How long does it take lewis hamilton travelling at 156 mph to cover 16 miles?

A block of mass m = 1.00 kg is attached to a spring of force constant k = 500 N/m. The block is pulled to a position xi= 5.00 cm

8_murik_8 [283]

Answer:

The speed of the block is 4.96 m/s.

Explanation:

Given that.

Mass of block = 1.00 kg

Spring constant = 500 N/m

Position $x_{i}=5.00\ cm$

Coefficient of friction = 0.350

(A). We need to calculate the speed the block has as it passes through equilibrium if the horizontal surface is friction less

Using formula of kinetic energy and potential energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2-\mu mgx$

Put the value into the formula

$\dfrac{1}{2}\times1.00\times v^2=\dfrac{1}{2}\times500\times(5.00\times10^{-2})-0.350\times1.00\times9.8\times5.00\times10^{-2}$

$v^2=\dfrac{2\times12.3285}{1.00}$

$v^2=24.657$

$v=4.96\ m/s$

Hence, The speed of the block is 4.96 m/s.

6 0

3 years ago

Help pls lol and tysm

Zanzabum

Answer:

umm section 2 and 4

Explanation:

because at section 2 it starts and at section 4 it moves again and stops at 3.

7 0

2 years ago

IBM has a fast computer that it calls the Blue Gene/L that can do '136.8

dmitriy555 [2]

Answer:

138.6 megacalculations

Explanation:

This is a pretty straightforward one.

All it needs is to convert the degree of measurement.

Dimensions in physics are attributed names, which state the power to which they're are raised. Just as how

Kilo and Mega means the numbers are raised to the power of 3 and 6 respectively. There also exists the ones that indicates how small, such as milli and micro, which are to the powers of -3 & -6.

The question says the IBM computer calculates at an astonishing 136.8 teracalculations.

Tera in physics means it's raised to the power of 12. Thus, the IBM calculates at an astonishing rate of

136.8*10^12 calculations per second.

We're then asked how many calculations it does in 1 micro second. Like I had highlighted earlier, 1 micro second is 1 raised to the power of -6. Or succinctly put,

1 micro second = 1*10^-6.

If the IBM does

138.6*10^12 = 1 second,

Then it does

x = 1*10^-6 second.

When we cross multiply, we have

138.6*10^12 * 1*10^-6, and that is

138.6*10^6 calculations, or say, 138.6 megacalculations.

The IBM does 138.6 megacalculations in 1 micro second, which is still astonishing, by the way

6 0

3 years ago

Which groups of organisms became extinct during the Paleozoic Era

guapka [62]

Dinosaurs but I need the whole groups yo tell you ;)

4 0

3 years ago

two circular loops of wire, each containing a single turn, have the same radius of 4.0 cm and a common center. the planes of the

REY [17]

The magnetic field at center of circular loops of wire is 3.78 x 10¯⁵ T.

We need to know about the magnetic field at the center of circular loops of wire to solve this problem. The magnetic field at the center can be determined as

B = μ₀ . I / 2r

where B is magnetic field, μ₀ is vacuum permeability (4π×10¯⁷ H/m), I is the current and r is radius.

From the question above, we know that:

r = 4 cm = 0.04 m

I = 1.7 A

By substituting the parameter, we get

B = μ₀ . I / 2r

B = 4π×10¯⁷ . 1.7 / (2.0.04)

B = 2.67 x 10¯⁵ T

Due to the perpendicular plane of loops, the total magnetic field at center will be

Btotal = √(2(B²))

Btotal = √(2(2.67 x 10¯⁵²))

Btotal = 3.78 x 10¯⁵ T

Find more on magnetic field at: brainly.com/question/7802337

#SPJ4

3 0

1 year ago