5. bending the electric wire a.change in size b.change in shape c.change in texture

A bus hits a bug and the bug splatters on the windshield, which force is greater?

grandymaker [24]

The forces acting on a body and the type of motion that results are given by Newton's three Laws of motion

The force of the bus is the same as reactive force of the bug

Reason:

According to Newton's third Law of motion, given that the bug collides

with the bus, the force with which the bus hits the bug is equal to the

reactive force of the bug on the on the bus

According to Newton's second Law of motion, force is equal to the rate of

change of the momentum produced

The impulse of the force of the bus on the bug is given as follows;

F·Δt = m·(v₂ - v₁)

Given that the force of the bus is large, the change in momentum, m·(v₂ - v₁),

is also large such that the parts of the bug are split by the rapid change in

velocity, and the bug splatters on the windshield, and is then carried along

on the trip,

The equally large reactive force of the bug, is such that the bug splatters

due its magnitude

Therefore, the correct response is that the forces are the same

Lean more here:

brainly.com/question/21279060

8 0

2 years ago

S/REF No. Date If the load distance of a level is 20 cm and effort distance is 6ocm, calculate the amount of effort required to

andrew11 [14]

The answer is 602 cause I added

7 0

3 years ago

FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

$\omega^2=0+2(1.47)(12.97)$ $w^2=38.12$ $\boxed{\omega=6.17.}$

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0

1 year ago

Question 4. A tuning fork ‘A’ produces 6 beats/sec with another fork ‘B’ of un-known frequency. On

8090 [49]

Clever problem.

We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?

Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.

If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.

The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.

4 0

3 years ago

Quel est le type de rayonnement produit par la radioactivité

Dmitriy789 [7]

Il existe troi types de rayons produits lors de la désintégration des éléments radioactifs:

-- "particules alpha" . . . noyaux d'hélium, composés chacun de 2 protons et 2 neutrons

-- "rayons bêta" ou "particules bêta" . . . flux d'électrons

-- "rayons gamma" . . . rayonnement électromagnétique avec les longueurs d'onde les plus courtes connues et l'énergie la plus élevée

8 0

3 years ago

5. bending the electric wire a.change in size b.change in shape c.change in texture​

5. bending the electric wire a.change in size b.change in shape c.change in texture