumps in the rug that make the ball slow down. The rubbing, or friction, between the ball and the rug is what makes the ball stop rolling. External Force is required.

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Yolanda is studying two waves. The first wave has an amplitude of 2 m, and the second has an amplitude of 3 m. Which statement a

Anna71 [15]

Answer:

a) Not Accurate

b) Not Accurate

c) Accurate

d) Accurate

Explanation:

Part a

Not Accurate, because destructive interference would lead to maximum possible magnitude of < 3 m

Part b

Not Accurate, because constructive interference would lead to minimum possible magnitude of > 2 m

Part c

Accurate, because destructive interference would lead to maximum possible magnitude of < 3 m by varying the phase difference between two waves she can achieve the desired results.

Part d

Accurate, because constructive interference would lead to minimum possible magnitude of > 2 m by varying the phase difference between two waves she can achieve the desired results.

8 0

3 years ago

A basketball is held over head at a height of 2.4 m. The ball is lobbed to a teammate at 8 m/s at an angle of 40'. If the ball i

cupoosta [38]

Explanation:

since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.

We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.

range can be calculated by the formula :-

$\boxed{\mathfrak{range = \frac{ u {}^{2} \sin 2\theta }{g} }}$

u is the velocity during its take off and $\theta$ is the angle at which its thrown

Given that

u = 8m/ s
$\theta$ = 40°

calculating range using the above formula

$= \frac{ {8}^{2} \sin2(40) }{10}$

$= \frac{64 \times \sin(80) }{10}$

value of sin 80 = 0. 985

$= \frac{64 \times 0.985}{10}$

$= \frac{63.027}{10}$

$= 6.3027$

Hence,

$\mathfrak { \blue{the \: teammate \: is \: \red{\underline{6.3027 \: meters} }\: away } }$

7 0

3 years ago

Genevieve was working on a science lab. In her lab, she was combining two chemicals and timing how long the reaction took to for

dexar [7]

Answer: Her test trials had a high level of accuracy but a low level of precision.

Explanation:

5 0

3 years ago

A 60 mAs results in an exposure of 85 mGya, with all factors remaining the same, what would the new exposure be if 120 mAs is us

Cloud [144]

Answer: d₂ = 170 mGya

Explanation:

the relationship between absonbed 'd' and exposure 'E' is given as;

D(Gv) = F . x (AS/xB)

F is a conversion coefficient depending on medium

so we can simply write

d₁/d₂ = x₁/x₂

Given that;

our x₁ = 60 mAs, x₂ = 120 mAs, d₁ = 85 mGya, d₂ = ?

from the given formula,

d₂ = (x₂d₁ / x₁)

now we substitute

d₂ = (120 × 85) / 60

d₂ = 170 mGya

∴ if 120 mAa is used, the new exposure will be 170 mGya

8 0

2 years ago

A particle of mass m moves under a conservative force with potential energy V ( x )= cx/(x2+a2),where c and a are positive const

Anvisha [2.4K]

Answer:

The position of stable equilibrium is -a

And the period of small oscillations must be: c/(ma^3)

Explanation:

Since the potential is:

$V(x) = \frac{c x}{a^2+x^2}$

We first look for a position of stable equilibrium. This posiiton must satisfy two considtions, that the first derivative of the potential must vanish at this point and the second derivative must be positive.

$V'(x) = \frac{c}{a^2+x^2}-\frac{2 c x^2}{\left(a^2+x^2\right)^2}$

Which vanish for

x = a ; x =-a

The second derivative of V(x) is:

$V''(x) = \frac{8 c x^3}{\left(a^2+x^2\right)^3}-\frac{6 c x}{\left(a^2+x^2\right)^2}$

And:

$V''(a) = -\frac{c}{2 a^3}\\V''(-a) = \frac{c}{2 a^3}\\$

Therefore:

The position of stable equilibrium is -a

And the period of small oscillations must be:

$\omega = \sqrt{2 V''(-a)/m} = \sqrt{\frac{c}{a^3 m}}$

(c/(ma^3))^1/2

Let's find the maximum amplitude if the particle starts at this point with velocity v

If this is the case, the total energy will be:

(mv^2)/2

And the maximum amplitude will be

x = a^3/c mv^2 = (m v^2 a^3)/ c

7 0

3 years ago