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SIZIF [17.4K]
4 years ago
13

While a roofer is working on a roof that slants at 39.0 degrees above the horizontal, he accidentally nudges his 88.0 N toolbox,

causing it to start sliding downward, starting from rest.
If it starts 4.90 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 18.0 N ?
Physics
1 answer:
Ostrovityanka [42]4 years ago
5 0

Answer:

V= 6.974 m/s

Explanation:

Component( box) weight acting parallel and down roof 88(sin39.0°)=55.4 N

Force of kinetic friction acting parallel and up roof = 18.0 N

Fnet force acting on tool box acting parallel and down roof

Fnet= 55.4 - 18.0

Fnet=37.4 N

acceleration of tool box down roof

a = 37.4(9.81)/88.0

a= 4.169 m/s²

d = 4.90 m

t = √2d/a

t= √2(4.90)/4.169

t= 1.662 s

V = at

V= 4.169(1.662)

V= 6.974 m/s

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3 years ago
A friend claims that throwing a baseball up towards the school roof illustrates gravitational potential energy transforming into
Cloud [144]

Answer:

The Statement is wrong because the reverse is the case as it is the  kinetic energy that is being transformed to gravitational potential energy.

Explanation:

As your friend throws  the baseball into the air the ball gains an initial velocity (u) and this makes the Kinetic energy to be equal to

                    KE = \frac{1}{2} mu^2

Here  m is the mass of the baseball

       Now as this ball moves further upward the that velocity it gained reduce due to the gravitational force and this in turn reduces the kinetic energy of the ball and this kinetic energy lost is being converted to gravitational potential energy which is mathematically represented as (m×g×h)

as energy can not be destroyed but converted to a different form according to the first law of thermodynamics

Looking a the formula for gravitational potential energy we see that the higher the ball goes the grater the gravitational potential energy.

4 0
4 years ago
Tarzan swings back and forth on a long vine with a period of 7.27 s. how long is the vine?(unit=m)
lana [24]

Answer:

Tarzan, who weighs 688N, swings from a cliff at the end of a convenient vine that is 18m long. From the top of the cliff to the bottom of the swing he descends by 3.2m.

Explanation:

3 0
4 years ago
One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i
kati45 [8]

Answer:

The force exerted on an electron is 7.2\times10^{-18}\ N

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance  = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}

E_{1}=1.0183\times10^{3}\ N/C

Using formula of electric field again

E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}

E_{2}=-1.064\times10^{3}\ N/C

We need to calculate the resultant electric field

Using formula of electric field

E=E_{1}+E_{2}

Put the value into the formula

E=1.0183\times10^{3}-1.064\times10^{3}

E=-0.045\times10^{3}\ N/C

We need to calculate the force exerted on an electron

Using formula of electric field

E = \dfrac{F}{q}

F=E\times q

Put the value into the formula

F=-0.045\times10^{3}\times(-1.6\times10^{-19})

F=7.2\times10^{-18}\ N

Hence, The force exerted on an electron is 7.2\times10^{-18}\ N

8 0
3 years ago
What would happen to the two balls if one of them were kept positively charged and the charge on the other ball were slowly incr
mariarad [96]

Answer:

the balls would move closer to each other

Explanation:

8 0
3 years ago
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