All categories

3 years ago

Numeria

Physics

2 answers:

Vinil7 [7]3 years ago

5 0

Answer:

hope it helps

plz mark as brainliest

polet [3.4K]3 years ago

3 0

Answer:

<h3> $\boxed{40 \: cm}$ </h3>

Explanation:

Load ( L ) = 800 N

Effort ( E ) = 200 N

Load distance ( LD ) = 10 cm

Effort distance ( ED ) = ?

now, Let's find the effort distance:

$\mathsf{L \times LD = E \times ED}$

Plug the values

$\mathsf{800 \times 10 = 200 \times ED}$

Multiply the numbers

$\mathsf{8000 = 200 \: ED}$

Swipe the sides of the equation

$\mathsf{200 \: ED = 8000}$

Divide both sides of the equation by 200

$\mathsf{ \frac{200 \: ED }{200} = \frac{8000}{200} }$

Calculate

$\mathsf{ED \: = \: 40 \: cm}$

Hope I helped!

Best regards!

You might be interested in

20 POINTS MULTIPLE CHOICE

Lynna [10]

If mass doubled and the force remained constant then the acceleration would also double

Hopes this helps

5 0

3 years ago

A ball is thrown up with a speed of 15m/s. How high will it go before it begins to fall? ( g = 10m/s2 )

LekaFEV [45]

Answer:

Height is 11.25m

Explanation:

<u>Given the following data;</u>

Initial velocity, u = 0

Final velocity, v = 15m/s

Acceleration due to gravity, g = 10m/s²

To find the height, we would use the third equation of motion;

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement (height) measured in meters.

$V^{2} = U^{2} + 2aS$

<em>Making S the subject, we have;</em>

$S = \frac {V^{2} - U^{2}}{2a}$

But a = g = 10m/s²

<em>Substituting into the equation, we have;</em>

$S = \frac {15^{2} - 0^{2}}{2*10}$

$S = \frac {225 - 0}{20}$

$S = \frac {225}{20}$

S = 11.25m

<em>Therefore, the ball will reach a height of 11.25m before it begins to fall. </em>

4 0

3 years ago

A Ray of light falling on rough surface follows the laws of reflection but no image of the object placed before it is C explain

stepladder [879]

Answer:

The light rays falling on a rough surface does follow the laws of reflection. The light rays are incident parallel on the rough surface but due to uneven surface the light rays are not reflected parallel rather they are reflected in different direction. Hence, no image is formed.

4 0

3 years ago

A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner, as t

salantis [7]

Answer:

2.07

Explanation:

Since you didn't supply the drawing, here is what I assumed:

A is the corner opposite the axis of rotation

B is one of the remaining two corners

L1 is the side between A & B

Centripetal acceleration is given by:

ac = v^2 / r = (v / r) * (v / r) * r…………1

Also angular speed is

w = v / r,………….2

Substituting (2) in (1) gives:

ac = (v / r) * (v / r) * r……….3

= (v / r)^2 * r

= w^2 * r

Therefore, the angular acceleration at A and at B are given by:

acA = w^2 * rA……..4

acB = w^2 * rB……..5

It is given that:

acA = n * acB…………6

Substituting (4) and (5) into (6) gives:

w^2 * rA = n * w^2 * rB ……….7==>

rA = n * rB……..8

In terms of the sides L1 and L2:

rA = sqrt (L1^2 + L2^2)…….9

and

rB = L2…………10

Considering (8):

n * L2 = sqrt (L1^2 + L2^2)………11

Squaring both sides:

n^2 * L2^2 = L1^2 + L2^2……….12

Dividing by L2^2:

n^2 = L1^2 / L2^2 + L2^2 / L2^2…….13

= (L1 / L2)^2 + 1 ==>

n^2 - 1 = (L1 / L2)^2 ………14==>

L1 / L2 = sqrt (n^2 - 1) ………15

= sqrt (2.30^2 - 1)

= 2.07. . . . . . <<<=== the value of the ratio L1 / L2 when n = 2.30

8 0

3 years ago

the police department determined that the force required to drag a 130 N(29 lb) car tire across the pavement at a constant veloc

ki77a [65]

The force of friction is given by:
f = μR, where μ is the friction coefficient and R is the reaction force, which will be equal to the weight.
100 = μ x 130
μ = 0.77

4 0

3 years ago