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3 years ago

If the car goes exits a freeway and goes from 65 mph to 35 mph is it accelerating?

Physics

1 answer:

Zolol [24]3 years ago

8 0

Answer:

No, the car is decelerating

Explanation:

No the car is decelerating if it exits a freeway and goes from 65

mph to 35 mph since the change in velocity is negative.

change in velocity = final - initial

change in velocity = 35 - 65

change in velocity = -30mph

Since the change in velocity is negative, hence the car is decelerating. Deceleration is a negative acceleration

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A 85.0kg man and a 65, 0kg woman are riding a Ferris wheel with a radius of 20.0m. What is the Ferris wheels tangential velocity

zvonat [6]

Answer:

The Ferris wheel's tangential (linear) velocity if the net centripetal force on the woman is 115 N is 3.92 m/s.

Explanation:

Let's use Newton's 2nd Law to help solve this problem.

F = ma

The force acting on the Ferris wheel is the centripetal force, given in the problem: $F_c=115 \ \text{N}$ .

The mass "m" is the sum of the man and woman's masses: $85+65= 150 \ \text{kg}$ .

The acceleration is the centripetal acceleration of the Ferris wheel: $a_c=\displaystyle \frac{v^2}{r}$ .

Let's write an equation and solve for "v", the tangential (linear) acceleration.

$\displaystyle 115=m(\frac{v^2}{r} )$
$\displaystyle 115 = (85+65)(\frac{v^2}{20})$
$\displaystyle 115=150(\frac{v^2}{20} )$
$.766667=\displaystyle(\frac{v^2}{20} )$
$15.\overline{3}=v^2$
$v=3.9158$

The Ferris wheel's tangential velocity is 3.92 m/s.

8 0

2 years ago

What is the definition of physics?

BigorU [14]

Answer:

your answer should be c

8 0

4 years ago

Read 2 more answers

A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It en

Virty [35]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The acceleration would be $a = 0.003* 6 =0.018\ m/s^2$

Explanation:

The objective of this solution is to obtain the acceleration of the particle

Now looking at Newton law which is mathematically represented as

$F = ma$

Where F is the force experience by a particle

m is the mass of the particle

a is the acceleration of the particle

And also this force is equivalent to magnetic force in a magnetic field which is mathematically represented as

$F = qvB$

Where q is the charge of the particle

v is the velocity of the charge

B is the magnetic field the charge is under it influence

Now equating this two formulas

$ma = qvB$

Making a the subject we have

$a = \frac{qvB}{m}$

In the question the direction of the is in the positive x-axis which is $i$ hence the direction would be in the $i$ direction

So substituting $(2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T$ for B

$a = \frac{q}{m} * v (2.0i +3.0j +4.0k)*10^{-3}$

Substituting $3.0 Km /s = 3.0*10^{3}\ m/s$ for v and $-6.0 \muC = -6.0*10^{-6} C$ for q

$a = \frac{-6.0*10^{-6}}{2.0*10^{-3}} * 3.0*10^{3} *(2i+3j+4k) *10^{-3}$

$a = 0.003 * 3i(2i+3j+4k)$

$a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j \ + (3*4)i \ \cdot \ k)$

According to vector multiplication

$i \cdot i = j \cdot j = k\cdot k = 1\\\\and \ i\cdot j = i\cdot k = 0$

$a = 0.003* 6 =0.018\ m/s^2$

8 0

3 years ago

Que es una ondaaaa?? no me deja escribiiiirrrrrr

UkoKoshka [18]

Answer:

yo queiro escribir mi hermano

7 0

3 years ago

A toy train rolls around a horizontal 1.0-m-diameter track.The coefficient of rolling friction is 0.10.a) What is the magnitude

valina [46]

Answer:

1.962 rad / s², 1.6 s

Explanation:

Radius of the part = 1.0m / 2 = 0.5 m

angular speed = 30 rpm = 30 rpm × (2πrad / rev) × 1 minutes / 60 seconds = 3.142 rads⁻¹

μk = frictional force / normal ( mg )

normal is the force acting upward against the force of gravity

frictional force = - μk mg

since the body came to rest then

Fnet + Ff = 0

Fnet = - Ff

Fnet = ma

ma = - μk mg

a = - μkg where g = 9.81

a = - 0.1 × 9.81 = 0.981 m/s²

magnitude of angular acceleration = tangential acceleration / radius = 0.981 / 0.5 = 1.962 rad / s²

b) time for the train to come to rest = angular velocity / angular angular acceleration = 3.142/ 1.962 = 1.6 s

The equation earlier derived answer this question

Fnet + Ff = 0 since the body came to a rest

Fnet = - Ff and Ff = - μk mg, Fnet = ma

ma = - μk mg

m cancel m on both side

a = - μkg since it magnitude

a = μkg

5 0

4 years ago