3 years ago

Which of the

Physics

1 answer:

djyliett [7]3 years ago

6 0

Answer:

Short THICC wire

Explanation:

You might be interested in

Two point charges are separated by 6.4 cm . The attractive force between them is 10 N . Suppose that the charges attracting each

LenaWriter [7]

Answer:

Two point charges are separated by 6.4 cm . The attractive force between them is 10 N .

units.

Explanation:

7 0

2 years ago

The second minimum in the diffraction pattern of a 0.11-mm-wide slit occurs at 0.72°. What is the wavelength of the light?

Anit [1.1K]

Answer:

691.13 nm

Explanation:

d = width of the slit = 0.11 x 10⁻³ m

θ = angle of diffraction pattern = 0.72° degree

λ = wavelength of the light = ?

m = order = 2 (since second minimum)

for the second minimum diffraction pattern we use the equation

d Sinθ = m λ

Inserting the values

(0.11 x 10⁻³) Sin0.72 = (2) λ

λ = 691.13 x 10⁻⁹ m

λ = 691.13 nm

6 0

3 years ago

Problem A spring was at its resting position where it is attached to a wall at its left side and a block at its right side as sh

puteri [66]

Answer:

F = 19.1 N

Explanation:

To find the force exerted by the string on the block you use the following formula:

$F=kx$ (1)

k: spring constant = 95.5 N/m

x: displacement of the block from its equilibrium position = 0.200 m

you replace the values of k and x in the equation (1):

$F=(95.5N/m)(0.200m)=19.1N$

Hence, the force exterted on the block is 19.1 N

8 0

3 years ago

Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.00 ✕ 105 kg on springs that ha

jasenka [17]

Explanation:

It is given that,

Mass of an object, $m=4\times 10^5\ kg$

(a) Time period of oscillation, T = 2.4 s

The formula for the time period of spring is given by :

$T=2\pi \sqrt{\dfrac{m}{k}}$

Where

k is the force constant

$k=\dfrac{4\pi ^2 m}{T^2}$

$k=\dfrac{4\pi ^2 \times 4\times 10^5}{(2.4)^2}$

$k=2.74\times 10^6\ N/m$

(b) Displacement in the spring, x = 2.2 m

Energy stored in the spring is given by :

$U=\dfrac{1}{2}kx^2$

$U=\dfrac{1}{2}\times 2.74\times 10^6\ N/m\times (2.2\ m)^2$

$U=6.63\times 10^6\ J$

Hence, this is the required solution.

7 0

3 years ago

How many Gigameters in 1600 meters

USPshnik [31]

1.6e-6 gl on the rest of your physics its a killer for me too

6 0

3 years ago

Read 2 more answers