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3 years ago

Initial velocity 10 m/s accelerates at 5 m/s for 2 seconds whats the final velocity

Physics

1 answer:

stiks02 [169]3 years ago

8 0

Answer:

<em>The final velocity is 20 m/s.</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, and t the time, the final speed can be calculated as follows:

$v_f=v_o+at$

The provided data is: vo=10 m/s, $a=5\ m/s^2$ , t=2 s. The final velocity is:

$v_f=10~m/s+5\ m/s^2\cdot 2\ s$

$v_f=20\ m/s$

The final velocity is 20 m/s.

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How long must a 0.70-mm-diameter aluminum wire be to have a 0.40 a current when connected to the terminals of a 1.5 v flashlight

Natalka [10]

By using Ohm's law, we can find what should be the resistance of the wire, R:
$R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega$

Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
$r=35 mm=0.35 \cdot 10^{-3} m$
So the area is
$A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2$

And by using the resistivity of the Aluminum, $\rho=2.65 \cdot 10^{-8} \Omega m$ , we can use the relationship between resistance R and resistivity:
$R= \frac{\rho L}{A}$
to find L, the length of the wire:
$L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m$

4 0

3 years ago

When is a zero not significant?

maria [59]

Answer:

Not between significant digits.

Explanation:

A zero not significant when it's not between significant digits.

8 0

3 years ago

Read 2 more answers

Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the

nydimaria [60]

Answer: $F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

$F_1=\frac{kq(-q)}{(L\sqrt{2})^2}$

where $L\sqrt{2}=$ Distance between the two charges

$F_1=-\frac{kq^2}{2L^2}$

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

$F_2=\frac{kq(-q)}{(L)^2}$

The magnitude of force by both the charge is same but at an angle of $90^{\circ}$

thus combination of two forces at 2 and 3 will be

$F'=\sqrt{2}\frac{kq^2}{2L^2}$

Now it will add with force due to 1 charge

Thus net force will be

$F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

6 0

3 years ago

A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d

Dmitry [639]

Answer:

the distance in meters traveled by a point outside the rim is 157.1 m

Explanation:

Given;

radius of the disk, r = 50 cm = 0.5 m

angular speed of the disk, ω = 100 rpm

time of motion, t = 30 s

The distance in meters traveled by a point outside the rim is calculated as follows;

$\theta = \omega t\\\\\theta = (100 \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m$

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m

6 0

3 years ago

A 6 m long, uniform ladder leans against a frictionless wall and makes an angle of 74.3 ◦ with the floor. The ladder has a mass

olganol [36]

Answer: µ=0.205

Explanation:

The horizontal forces acting on the ladder are the friction(f) at the floor and the normal force (Fw) at the wall. For horizontal equilibrium,

f=Fw

The sum of the moments about the base of the ladder Is 0

ΣM = 0 = Fw*L*sin74.3º - (25.8kg*(L/2) + 67.08kg*0.82L)*cos74.3º*9.8m/s²

Note that it doesn't matter WHAT the length of the ladder is -- it cancels.

Solve this for Fw.

0= 0.9637FwL - (67.91L)2.652

Fw=180.1/0.9637

Fw=186.87N

f=186.81N

Since Fw=f

We know Fw, so we know f.

But f = µ*Fn

where Fn is the normal force at the floor --

Fn = (25.8 + 67.08)kg * 9.8m/s² =

910.22N

µ = f / Fn

186.81/910.22

µ= 0.205

4 0

3 years ago