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jarptica [38.1K]

3 years ago

7

A box weighing 10 N is 4 m away from the fulcrum. How much force must be applied 1 m away on the opposite end of the fulcrum in

order for the sides to be balanced? What is the actual mechanical advantage of the machine?

1 answer:

julia-pushkina [17]3 years ago

4 0

Answer: Socratic

Explanation: i’m not sure but you can use Socratic it’s a good app that helps

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Hello guys! Can u please help me with physics. I translated it in English. Can yall help me please how much u can!!

DedPeter [7]

1. Since the body is thrown vertically upward, the only force acting on it as it rises and falls is gravity, which causes a constant downward acceleration with magnitude g = 9.8 m/s². Because this acceleration is constant, we can use the formula

v² - u² = 2a ∆x

where

u = initial speed

v = final speed

a = acceleration

∆x = displacement

At its maximum height, some distance y above the point where the body is launched, the body has zero velocity, so

0² - (20 m/s)² = 2 (-9.8 m/s²) y

Solve for y :

y = (20 m/s)² / (2 (9.8 m/s²)) ≈ 20.4 m

2. Relative to the ground, the body's maximum height is 60 m + 20.4 m ≈ 80.4 m.

3. At any time t ≥ 0, the body's vertical velocity is given by

v = 20 m/s - gt

At the highest point, we have

0 = 20 m/s - (9.8 m/s²) t

and solving for t gives

t = (20 m/s) / (9.8 m/s²) ≈ 2.04 s

4. The body's height y above the ground at any time t ≥ 0 is given by

y = 60 m + (20 m/s) t - 1/2 gt²

Solve for t when y = 0 :

0 = 60 m + (20 m/s) t - 1/2 (9.8 m/s²) t²

Using the quadratic formula,

t = (-b + √(b² - 4ac)) / (2a)

(and omitting the negative root, which gives a negative solution) where a = -1/2 (9.8 m/s²), b = 20 m/s, and c = 60 m. You should end up with

t ≈ 6.09 s

5. At the time found in (4), the body's velocity is

v = 20 m/s - g (6.09 s) ≈ -39.7 m/s

Speed is the magnitude of velocity, so the speed in question is 39.7 m/s.

6 0

3 years ago

A room with 3.1-m-high ceilings has a metal plate on the floor with V = 0V and a separate metal plate on the ceiling. A 1.1g gla

miss Akunina [59]

Answer:

The ball traveled 0.827 m

Explanation:

Given;

distance between the metal plates of the room, d = 3.1 m

mass of the glass, m = 1.1g

charge on the glass, q = 4.7 nC

speed of the glass ball, v = 4.8 m/s

voltage of the ceiling, V = +3.0 x 10⁶ V

The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;

F = qV/d

|F| = (4.7 x 10⁻⁹ x 3 x 10⁶) / (3.1)

|F| = 4.548 x 10⁻³ N

F = - 4.548 x 10⁻³ N

The net horizontal force experienced by this ball is;

$F_{net} = F_c - mg\\\\F_{net} = -4.548 *10^{-3} - (1.1*10^{-3} * 9.8)\\\\F_{net} = -15.328*10^{-3} \ N$

The work done between the ends of the plate is equal to product of the magnitude of net force on the ball and the distance traveled by the ball.

$W = F_{net} *h\\\\W = 15.328 *10^{-3} * h$

W = K.E

$15.328*10^{-3} *h = \frac{1}{2}mv^2\\\\ 15.328*10^{-3} *h = \frac{1}{2}(1.1*10^{-3})(4.8)^2\\\\ 15.328*10^{-3} *h =0.0127\\\\h = \frac{0.0127}{15.328*10^{-3}}\\\\ h = 0.827 \ m$

Therefore, the ball traveled 0.827 m

4 0

3 years ago

a car advertisement claims their car can go from a stopped position to move 60 miles per hour in 5 seconds the advertisement is

zysi [14]

What do you want us to anwser

7 0

3 years ago

Ted Clubber Lang. A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at t

il63 [147K]

Answer:

15 m/s or 1500 cm/s

Explanation:

Given that

Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s

Distance moved during the hook, d(h) = 5 cm = 0.05 m

Distance moved by the fist, d(f) = 100 cm = 1 m

Average speed of the fist during the hook, v(f) = ? cm/s = m/s

This can be solved by a very simple relation.

d(f) / d(h) = v(f) / v(h)

v(f) = [d(f) * v(h)] / d(h)

v(f) = (1 * 0.75) / 0.05

v(f) = 0.75 / 0.05

v(f) = 15 m/s

Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s

6 0

3 years ago

A dog is running at an initial speed of 10 m/s. He covers 50 m in 4 seconds. What is the acceleration of the dog?

nikdorinn [45]

D i hope this helps
:))

8 0

3 years ago