6th grade science pls help

When a wheel is rotated through an angle of 21◦ , a point on the circumference travels through an arc length of 2.1 m. When the

LiRa [457]

Answer:

R = 5.73 m

Explanation:

For an angle of rotation through 21 degree we know that

arc length is given as

$angle \times Radius = Arc$

now we know that

Arc = 2.1 m

Angle = 21 degree

$Angle = 21\times \frac{\pi}{180}$

so now we have

$21\times \frac{\pi}{180} = \frac{2.1}{R}$

$R = \frac{2.1 \times 180}{21 \times \pi}$

$R = 5.73 m$

6 0

3 years ago

PLEASE HELP

Sergeu [11.5K]

The vertical component of the initial velocity is $v_0_y = \frac{y}{t} + \frac{1}{2} gt$

The horizontal component of the initial velocity is $v_0_x = \frac{x}{t}$

The horizontal displacement when the object reaches maximum height is $X = \frac{xy}{gt^2} + \frac{x}{2}$

The given parameters;

the horizontal displacement of the object, = x

the vertical displacement of the object, = y

acceleration due to gravity, = g

time of motion, = t

The vertical component of the initial velocity is given as;

$y = v_0_yt - \frac{1}{2} gt^2\\\\v_0_yt = y + \frac{1}{2} gt^2\\\\v_0_y = \frac{y}{t} + \frac{1}{2} gt$

The horizontal component of the initial velocity is calculated as;

$x = v_0_xt\\\\v_0_x = \frac{x}{t}$

The time to reach to the maximum height is calculated as;

$T = \frac{v_f_y -v_0_y}{-g} \\\\T = \frac{-v_0_y}{-g} \\\\T = \frac{v_0_y}{g} \\\\T = \frac{1}{g} (v_0_y)\\\\T = \frac{1}{g} (\frac{y}{t} + \frac{1}{2} gt)\\\\T = \frac{y}{gt} + \frac{1}{2} t$

The horizontal displacement when the object reaches maximum height is calculated as;

$X= v_0_x \times T\\\\X= \frac{x}{t} \times (\frac{y}{gt} + \frac{1}{2} t)\\\\X = \frac{xy}{gt^2} + \frac{x}{2}$

Learn more here: brainly.com/question/20689870

8 0

3 years ago

Explain integration and differentiation in simple method!

Anika [276]

Differentiation in its simplest of terms means breaking something into small parts. On the other hand, integration is taking those really small parts and gluing them in the right order. In short, these terms are the direct opposite or inverses of each other. The term which can tell you how fast you are going at a moment in time at ones current location is called a derivative. The term on the other hand, which can tell you how far you have travelled if you have been keeping track of your location and your time is what an integral is referred to. It is like differentiation only needs knowledge on the local neighbourhood while integration will need the knowledge on a global knowledge.

5 0

3 years ago

On a balanced seesaw, a boy three times as heavy as his partner sits

slega [8]

Answer:

1/3 the distance from the fulcrum

Explanation:

On a balanced seesaw, the torques around the fulcrum calculated on one side and on another side must be equal. This means that:

$W_1 d_1 = W_2 d_2$

where

W1 is the weight of the boy

d1 is its distance from the fulcrum

W2 is the weight of his partner

d2 is the distance of the partner from the fulcrum

In this problem, we know that the boy is three times as heavy as his partner, so

$W_1 = 3 W_2$

If we substitute this into the equation, we find:

$(3 W_2) d_1 = W_2 d_2$

and by simplifying:

$3 d_1 = d_2\\d_1 = \frac{1}{3}d_2$

which means that the boy sits at 1/3 the distance from the fulcrum.

8 0

3 years ago

Develop an equation (with a proportionality constant) that describes the relationship between the gravitational force (fgrav), t

ivolga24 [154]

According to newton's law of gravitation, the gravitational force(F) is directly proportional to the product mass of the moon(Mm) and the mass of the planet (Mp) and it is inversely proportional to the square of the separation between them.

Fg ∝ (Mp)(Mm) →(1)

Fg ∝ 1/d²→(2)

Combining equation (1) and (2),

Fg ∝ (Mp)(Mm)/d²

Fg = G(Mp)(Mm)/d²

This is an equation that describes the relation between mass of moon (Mm) and mass of planet (Mp) and separation(d) between them.

To support the claim in favuor of this equation we use this equation to obtain the value of acceleration due to gravity on earth.

Let m be the mass of an object on earth then Fg between earth (Mp) and mass of an object is obtained by:

Fg = G(Mp)(m)/R², where R= Radius of earth

This force is equal to the weight of an object i.e.,

g= G(Mp)/R²

Putting the values of G, Mp and R , we get, g=9.81 m/s²

which is the value we obtained on earth for acceleration due to gravity.

To know more about gravitational constant, visit:

brainly.com/question/13959861

#SPJ4

5 0

1 year ago