What is the centripetal force necessary to keep a 0.4 kg object rotating in a circular path

A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1

aivan3 [116]

Answer:

Part a)

$F_n = 306 N$

Part b)

$v = 12.1 m/s$

So this speed is independent of the mass of the rider

Explanation:

Part a)

By force equation on the rider at the position of the hump we can say

$mg - F_n = ma_c$

now we will have

$mg - F_n = \frac{mv^2}{R}$

$F_n = mg - \frac{mv^2}{R}$

now we have

$F_n = 100(9.81) - \frac{100(9^2)}{12}$

$F_n = 981 - 675$

$F_n = 306 N$

Part b)

At the top of the loop if the minimum speed is required so that it remains in contact so we will have

$F_n + mg = ma_c$

$F_n = 0$ at minimum speed

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = \sqrt{15 \times 9.81}$

$v = 12.1 m/s$

So this speed is independent of the mass of the rider

5 0

3 years ago

5. An electrical power plant generates electricity with a current of 50 A and a potential difference of 20 000 V. In order to mi

lakkis [162]

Answer: Current = 2 A

Explanation:

Given that an electrical power plant generates electricity with a

current I = 50 A

Potential difference V = 20 000 V

The resistance R will be achieved by Ohms law formula which state that

V = IR

But the power generated will be the product of potential difference and the current

Power P = IV

P = 50 × 20000

P = 1, 000000 W

When the transformer steps up the potential difference to 500 000 V before it is transmitted

Power is always constant.

Using the formula for power again with

V = 500000

1000000 = 500000× I

Make I the subject of formula

Current I = 1000000/500000

Current I = 2 A

3 0

3 years ago

After the box comes to rest at position x1, a person starts pushing the box, giving it a speed v1. when the box reaches position

KiRa [710]

As we know by work energy theorem

total work done = change in kinetic energy

so here we can say that wok done on the box will be equal to the change in kinetic energy of the system

$W_p = KE_f - KE_i$

initial the box is at rest at position x = x1

so initial kinetic energy will be ZERO

at final position x = x2 final kinetic energy is given as

$KE_f = \frac{1}{2}mv_1^2$

now work done is given as

$W_p = \frac{1}{2}mv_1^2 - 0$

so we can say

$W_p = \frac{1}{2}mv_1^2$

so above is the work done on the box to slide it from x1 to x2

3 0

3 years ago

Why the bulb of thermometer in cylindrical

crimeas [40]

I'm not accurately sure if you're asking for why the bulb of a thermometer is in a cylindrical shape. So let me continue. The shape of the which is thin and cylindrical in the shape is the increase of the effect of mercury in the tube to rise and fall depending on the contact temperature.

4 0

3 years ago

A person kicks a ball off of a 50m high cliff with a speed of 10 m/s. How long will it take the ball to hit the ground? * 7 poin

Musya8 [376]

Presumably, the ball is kicked parallel to the ground below the cliff, so its altitude y at time t is

$y(t)=50\,\mathrm m-\dfrac12gt^2$