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3 years ago

A light microscope allows for more magnification than electron microscope because it uses a beam of visible light. true false

Physics

1 answer:

Igoryamba3 years ago

4 0

False because light microscopes have low resolve and magnification.

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All mutations are favorable and increase the ability for the

masya89 [10]

False sometimes you get a bad mutation

3 0

2 years ago

What damage a tornado can cause

Ksivusya [100]

Alrighty, so, houses can be destroyed, people can be killed, animals can be killed, habitats can be destroyed, and trees can be uprooted. Tornadoes can create massive damage.

Hope this helps! (:

5 0

2 years ago

A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the grou

True [87]

Answer:

At the highest point the velocity is zero, the acceleration is directed downward.

Explanation:

This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.

I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.

At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.

6 0

3 years ago

6.6 kg block initially at rest is pulled to theright along a horizontal, frictionless surfaceby a constant, horizontal force of

neonofarm [45]

Answer:

v = 3.04 m/s

Explanation:

given,

mass of the block, M = 6.6 Kg

horizontal force, F = 12.2 N

distance, L = 2.5 m

initial speed = 0 m/s

speed of the block,v = ?

we now

Work done is equal to change in Kinetic energy.

Work done = Force x displacement

W = Δ K E

Δ K E = Force x displacement

$\dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2= F .s$

$\dfrac{1}{2}\times 6.6 \times v^2 - 0= 12.2\times 2.5$

3.3 v² = 30.5

v² = 9.242

v = 3.04 m/s

speed of the block is equal to 3.04 m/s

4 0

2 years ago

Work of 5 Joules is done in stretching a spring from its natural length to 19 cm beyond its natural length. What is the force (i

densk [106]

Answer:

Explanation:

Given that,

5J work is done by stretching a spring

e = 19cm = 0.19m

Assuming the spring is ideal, then we can apply Hooke's law

F = kx

To calculate k, we can apply the Workdone by a spring formula

W=∫F.dx

Since F=kx

W = ∫kx dx from x = 0 to x = 0.19

W = ½kx² from x = 0 to x = 0.19

W = ½k (0.19²-0²)

5 = ½k(0.0361-0)

5×2 = 0.0361k

Then, k = 10/0.0361

k = 277.008 N/m

The spring constant is 277.008N/m

Then, applying Hooke's law to find the applied force

F = kx

F = 277.008 × 0.19

F = 52.63 N

The applied force is 52.63N

6 0

2 years ago