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2 years ago

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an object weighting 100g is thrown upwards from the ground at a speed of 100 m/s.where will the potential energy of the object b

e two thirds of the kinetic energy?

1 answer:

Kay [80]2 years ago

3 0

Answer:

333.3 m

Explanation:

Given

$m =100g\ =\ 0.1kg\\v = 100 m/s\\g = 10 m/s ^2$

Potential energy = $\frac{2}{3}\ of\ Kinetic\ energy$ ......Equation(1)

We know that

Potential energy=mgh

Kinetic energy = $\frac{1}{2} mv^{2}$

Now From the Equation(1)

$mgh=\frac{2}{3}*\frac{1}{2} mv^{2}\\ gh=\frac{v^{2} }{3} \\10 * h=\ \frac{10000}{3}\\ h=\ \frac{1000}{3} \\h=333.3\ m$

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A firefighter of mass 81 kg slides down a vertical pole with an acceleration of 3 m/s 2 . The acceleration of gravity is 10 m/s

natima [27]

Answer:

The force of friction that acts on him is

$F_k=567N$

Explanation:

The firefighter with an acceleration of 3m/s^2 take the gravity acceleration as 10m/s^2 isn't necessary to know the coefficient of friction just to know the force of friction:

$F=m*a$

$F=F_w-F_k$

$m*a=F_w-F_k$

$F_w=81kg*10m/s^2=810N$

Sole to Fk

$81kg*3m/s^2=810N-F_k$

$F_k=810N-243N$

$F_k=567N$

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2 years ago

10.A car is travelling at a constant speed of 27m/s. The driver looks away from the road for a 2.0s to tune in a station on the

Korvikt [17]

Explanation:

Distance = speed × time

d = (27 m/s) (2.0 s)

d = 54 m

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2 years ago

A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o

iVinArrow [24]

There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: $v=15 m/s$
- radius of the hill: $r=100 m$

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car $W=mg$ (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
$mg=m \frac{v^2}{r}$
from which we find
$v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s$

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3 years ago

A girl is shown at position A on a swing when the seat is directly below the support bar. The seat is then at height A as shown

MrRa [10]

Answer:

<u></u>

<u>1. The potential energy of the swing is the greatest at the position B.</u>

<u>2. As the swing moves from point B to point A, the kinetic energy is increasing.</u>

Explanation:

Even though the syntax of the text is not completely clear, likely because it accompanies a drawing that is not included, it results clear that the posittion A is where the seat is at the lowest position, and the position B is upper.

The gravitational <em>potential energy </em>is directly proportional to the height of the objects with respect to some reference altitude. Thus, when the seat is at the position A the swing has the smallest potential energy and when the seat is at the <em>position B the swing has the greatest potential energy.</em>

Regarding the forms of energy, as the swing moves from point B to point A, it is going downward, gaining kinetic energy (speed) at the expense of the potential energy (losing altitude). When the seat passes by the position A, the kinetic energy is maximum and the potential energy is miminum. Then the seat starts to gain altitude again, losing the kinetic energy and gaining potential energy, up to it gets to the other end,

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3 years ago

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mihalych1998 [28]

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