Is an electrical current which comes from a battery

uniform ladder of length 6.0 m and weight 300 N leans against a frictionless vertical wall. The foot of the ladder isplaced 3.0

olganol [36]

Answer:

Fx1 (6 m) sin 60 = 300 (3 m) cos 60 balancing torques about floor

Fx1 = 900 * 1/2 / 5.20 = 86.6 N this is the horizontal force that must be supplied by the wall to balance torques about the floor

This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.

Fx1 = Fx2 = 86.6 N

3 0

3 years ago

A van accelerates from 4m/s to 20m/s in 8s. How far does it travel in this time?

zubka84 [21]

Answer:

96 m

Explanation:

Given,

Initial velocity ( u ) = 4 m/s

Final velocity ( v ) = 20 m/s

Time ( t ) = 8 s

Let Acceleration be " a ".

Formula : -

a = ( v - u ) / t

a = ( 20 - 4 ) / 8

= 16 / 8

a = 2 m/s²

Let displacement be " s ".

Formula : -

s = ut + at² / 2

s = ( 4 ) ( 8 ) + ( 2 ) ( 8² ) / 2

= 32 + ( 2 ) ( 64 ) / 2

= 32 + ( 2 ) ( 32 )

= 32 + 64

s = 96 m

Therefore, it travels 96 m in time 8 s.

3 0

3 years ago

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between

Anni [7]

Answer:

0.074m/s

Explanation:

We need the formula for conservation of momentum in a collision, this equation is given by,

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$

Where,

$m_1$ = mass of ball

$m_2$ = mass of the person

$u_1$ = Velocity of ball before collision

$u_2$ = Velocity of the person before collision

$v_1$ = velocity of ball afer collision

$v_2$ = velocity of the person after collision

We know that after the collision, as the person as the ball have both the same velocity, then,

$v_1 = v_2$

$m_1u_1 + m_2u_2 = (m_1+m_2)v_2$

Re-arrenge to find $v_2$ ,

$v_2 = \frac{m_1u_1+m_2u_2}{m_1+m_2}$

Our values are,

$m_1$ = 0.425kg

$u_1$ = 12m/s

$m_2$ = 68.5kg

$u_2$ = 0m/s

Substituting,

$v_2 = \frac{(0.425)(12)+(68.5)(0)}{0.425+68.5}$

$v_2 = 0.074m/s$

<em />

<em>The speed of the person would be 0.074m/s after the collision between him/her and the ball</em>

7 0

3 years ago

represents the space-time, speed-time and acceleration-time graphs for a ball pulled upwards with a speed of 10 m / s from 1 met

alexandr1967 [171]

Gyhugugygubuhuhuhubu

5 0

3 years ago

A bird flying straight upward at 5 m/s drops a berry when it is 300 m above the ground. How fast is the berry going when it hits

Nikitich [7]

Answer:

v=77.62 m/s

Explanation:

Given that

h= - 300 m

speed of the bird ,u= 5 m/s

Lets take Speed of the berry when it hit the ground = v m/s

we know that ,if object is moving upward

v² = u² - 2 g h

u=Initial speed

v=Final speed

h=Height

Now by putting the values

v² = u² - 2 g h

v² = 5² - 2 x 10 x (-300) ( take g = 10 m/s²)

v² =25 + 20 x 300

v² ==25 + 6000

v² =6025

v=77.62 m/s

Therefore the final speed of the berry will be 77.62 m/s.

5 0

3 years ago