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3 years ago

When the velocity of a moving object stays the same, it has a what speed

1 answer:

8 0

Same rate of change of distance but velocity include circular motion and known as vector quantity but speed is scalar quantity only.

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A 25 kg child is riding on a swing. If the child travels 8.9 m/s at the bottom of their swing, how high into the air is the chil

Setler [38]

Answer:

h = 4.04 m

Explanation:

Given that,

Mass of a child, m = 25 kg

The speed of the child at the bottom of the swing is 8.9 m/s

We need to find the height in the air is the child is able to swing. Let the height is h. Using the conservation of energy such that,

$mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}$

Put all the values,

$h=\dfrac{(8.9)^2}{2\times 9.8}\\\\h=4.04\ m$

So, the child is able to go at a height of 4.04 m.

7 0

3 years ago

Please help!!!!!!!!!

Inessa [10]

A, D is the correct answers

6 0

3 years ago

A man is dragging a trunk up the loading ramp of a mover's

FinnZ [79.3K]

Answer:

Horizontal component = 241 N

Vertical component = 287 N

Explanation:

Given:

Force = F = 375 N

Referring to diagram attached, the force F is making an angle

theta = 20+30 = 50 with the horizontal.

Horizontal component = F*cos(theta) = 375*0.64278 = 241 N

Vertical component = F*sin(theta) = 375*0.76604 = 287 N

4 0

3 years ago

Can somone pls help asap???!!

joja [24]

I think its 1,2 sorry if wrong

7 0

3 years ago

Read 2 more answers

A photographer uses his camera, whose lens has a 50 mm focal length, to focus on an object 1.5 m away. He then wants to take a p

allsm [11]

Answer:

The distance is $z = 0.008 \ m$

Explanation:

From the question we are told that

The focal length is $f = 50 \ mm = 50*10^{-3} \ m$

Generally the lens equation is mathematically represented as

$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$

At image distance u = 1.5 m

$\frac{1}{1.5} + \frac{1}{v} = \frac{1}{50 *10^{-3}}$

=> $\frac{1}{50 *10^{-3}} - \frac{1}{1.5} = \frac{1}{v}$

=> $v = 0.052 \ m$

At image distance $u = 30\ cm = 0.30 \ m$

$\frac{1}{0.3} + \frac{1}{v_1} = \frac{1}{50 *10^{-3}}$

=> $\frac{1}{50 *10^{-3}} - \frac{1}{0.30 } = \frac{1}{v_1}$

=> $v_1 = 0.06 \ m$

The distance the lens need to move is evaluate as

$z = |v - v_1|$

$z = |0.052 - 0.06|$

$z = 0.009 \ m$

8 0

3 years ago