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2 years ago

7

An object slides down a frictionless inclined plane. At the bottom, it has a speed of 9.80 m/s. What is the vertical height of t

he plane

1 answer:

Papessa [141]2 years ago

7 0

The vertical height of the given plane is 4.9 m.

The given parameters:

<em>speed of the object at the bottom of the ramp, v = 9.8 m/s</em>

The vertical height of the plane is calculated by applying principle of conservation of mechanical energy as follows;

$P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{9.8 \times 9.8}{2 \times 9.8} \\\\h = 4.9 \ m$

Thus, the vertical height of the given plane is 4.9 m.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

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Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each o

zhannawk [14.2K]

Answer:

The magnitude of the charge on each sphere is 0.135 μC

Explanation:

Given that,

Mass = 1.0

Distance = 2.0 cm

Acceleration = 414 m/s²

We need to calculate the magnitude of charge

Using newton's second law

$F= ma$

$a=\dfrac{F}{m}$

Put the value of F

$a=\dfrac{kq^2}{mr^2}$

Put the value into the formula

$414=\dfrac{9\times10^{9}\times q^2}{1.0\times10^{-3}\times(2.0\times10^{-2})^2}$

$q^2=\dfrac{414\times1.0\times10^{-3}\times(2.0\times10^{-2})^2}{9\times10^{9}}$

$q^2=1.84\times10^{-14}$

$q=0.135\times10^{-6}\ C$

$q=0.135\ \mu C$

Hence, The magnitude of the charge on each sphere is 0.135μC.

7 0

3 years ago

Read 2 more answers

A 48.0-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocit

Kipish [7]

Answer:

F = 47.6 N

Explanation:

Newton's 2nd law can be expressed as the rate of change of the total momentum, respect of time, as follows:

$F = \frac{\Delta p}{\Delta t}$

So, in order to find the average force exerted by the skater on the wall, we can find the change in momentum due to the force exerted by the wall (which is equal and opposite to the one exerted by the skater), and divide it by the time interval , as follows:

$F_{wall} = \frac{\Delta p}{\Delta t} =\frac{(48.0 kg*(-1.06m/s)}{1.07s} = -47.6 N$

⇒ Fsk = 47.6 N (normal to the wall)

3 0

3 years ago

A cow wanders 30 m North, turns 22 degrees right of its original path, and wanders another 40 m. Find its total displacement.

scoundrel [369]

Answer:OB=58.3m

Explanation:

So here cow wanders 30m in north and turns 22 degrees in right side and moves 40m more, as shown in figure given.

now take the starting point as a origin such that cow moves in x-y co-ordinate axis.

As shown in figure length OA is the length when cow moves in north or y direction. Later she takes 22 degrees turn to right and moves 40m more.

So the final displacement is the length of cow from the origin that is length OB.

now co-ordinates of B are [40cos22°,40sin22°+30] i.e [37.084,44.984]

now displacement of cow= length of OB

= $\sqrt{[37.084]^{2}+[44.984]^{2} }$

= $\sqrt{3398.78}$

OB = $58.3$

7 0

8 months ago

Static Friction

kkurt [141]

Name the element and explain why it is unusual:
a)A liquid metal
b)A non-metal that conducts electricity

3 0

2 years ago

When the displacement in SHM is equal to 1/3 of the amplitude xm, what fraction of the total energy is (a) kinetic energy and (b

Nesterboy [21]

Answer:

Explanation:

Given

Displacement is $\frac{1}{3}$ of Amplitude

i.e. $x=\frac{A}{3}$ , where A is maximum amplitude

Potential Energy is given by

$U=\frac{1}{2}kx^2$

$U=\frac{1}{2}k(\frac{A}{3})^2$

$U=\frac{1}{18}kA^2$

Total Energy of SHM is given by

$T.E.=\frac{1}{2}kA^2$

Total Energy=kinetic Energy+Potential Energy

$K.E.=\frac{1}{2}kA^2 -\frac{1}{18}kA^2$

$K.E.=\frac{8}{18}kA^2$

Potential Energy is $\frac{1}{8}$ th of Total Energy

Kinetic Energy is $\frac{8}{9}$ of Total Energy

(c)Kinetic Energy is $0.5\times \frac{1}{2}kA^2$

$P.E.=\frac{1}{4}kA^2$

$\frac{1}{2}kx^2=\frac{1}{4}kA^2$

$x=\frac{A}{\sqrt{2}}$

7 0

3 years ago

Read 2 more answers