Will give correct answer brainliest please help

The suspended load of a stream _____.

denis23 [38]

<span>the first one ( usually consists of fine sand, silt, and clay particles )

Hope im right!(: </span>

4 0

3 years ago

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A stone falls freely from rest for 8.0s what is it final velocity

NemiM [27]

Is that the full question?

7 0

3 years ago

Cassy shoots a large marble (Marble A, mass: 0.06kg) at a smaller marble (Marble B, mass: 0.03kg) that is sitting still. Marble

4vir4ik [10]

The question can be solved using conservation of linear momentum.

$M_{a}$ = 0.06kg and $M_{b}$ = 0.03kg

Let the initial velocity of Marble A be , $V_{a1}$ = 0.7m/s

Let the initial velocity of Marble B be, $V_{b1}$ = 0m/s

Let the velocity of Marble A after collisiong , $V_{a2}$ = -0,02m/s

Let the velocity of Marble B after collision be $V_{b2}$

From the conservation of linear momentum equation. We get,

$M_{a}V_{a1}+M_{b}V_{b1}=M_{a}V_{a2}+M_{b}V_{b2}$

Substituting the values we get,

(0.06)(0.7) + 0 = (0.06)(-0.02) + (0.03) $V_{b2}$

we get, $V_{b2}$ = 1.44m/s

6 0

3 years ago

If Scoobie could drive a Jetson's flying car at a constant speed of 160.0 km/hr across oceans and space, approximately how long

lara [203]

Your first step is to find the circumference of the earth, with the numbers given. You can do that by putting the radius of 6200 kilometres into the 2πr equation. That should get you a circumference of 12400π, or about 38,955.75 kilometres.

Next, you can use the rate the Jetson's car is going (180km/h) and divide the 38,955.75 by it to see how many hours it would take at that constant speed.

38,955.75 / 180 = 216.42 hours

Then you can divide that by 24 to get how many days

6 0

3 years ago

If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th

mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

$x=ut$

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

$t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s$

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

$y=\frac{1}{2} gt^2$

Substitute 9.8 m/s² for g and 0.461 s for t.

$y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m$

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by, $(2.0 m)-(1.04 m)=0.96 m$

The catcher needs to hold his glove at a height of <u>0,96 m from the ground.</u>

8 0

3 years ago