A car takes 10 s to go from vi=0 m/s to vf=25 m/s. How far does the car travel in that time?

What instrument uses a magnetic field to magnify imags up to 100000000?

Mariana [72]

Transmission electron microscope

5 0

3 years ago

The auto in the sketch moves forward as the brakes are applied. A bystander says that during the interval of braking, the auto's

Ivan

Answer:

The statement is true: velocity and acceleration have opposite directions in the interval of braking.

Explanation:

Let's say we have a velocity $v>0$ .

The acceleration $a$ is the rate of change of the velocity $v$ . This means that if $v$ is increasing during time, then $a$ must be positive. But if $v$ is decreasing over time, then $a$ will be negative (even though the velocity is positive).

Mathematically:

$a=\frac{dv}{dt}$

$v$ decreases ⇒ $\frac{dv}{dt}$

⇒ $a$ .

Example:

$v(t)=e^{-t}>0 \\\\\frac{dv}{dt}=-te^{-t}$

3 0

3 years ago

A block of mass m = 150 kg rests against a spring with a spring constant of k = 880 N/m on an inclined plane which makes an angl

weeeeeb [17]

Answer:

b) k Δx - W cos θ - μ mg cos θ = m a , c) θ = 86.6º, d) Δx = 1.18 m

Explanation:

a) In the attachment we can see a diagram of the forces in this problem and the coordinate axes for its decomposition.

F is the force applied by the spring, while it is compressed, this force disappears when the block leaves the spring

b) Let's apply Newton's second law for when the spring is compressed

let's use trigonometry to break down the weight

sin θ = Wₓ / W

cos θ = W_y / W

Wₓ = W sin θ

W_y = W cos θ

Y axis

N - W_y = 0

N = W_y

N = W cos θ

X axis

F -Wₓ -fr = ma

the force applied by the spring is given by hooke's law

F = k Δx

friction force has the expression

fr = μ N

fr = μ W cos θ

we substitute

k Δx - W cos θ - μ mg cos θ = m a ( 1)

c) If the plane has no friction, what is the angle so that Δx = 0.1m

We write the equation 1, with fr = 0 and since the system is still a = 0

k Δx - W cos θ -0 = 0

cos θ = $\frac{k \Delta x}{ m g}$

cos θ = $\frac{880 \ 0.1}{ 150 \ 9.8}$

cos θ = 0.0598

θ = cos⁻¹ 0.0598

θ = 86.6º

d) In this part they give the angle θ = 45º and there is no friction, they ask the compression

the acceleration is zero, we substitute in 1

k Δx - W cos θ - 0 = 0

Δx = $\frac{mg \ cos \ \theta}{k}$

Δx = $\frac{ 150 \ 9.8 \ cos45}{880}$

Δx = 1.18 m

7 0

3 years ago

What is the average emf induced in a circular wedding ring of diameter 2.2 cm if the ring is at the center of the MRI magnet coi

Minchanka [31]

Missing information

Consider a superconducting MRI magnet for which the magnet field decreases from 8.0 T to nearly 0 in 20 s

Answer:

0.152 mV

Explanation:

Change in magnetic field, dB=8-0=8 T

Time interval, dt=20 s

Diameter of ring= 2.2cm hence radius is 1.1 cm equivalent to 0.011

$\epsilon=nA\frac {dB}{dt}$ where n is the number of terms and for this question n=1 and A is area which is given by $\pi r^{2}$ hence

$\epsilon=n\pi r^{2} \frac {dB}{dt}$ and substituting the values

$\epsilon=1\pi 0.011^{2}\times \frac {8}{20}=1.52\times 10^{-4} V=0.152 mV$

Therefore, induced emf is 0.152 mV

7 0

4 years ago

You do 30 J of work to load a toy dart gun. However, the dart is 10 cm long and feels a frictional force of 10 N while going thr

shepuryov [24]

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

$K.E_{1}$ = 30 J, l = 10 cm = 0.1 m (as 1 m = 100 cm)

F = 10 N

The formula to calculate the work done is as follows.

W = $F \times l$

Hence, putting the given values into the above formula as follows.

W = $F \times l$

= $10 N \times 0.1 m$

= 1 J

As the dart loses 1 joule of energy when it goes through the barrel. Therefore, the kinetic energy it has when it comes out is calculated as follows.

K.E = 30 - 1

= 29 J

Thus, we can conclude that the kinetic energy of the fired dart is 29 J.

6 0

3 years ago