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3 years ago

What happens to the car if it is traveling in a circular path suddenly encounters ice?

Physics

2 answers:

pychu [463]3 years ago

8 0

the car's friction is reduced to zero

the car's centripetal force drops to zero

the car continues in a straight path from the point at which it encountered the ice

dimulka [17.4K]3 years ago

5 0

When a car is travelling in a circular path then for the circular motion of the the centripetal force is provided by the friction force

This friction force will help to take turn in circular path

Now during the motion of car if friction suddenly disappear due to which there is no centripetal force will act on it

Since there is no centripetal force on the car so now it will not be able to take turn

so here is the possible outcomes

A.) the car's friction is reduced to zero.

B.) the car's centripetal force drops to zero.

C.) the car continues in a straight path from the point at which it encountered the ice.

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A 2.50 gram rectangular object has measurements of 22.0 mm, 13.5 mm, and 12.5 mm. what is the object's density in units of g/ml?

Zinaida [17]

G/mL is equivalent to g/cm^3, so we first convert the dimensions into cm:
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4 years ago

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Answer:

Explanation:

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3 years ago

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3 0

3 years ago

An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e

White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

$x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)$

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

$\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}$

with this value you can compute the frequency:

$f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz$

the mass of the block is given by the formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg$

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

$\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s$

Finally, the amplitude is:

$x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m$

5 0

3 years ago

M=3000km v=25m/s what’s the momentum

valina [46]

<h3>Answer:</h3>

Momentum of the given body will be : 75000 Kg m/s

<h3>Explanation:</h3>

According to Newton's first law of motion, all bodies continue to be in the state of rest or motion unless an external force is applied on the body. We can use this in the case of momentum also

The formula of momentum is given by :

$:\implies \sf\quad \sf \: P = mv$

Here, we are given the mass of the body ( m ) as 3000kg and the velocity of the body ( v ) as 25 m/s. On putting the values in the formula:

$\begin{aligned}&:\implies \sf\quad \sf \: P = mv \\& :\implies \sf\quad \sf \: P = 3000 \times 25 \\ & :\implies \sf\quad \sf \: \boxed{ \sf \: P = 75000kgm {s}^{ - 1} } \end{aligned}$

Momentum is associated with the mass of the moving body and can be defined as the quantity of motion measured as a product of mass and velocity.

8 0

2 years ago