The increase in resistant strains of insects to chemicals is a result of:

For a vertical spring-mass oscillator that is moving up and down, which of the following statements are true? (more than one sta

omeli [17]

Answer:

At the lowest point in the oscillation, the momentum is zero.

At the lowest point in the oscillation, $mg < ks_s$

Explanation:

Since spring block system is performing to and fro motion along straight line

So here we can say at the lowest position of its path the velocity will become zero.

So we can say that momentum of the spring block system is given as

$P = mv$

$P = 0$

Also we know that after reaching the lowest point the block will again go up towards its mean position

So at the lowest point of the spring block system the block will move upwards again

So this will accelerate upwards hence

$F_{spring} > mg$

$Ks > mg$

6 0

2 years ago

A heliocentric system is _____-centered.

aniked [119]

A heliocentric system is a sun-centered

3 0

3 years ago

Read 2 more answers

HELP

dem82 [27]

Answer:

Explanation:

28 / 70 = 0.3857142... = 0.39 hr

280 / 100 = 2.8 hrs.

(100 - 0) / 10 = 10 m/s²

(60 - 20) / 4 = 10 m/s²

3 0

2 years ago

An ideal spring hangs from the ceiling. A 1.95 kg mass is hung from the spring, stretching the spring a distance d=0.0865 m from

uranmaximum [27]

Answer:

kinetic energy = 0.1168 J

Explanation:

From Hooke's law, we know that ;

F = kx

k = F/x

We are given ;

Mass; m = 1.95 kg

Spring stretch; d = x = 0.0865

So, Force = mg = 1.95 × 9.81

k = 1.95 × 9.81/0.0865 = 221.15 N/m

Now, initial energy is;

E1 = mgL + ½k(x - L)²

Also, final energy; E2 = ½kx² + ½mv²

From conservation of energy, E1 = E2

Thus;

mgL + ½k(x - L)² = ½kx² + ½mv²

Making the kinetic energy ½mv² the subject, we have;

½mv² = mgL + ½k(x - L)² - ½kx²

We are given L=0.0325 m

Plugging other relevant values, we have ;

½mv² = (1.95 × 9.81 × 0.0325) + (½ × 221.15(0.0865 - 0.0325)² - ½(221.15 × 0.0865²)

½mv² = 0.62170875 + 0.3224367 - 0.82734979375

½mv² = 0.1168 J

7 0

2 years ago

A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find

pshichka [43]

Answer:

0.050 m

Explanation:

The strength of the magnetic field produced by a current-carrying wire is given by

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0=4\pi \cdot 10^{-7} H/m$ is the vacuum permeability

I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

$I=1.41 A$ (current in the wire)

$B=5.61\mu T=5.61\cdot 10^{-6} T$ (strength of magnetic field)

Solving for r, we find the distance from the wire:

$r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m$

4 0

3 years ago