If the force is applied to a car, then its acceleration will ___________ because of Newton's ____________

Why should we convert units within the metric system

irga5000 [103]

It is often necessary to convert one metric measurement to another unit—this happens frequently in the medical, scientific, and technical fields, where the metric system is commonly used.

7 0

4 years ago

A circular loop of flexible iron wire has an initial circumference of 164cm , but its circumference is decreasing at a constant

Travka [436]

Answer:

emf = 0.02525 V

induced current with a counterclockwise direction

Explanation:

The emf is given by the following formula:

$emf=-\frac{\Delta \Phi_B}{\Delta t}=-B\frac{\Delta A}{\Delta t}$ $\ \ =-B\frac{A_2-A_1}{t_2-t_1}$ (1)

ФB: magnetic flux = BA

B: magnitude of the magnetic field = 1.00T

A2: final area of the loop; A1: initial area

t2: final time, t1: initial time

You first calculate the final A2, by taking into account that the circumference of loop decreases at 11.0cm/s.

In t = 4 s the final circumference will be:

$c_2=c_1-(11.0cm/s)t=164cm-(11.0cm/s)(4s)=120cm$

To find the areas A1 and A2 you calculate the radius:

$r_1=\frac{164cm}{2\pi}=26.101cm\\\\r_2=\frac{120cm}{2\pi}=19.098cm$

r1 = 0.261 m

r2 = 0.190 m

Then, the areas A1 and A2 are:

$A_1=\pi r_1^2=\pi (0.261m)^2=0.214m^2\\\\A_2=\pi r_2^2=\pi (0.190m)^2=0.113m^2$

Finally, the emf induced, by using the equation (1), is:

$emf=-(1.00T)\frac{(0.113m^2)-(0.214m^2)}{4s-0s}=0.0252V=25.25mV$

The induced current has counterclockwise direction, because the induced magneitc field generated by the induced current must be opposite to the constant magnetic field B.

4 0

3 years ago

A small particle with positive charge q = +4.25 x 10^-4C and mass m = 5.00 x 10^-5 kg is moving in a region of uniform electric

Tcecarenko [31]

Answer:

a) r = (0.6 i- 2039 j ^ + 0.102 k⁾ m and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s c) v_{z} = 1.02 10⁻¹m / s

Explanation:

a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration

Fe + Fm = m a

a = (Fe + Fm) / m

the electric force is

Fe = q E k ^

Fe = 4.25 10-4 60 k ^

Fe = 2.55 10-2 k ^

the magnetic force is

Fm = q v x B

Fm = 4.25 10⁻⁴ $\left[\begin{array}{ccc}i&j&k\\30&0&0\\0&0&49\end{array}\right]$

fm = 4.25 10⁻⁴ (-j ^ 30 4)

fm 0 = ^ -5,10 10⁻² j

We look for every component of acceleration

X axis

aₓ = 0

there is no force

Axis y

ay = -5.10 10²/5 10⁻⁵ j ^

ay = -1.02 107 j ^ / s2

z axis

az = 2.55 10⁻² / 5 10⁻⁵ k ^

az = 5.1 10² k ^ m / s²

Having the acceleration in each axis we can encocoar the position using kinematics

X axis

the initial velocity is vo = 30 m / s and an initial position xo = 0

x = vo t + ½ aₓ t₂2

x = 30 0.02 + 0

x = 0.6m

Axis y

acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m

y = I + go t + ½ ay t²

y = 1 + 0 + ½ (-1.02 10⁷) 0.02²

y = 1 - 2.04 10³

y = -2039 m j ^

z axis

acceleration is aza = 5.1 10² m / s², the position and initial speed are zero

z = zo + v₀ t + ½ az t²

z = 0 + 0 + ½ 5.1 10² 0.02²

z = 1.02 10⁻¹ m k ^

therefore the position of the bodies is

r = (0.6 i- 2039 j ^ + 0.102 k⁾ m

b) x axis

since there is no acceleration the speed remains constant

vₓ = 30.0 m / s

Axis y

let's use the equation v = v₀ + $a_{y}$ t

$v_{y}$ = 0 + -1.02 10⁷ 0.02

v_{y} = 2.04 10⁵ m / s

z axis

v_{z} = vo + az t

v_{z} = 0 + 5.1 10² 0.02

v_{z} = 1.02 10⁻¹m / s

8 0

4 years ago

Please help me find b) and d):

maks197457 [2]

Answer:

wow it's so hard

8 0

3 years ago

A worker pushes horizontally on a 35.0 kg crate with a force of magnitude 112 N. The coefficient of static friction between the

OLga [1]

Answer: a) -127 N b) No. c) -112 N d) 40 N e) 15 N

Explanation:

a) Friction force always oppose to the relative movement between two surfaces, and, provided that be less than the fs max, adopt any value to counteract the applied force.

The fs max, is the horizontal component of the contact force, and can be written as follows:

Fs max = us . Fn

As the block is at rest in the vertical direction, this means that Fn must be numerically equal to the weight of the object:

Fn = m g = 35 kg. 9.8 m/s2 = 343 N → Fs max = 0.37. 343 N = 127 N

b) Now, as the applied force is smaller than Fs max, this means that the friction force, is equal and opposite to the applied forcé, i.e., -112 N, so the crate doesn´t move.

c) Please see above.

d) As explained above, the maximum friction force, is proportional to the normal force, which adopts any value needed to satisfy the Newton´s 2nd Law.

So, if we diminish the normal force, we can lower the máximum friction force, helping to the worker to move the crate.

The mínimum needed normal force, will be the one that satisfies the following:

Fs max = F applied = us Fn = 112 N = 0.37. Fn.

Solving for Fn, we get Fn= 303 N

So, the difference between the original normal force and the new one, will be the mínimum upward force needed to make the crate to move, as follows:

Fup = 343 N -303 N = 40 N

e) If we keep the normal force unchanged, but add an horizontal force to help the worker, we will need that the sum of both forces, will be equal to the Fs max, as follows:

Fh + Fapp = 127 N → Fh = 127 N – 112 N = 15 N

5 0

3 years ago

If the force is applied to a car, then its acceleration will ___________ because of Newton's _____________.

If the force is applied to a car, then its acceleration will _ because of Newton's ___.