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2 years ago

Why does it take more force to move a heavy box up a ramp than to move the same box down the ramp?

Physics

1 answer:

ICE Princess25 [194]2 years ago

8 0

D I think. Tell me if this helped

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Consumers that eat both producers and other consumers are known as

matrenka [14]

Omnivores...————Fhfhfjdhfh

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3 years ago

A wooden plaque is in the shape of an ellipse with height 30 centimeters and width 22 centimeters. Find an equation for the elli

Volgvan

Answer:

Explanation:

height of Ellipse $=30 cm$

i.e. $2 a=30$

Width of Ellipse $=22 cm$

i.e. $2 b=22$

Equation of a vertical Ellipse is

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

$a=15 ,b =11$

$\frac{x^2}{11^2}+\frac{y^2}{15^2}=1$

at $y=4 cm$

$\frac{x^2}{11^2}+\frac{4^2}{15^2}=1$

$x=\frac{11}{15}\times \sqrt{15^2-4^2}$

$x=10.6 cm$

5 0

2 years ago

1) The smallest part of an element that behaves like the element is the atom.

Mandarinka [93]

Here are the answers:

1. False - Molecules is the smallest part of an element that behaves like the element.

2. False - The nucleus contains both protons and neutrons

3. True

4. True

5. A. Nucleus

6. D. Neutron

7. B. Protons and Neutrons

8. C. Electron

9. C. 6

10. C.6

5 0

3 years ago

Which factors could be potential sources of error in the experiment? check all that apply.

Vadim26 [7]

(A)energy lost in the lever due to friction

(C) visual estimation of height of the beanbag

(E)position of the fulcrum for the lever affecting transfer of energy

6 0

3 years ago

Read 2 more answers

a baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45 with respect to the ground. (g=32 feet/

LiRa [457]

Answer:

hmax=81ft

Explanation:

Maximum height of the object is the highest vertical position along its trajectory.

The vertical velocity is equal to 0 (Vy = 0)

$0=V_{y}-g*t=v_{0}*sin(\alpha)-g*th\\$

we isolate th (needed to reach the maximum height hmax)

$th = \frac{v_{0}*sin(\alpha)}{g}$

The formula describing vertical distance is:

$y = Vy * t-g* t^{2} / 2$

So, given y = hmax and t = th, we can join those two equations together:

$hmax = Vy* th-g*th^{2}/2$

$hmax =Vo^{2}*sin(\alpha )^{2}/(2*g)$

if we launch a projectile from some initial height h all you need to do is add this initial elevation

$hmax =h+Vo^{2}*sin(\alpha)^{2}/(2*g)$

$hmax =3+100^{2}*sin(45)^{2}/(2 * 32)=81 ft$

6 0

3 years ago