All categories

3 years ago

Steam types of forces in nature

Physics

1 answer:

max2010maxim [7]3 years ago

7 0

Answer:

Gravity, Weak, Electromagnetic and Strong.

You might be interested in

Match the particles with their characteristics,

In-s [12.5K]

Positive charge=proton
Negative charge=electron
No charge/neutral=neutron

6 0

3 years ago

Read 2 more answers

What formula relstes work n power

777dan777 [17]

<span>Raj is trying to make a diagram to show what he has learned about nuclear fusion.
</span>

8 0

3 years ago

The fact that your eyes rotate in their sockets, either getting closer or farther from each other, as an object changes its dist

marta [7]

Answer:

convergence

Explanation:

5 0

3 years ago

A plastic film moves over two drums. During a 4-s interval the speed of the tape is increased uniformly from v0 = 2ft/s to v1 =

Ratling [72]

Answer:

Question 1)

a) The speed of the drums is increased from 2 ft/s to 4 ft/s in 4 s. From the below kinematic equations the acceleration of the drums can be determined.

$v_1 = v_0 + at \\4 = 2 + 4a\\a = 0.5~ft/s^2$

This is the linear acceleration of the drums. Since the tape does not slip on the drums, by the rule of rolling without slipping,

$v = \omega R\\a = \alpha R$

where α is the angular acceleration.

In order to continue this question, the radius of the drums should be given.

Let us denote the radius of the drums as R, the angular acceleration of drum B is

α = 0.5/R.

b) The distance travelled by the drums can be found by the following kinematics formula:

$v_1^2 = v_0^2 + 2ax\\4^2 = 2^2 + 2(0.5)x\\x = 12 ft$

One revolution is equal to the circumference of the drum. So, total number of revolutions is

$x / (2\pi R) = 6/(\pi R)$

Question 2)

a) In a rocket propulsion question, the acceleration of the rocket can be found by the following formula:

$a = \frac{dv}{dt} = -\frac{v_{fuel}}{m}\frac{dm}{dt} = -\frac{13000}{2600}25 = 125~ft/s^2$

b) $a = -\frac{v_{fuel}}{m}\frac{dm}{dt} = - \frac{13000}{400}25 = 812.5~ft/s^2$

5 0

3 years ago

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.59 times a second. A tack is stuck in the tire a

Vesna [10]

Answer:

Tangential speed=5.4 m/s

Radial acceleration= $88.6m/s^2$

Explanation:

We are given that

Angular speed=2.59 rev/s

We know that

1 revolution= $2\pi rad$

2.59 rev= $2\pi\times 2.59=5.18\pi=5.18\times 3.14=16.27 rad/s$

By using $\pi=3.14$

Angular velocity= $\omega=16.27rad/s$

Distance from axis=r=0.329 m

Tangential speed= $r\omega=16.27\times 0.329=5.4m/s$

Radial acceleration= $\frac{v^2}{r}$

Radial acceleration= $\frac{(5.4)^2}{0.329}=88.6m/s^2$

6 0

3 years ago