Help me!!! Due todayyyyy!!!!!!!!!!!!!!! ❤️ :)

What makes steel durable

sdas [7]

It is durable because it is one of the strongest metals and doesn't corrode easily.

4 0

3 years ago

Do you know any good books about physics and math?

horrorfan [7]

What Is Mathematics? Is a book about math

A book about physics /A Brief History of Time

5 0

2 years ago

Match the letters to the numbers for 30 points!

liq [111]

-- 'Ca' (Calcium) is an element.
-- The proton has a positive charge.
-- Nuclear fusion results in the synthesis of atoms of new elements.
-- H₂O (water) is a chemical compound.
-- Nuclear fission is a decay of the nucleus.
-- The atomic number of an element is the number of protons
in each atom of it.
-- I suppose you're using the Greek letter <span>η ('eta', not 'nu')
to represent the neutron.
-- I suppose you're using ' e ' to represent the electron.
</span>

7 0

3 years ago

Read 2 more answers

A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch

ANTONII [103]

Answer:

A. A = 0.913 m

B. amax = 132.24m/s^2

C. Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

$v_{max}=A\omega$ (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

$\omega=\sqrt{\frac{k}{m}}$ (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

$\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}$

Next, you solve the equation (1) for A and replace the values of vmax and w:

$A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m$

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

$a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}$

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

$F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N$

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

4 0

3 years ago

An object is pulled up an incline plane at a constant velocity as the picture above shows. Calculate the tension on the rope.

WARRIOR [948]

Hi there!

$\large\boxed{545N}$

Since the object is being pulled at a constant velocity, the forces must be balanced.

Since there is no movement vertically, we must take into account the horizontal forces. We can also assume a positive acceleration to be in the direction of motion.

The acceleration and force due to gravity on an incline is:

a = gsinФ

F = MgsinФ

∑F = -MgsinФ + T

Since it is getting pulled at a constant velocity, ∑F = 0. So:

0 = -MgsinФ + T

MgsinФ = T

Solve for T by plugging in values. Let g = 10 m/s²

T = (120)(10)sin(27) ≈ 545 N

6 0

3 years ago