I need the answer to number #23 please will give BRAINLIEST

Determine the amount of torque

malfutka [58]

Answer:

250Nm

Explanation:

Given parameters:

Length of the long pry bar = 1m

Force acting on it = 250N

Angle = 90°

Unknown:

Amount of torque applied = ?

Solution:

Torque is the turning force on a body that causes the rotation of the body.

The formula is given as:

Torque = Force x r Sin Ф

r is the distance

So;

Torque = 250 x 1 x sin 90 = 250Nm

7 0

3 years ago

Which of the following equations is balanced correctly? A. 3 H2O → H2 + 3 O2 B. Cl2 + 2 KBr → KCl + Br2 C. 2 C2H2 + 5 O2 → 4 CO2

posledela

A. the carbons are unbalanced B. the hydrogens are unbalanced. D. the chlorines are unbalanced. That leaves C. to be correctly balanced.

8 0

3 years ago

Read 2 more answers

Find the equilibrant of two 10.0-N forces acting upon a body when the angle between the forces is 90° Solve graphically using a

bazaltina [42]

The equilibrant force of the two given forces is 14.14 N.

<h3 /><h3 /><h3>What is equilibrant force?</h3>

This is a single force that balances other given forces.

The given parameters:

First force, F₁ = 10 N
Second force, F₂ = 10 N
Angle between the forces, θ = 90⁰

The equilibrant force of the two given forces is calculated as follows;

$F = \sqrt{F_1 ^2 + F_2 ^2} \\\\F = \sqrt{(10)^2 + (10)^2} \\\\F = 14.14 \ N$

Thus, the equilibrant force of the two given forces is 14.14 N.

Learn more about equilibrant force here: brainly.com/question/8045102

5 0

2 years ago

I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite

e-lub [12.9K]

Answer:

The range of powers is $- 5 \ D \le P \le - 2.667\ D$

Explanation:

From the question we are told that

The far point of the left eye is $n_f = 20 cm$

The near point of the left eye is $n = 15cm$

The near point with the glasses on is $n_g =25 \ cm$

From these parameter we can see that with the glass on that for near point the

Object distance would be $u = -25 \ cm$

Image distance would be $v = -15 \ cm$

To obtain the focal length we would apply the lens formula which is mathematically represented as

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

substituting values

$\frac{1}{f} = \frac{1}{-15} - \frac{1}{-25}$

$f = - \frac{75}{2} cm$

converting to meters

$f = - \frac{75}{2} * \frac{1}{100}$

$f = - \frac{75}{200} \ m$

Generally the power of the lens is mathematically represented as

$P = \frac{1}{f}$

Substituting values

$P = - \frac{200}{75} m$

$P = - 2.667 \ D$

From these parameter we can see that with the glass on that for far point the

Object distance would be $u_f = - \infty \ cm$

Image distance would be $v_f = -20 \ cm$

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

$\frac{1}{f_f} = \frac{1}{v_f} - \frac{1}{u_f}$

substituting values

$\frac{1}{f} = \frac{1}{-20} - \frac{1}{- \infty}$

$\frac{1}{f} = \frac{1}{-20} - 0$

$f_f = \frac{20}{1} \ cm$

converting to meters

$f_f = - \frac{20}{1} * \frac{1}{100}$

Generally the power of the lens is mathematically represented as

$P = \frac{1}{f_f}$

Substituting values

$P = - \frac{100}{20} m$

$P = - 5 \ D$

This implies that the range of powers of the lens in his glass is

$- 5 \ D \le P \le - 2.667\ D$

3 0

3 years ago

This whole worksheet. I am not good at this and I am completely lost

Ket [755]

1.) potential energy
2.)potential and kinetic
3.)The roller coaster car has the most kinetic energy at point X i know this because the car is moving and kinetic energy has the power to move or change things therefore point X is when the roller coaster car has the most energy.
4.)potential energy
5.)kinetic energy
6.) potential and kinetic energy

3 0

3 years ago