Who developed the universal law of gravitation

you slide abox of books at constant speed up a30 degree ramp, applying a force of 200 newtons directed up the slope. The coeffic

Tomtit [17]

The work done is 400 J

Explanation:

The work done by you in pushing the box along the slope is given by

$W=Fd$

where

F is the magnitude of the force applied

d is the distance covered by the box along the slope

Here we have the following:

F = 200 N is the magnitude of the force applied

$d=\frac{h}{sin \theta}$ is the distance covered, where

h = 1 m is the vertical rise

$\theta=30^{\circ}$ is the slope of the plane

Substituting and solving, we find

$W=F\frac{h}{sin \theta}=\frac{(200)(1)}{sin 30^{\circ}}=400 J$

Learn more about work:

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5 0

3 years ago

Please help I cannot fail!

alekssr [168]

Reactants
C8H18
O2

Products
CO2
H2O

4 0

2 years ago

Which statement correctly describes gravity?

Advocard [28]

Answer:

a force that attracts matter to the earth

Explanation:

depends on where you are the gravity can be different in space there is no gravity on Earth there is , that's why when you jump you come back down

5 0

3 years ago

Question 3 of 10

Eva8 [605]

Answer:

D. Newton's second law

Explanation:

Newton's second law of motion states that force of an object is a product of its mass and its acceleration.

Mathematically, F= ma where m is mass and a is acceleration

So from the statement above : The acceleration of an object is proportional to the force applied to it and inversely proportional to its mass , it can be seen from the formula variation as;

F= ma -----making a the subject of the formula

a= F/ m

a= 1/m * F --------- a is inversely related to m as you can see from 1/m but directly related to F hence;

Increase in mass with the same force applied causes the body to accelerate slower where as when force increases, the body accelerates faster.

5 0

2 years ago

A rocket moves upward, starting from rest with an acceleration of 25.4 m/s^2 for 3.39 s. It runs out of fuel at the end of the 3

kolbaska11 [484]

Explanation:

Initial speed of the rocket, u = 0

Acceleration of the rocket, $a=25.4\ m/s^2$

Time taken, t = 3.39 s

Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :

$v=u+at$

$v=25.4\times 3.39=86.10\ m/s$

Let x is the initial position of the rocket. Using third equation of kinematics as :

$v^2=u^2+2ax_o$

$x_o=\dfrac{v^2}{2a}$

$x_o=\dfrac{86.10^2}{2\times 25.4}=145.92\ m$

Let $x_o$ is the position at the maximum height. Again using equation of motion as :

$v^2-u^2=2a(x-x_o)$

Now $a=-g$ and v and u will interchange

$u^2=2g(x-x_o)$

$x=x_o+\dfrac{u^2}{2g}$

$x=145.92+\dfrac{(86.10)^2}{2\times 9.8}$

x = 524.14 meters

Hence, this is the required solution.

5 0

3 years ago