All categories

3 years ago

Forces in Motion:

Physics

1 answer:

Maurinko [17]3 years ago

3 0

<h3>Answer:</h3>

29.4 m/s

<h3>Explanation:</h3>

Acceleration due to gravity is generally assumed to be about 9.8 m/s², so after 3 s, the velocity will be ...

(3 s)×(9.8 m/s²) = 29.4 m/s . . . . downward

You might be interested in

Millikan is doing his oil drop experiment. He has a droplet with radius 1.6 µm suspended motionless in a uniform electric field

swat32

Answer:

The charge on the droplet is $3.106\times10^{-16}\ C$ .

Yes, quantization of charge is obeyed within experimental error.

Explanation:

Given that,

Radius = 1.6μm

Electric field = 46 N/C

Density of oil = 0.085 g/cm³

We need to calculate the charge on the droplet

Using formula of force

$F= qE$

$mg=qE$

$V\times\rho\times g=qE$

$q=\dfrac{V\times\rho}{E}$

$q=\dfrac{\dfrac{4}{3}\pi\times r^3\times\rho\times g}{E}$

Put the value into the formula

$q=\dfrac{\dfrac{4}{3}\times\pi\times(1.6\times10^{-6})^3\times85\times9.8}{46}$

$q=3.106\times10^{-16}\ C$

We need to calculate the quantization of charge

Using formula of quantization

$n = \dfrac{q}{e}$

Put the value into the formula

$n=\dfrac{3.106\times10^{-16}}{1.6\times10^{-19}}$

$n=1941.25$

Yes, quantization of charge is obeyed within experimental error.

Hence, The charge on the droplet is $3.106\times10^{-16}\ C$ .

Yes, quantization of charge is obeyed within experimental error.

7 0

3 years ago

Read 2 more answers

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

8 0

4 years ago

If an object stay still or could 10 years moving and the same speed in the same direction it is a __ force

sveta [45]

Answer:

infinite

Explanation:

8 0

3 years ago

Read 2 more answers

What is role of force on the speed of moving object?

aivan3 [116]

Explanation:

this is the ans hope it works

8 0

3 years ago

Read 2 more answers

kykrilka [37]

Answer:

<h2>n+5 is ur answer</h2>

hope it helps uh.

6 0

3 years ago