HELP ME PLEASE WITH MY HOMEWORK ILL GIVE BRAINLIEST

You are at the top of the Empire State Building on the 102nd floor, which is located 373 m above the ground, when your favorite

Liula [17]

Answer:

1.5 m

Explanation:

H = actual height of the superhero = ?

H₀ = height of the superhero as observed = 1.73 m

v = speed of the superhero = 0.50 c

Using the equation

$H = H_{o} \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}$

Inserting the values

$H = 1.73 \sqrt{1 - \left ( \frac{0.50 c}{c} \right )^{2}}$

H = 1.5 m

3 0

3 years ago

Vector A has a magnitude of 30 units. Vector B is perpendicular to vector Aand has a magnitude of 40 units. What would the magni

Fudgin [204]

Answer:

$|\vec A + \vec B| = 50 units$

Explanation:

As we know that magnitude of two vectors is given as

$|\vec A + \vec B| = \sqrt{A^2 + B^2 + 2AB cos\theta}$

here we know that

A = magnitude of vector A

B = magnitude of vector B

$\theta$ = angle between two vectors

so here we know that

A = 30 units

B = 40 units

angle = 90 degree

so we have

$|\vec A + \vec B| = \sqrt{30^2 + 40^2 + 2(30)(40)cos90}$

$|\vec A + \vec B| = \sqrt{30^2 + 40^2}$

$|\vec A + \vec B| = 50 units$

3 0

2 years ago

A construction worker is pushing a 50.0-kg box with a force of 150.0 N to the right. If the box is moving at a constant velocity

9966 [12]

Yes omg yes I literally have the same question and need to find the answer

7 0

2 years ago

A hot plume of mantle material, which may extend to the core-mantle boundary, produces a volcanic region a few hundred kilometer

Fantom [35]

Answer: Hot spot

The volcanic hot spot is an area in the mantle from which heat rises as a thermal plume from deep in the Earth.

5 0

2 years ago

Read 2 more answers

If it requires 8.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re

jenyasd209 [6]

Answer:

The amount of work done required to stretch spring by additional 4 cm is 64 J.

Explanation:

The energy used for stretching spring is given by the relation :

$E = \frac{1}{2}kx^{2}$ .......(1)

Here k is spring constant and x is the displacement of spring from its equilibrium position.

For stretch spring by 2.0 cm or 0.02 m, we need 8.0 J of energy. Hence, substitute the suitable values in equation (1).

$8 = \frac{1}{2}\timesk\times k \times(0.02)^{2}$

k = 4 x 10⁴ N/m

Energy needed to stretch a spring by 6.0 cm can be determine by the equation (1).

Substitute 0.06 m for x and 4 x 10⁴ N/m for k in equation (1).

$E = \frac{1}{2}\times4\times10^{4}\times (0.06)^{2}$

E = 72 J

But we already have 8.0 J. So, the extra energy needed to stretch spring by additional 4 cm is :

E = ( 72 - 8 ) J = 64 J

7 0

2 years ago