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3 years ago

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an 8.10 kilogram box of mass is kept at a height of 0.99 meters above ground level. what is the potential energy of the box? (gi

ven g = 9.8 meters/second2

2 answers:

zepelin [54]3 years ago

8 0

Ep=mgh=78,5862J...........

kogti [31]3 years ago

8 0

Answer:

79 joules

Explanation:

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A 4 kilogram box is at rest on a frictionless floor. A net force of 12 newtons acts on it. What is the acceleration of the box?

Rashid [163]

Answer:

<h2>3 m/s^2</h2>

Explanation:

Step one:

given

Mass m= 4kg

Force F= 12N

Required

Acceleration the relation between force, acceleration, and mass is Newton's first equation of motion, which says a body will continue to be at rest or uniform motion unless acted upon by an external force

F=ma

a=F/m

a=12/4

a=3 m/s^2

4 0

3 years ago

Interest groups representing businesses and investors are often among the most successful lobbying groups in foreign policy. Why

alisha [4.7K]

The members of these groups make up the majority of voters in many districts thus this be considered a problem.

<u>Option: D</u>

<u>Explanation:</u>

Interest groups play a key role in US politics. Such organizations are made up of wealthy and powerful members who often seek to impose some form of leverage in politicians to promote their goals and agendas. Across the years via many campaigns, they have understood how to speak and manipulate elected leaders and apply leverage to get the kind of legislation that is in their favor. Here the majority of voters in several districts are standing due to group members, as we recognize the interest group belongs to a body in which it uses different methods of lobbying to influence others.

7 0

3 years ago

A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg

alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = - $\frac{v}{u}$

where

u = object distance

v = image distance

Now,

1.90 = $\frac{v}{u}$

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

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

$\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}$

$\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}$

$\frac{1}{16.5} = \frac{ 0.90}{1.90u}$

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0

4 years ago

A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A.

Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

$\Delta K = \Delta E_{p} = q\Delta V$

<u>Where</u>:

K: is the kinetic energy

$E_{p}$ : is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV

$\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J$

Now, the kinetic energy of the particle as it moves through point B is:

$\Delta K = K_{f} - K_{i}$

$K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J$

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!

8 0

3 years ago

a projectile is launched at an angle of 30 degrees and lands later at the same level. if it's initial speed is 50 m/s, solve for

Mrrafil [7]

$using \: the \: formula \\ t = \frac{2u \sin( \alpha ) }{g} where \: u = initial \: speed \: \\ \alpha = angle \: of \: projection \\ g = acceleration \: due \: to \: gravity \\ \frac{2 \times 50 \times \sin(30) }{10} \\ \frac{100 \times 0.5}{10} = \frac{50}{10} = 5seconds$

Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
horizontal distance = Range= [U^2 * sin2α]/g
[50^2 * sin60]/10
2500 * 0.8660/10
2165/10=216.5metres

3 0

3 years ago