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3 years ago

Help for brainliest!

1 answer:

3 0

A “real” image occurs when light rays actually intersect at the image, and become inverted, or turned upside down. ... In flat, or plane mirrors, the image is a virtual image, and is the same distance behind the mirror as the object is in front of the mirror. The image is also the same size as the object.

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(b) 360 days into seconds.

Angelina_Jolie [31]

Explanation:

(b) We know that,

1 day = 24 hours

1 hour = 3600 s

So, we found that, 1 day = 86400 s

We need to find the 360 days into seconds. So,

1 day = 86400 s

360 days = 86400×360

360 days = 31104000 seconds

(d) Weight of a body, W = 600 N

Acceleration due to gravity on mars is 3.7 m/s²

Weight, W = mg

m is mass of body

$m=\dfrac{W}{g}\\\\m=\dfrac{600}{3.7}\\\\m=162.16\ kg$

(e) Mass of body, m = 100 kg

Acceleration due to gravity on the moon, 1.6 m/s²

Weight, W = 100 × 1.6

W = 160 N

8 0

3 years ago

A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse

Elanso [62]

Answer:

$v_{f} = 0.51 \frac{m}{s}$

Explanation:

We apply Newton's second law at the crate :

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=90kg : crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Crate weight (W)

W= m*g

W= 90kg*9.8 m/s²

W= 882 N

Friction force : Ff

Ff= μk*N Formula (2)

μk: coefficient of kinetic friction

N : Normal force (N)

Problem development

We apply the formula (1)

∑Fy = m*ay , ay=0

N-W = 0

N = W

N = 882 N

We replace the data in the formula (2)

Ff= μk*N = 0.351* 882 N

Ff= 309.58 N

We apply the formula (1) in x direction:

∑Fx = m*ax , ax=0

282 N - 309.58 N = 90*a

a= (282 N - 309.58 N ) / (90)

a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:

d:displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

$v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}$

$v_{f} = 0.51 \frac{m}{s}$

8 0

3 years ago

Use the drop-down menus to complete the statement.

VashaNatasha [74]

Answer: A is your answer i am sorry if i am wrong

Explanation:

he first PLCs were programmed with a technique that was based on relay logic wiring schematics. This eliminated the need to teach the electricians, technicians and engineers how to program a computer - but, this method has stuck and it is the most common technique for programming PLCs today.

7 0

3 years ago

Read 2 more answers

The law of conversation of energy states that energy cannot be created or

irinina [24]

Answer:

It says energy can't be created or destroyed

Explanation:

5 0

3 years ago

An old millstone, used for grinding grain in a gristmill, is a solid cylindrical wheel that can rotate about its central axle wi

MAXImum [283]

Answer:

The answer is I=70,513kgm^2

Explanation:

Here we will use the rotational mechanics equation T=Ia, where T is the Torque, I is the Moment of Inertia and a is the angular acceleration.

When we speak about Torque it´s basically a Tangencial Force applied over a cylindrical or circular edge. It causes a rotation. In this case, we will have that T=Ft*r, where Ft is the Tangencial Forge and r is the radius

Now we will find the Moment of Inertia this way:

$Ft*r=I*a$ -> $(Ft*r)/(a) = I$

Replacing we get that I is:

$I=(200N*0,33m)/(0,936rad/s^2)$

Then $I=70,513kgm^2$

In case you need to find extra information, keep in mind the Moment of Inertia for a solid cylindrical wheel is:

$I=(1/2)*(m*r^2)$

4 0

3 years ago