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goldfiish [28.3K]

3 years ago

7

A Viking ship roller coaster at the fair has a mass of 36,000 kg. If at its peak it reaches a height of 20 m off the ground, how

much gravitational potential energy does it have?

1 answer:

adell [148]3 years ago

8 0

Answer:

7056 kJ

Explanation:

Given that,

Mass of a ship roller coaster is 36,000 kg.

It reaches a height of 20 m off the ground

We need to find the gravitational potential energy does it have. The formula for the gravitational potential energy ios given by :

E = mgh

g is acceleration due to gravity

E = 36,000 kg × 9.8 m/s² × 20 m

= 7056000 J

or

E = 7056 kJ

So, it will have 7056 kJ of gravitational potential energy.

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What is the period of a pendulum that takes 5 seconds to make a complete back and forth vibration

dezoksy [38]

Answer:

5s

Explanation:

-A period is the time it takes to make one full back-and-forth swing.

-A period is calculated as the time taken for a complete swing divided by one cycle:

$T=t/1c\\\\=5s/1\\\\=5s$

Hence, the period of the pendulum is 5s

8 0

4 years ago

A 65 kg cart travels at a constant speed of 4.6 m/s. What is its kinetic energy?

Talja [164]

Answer:

Explanation:

mass (m) = 65 kg

velocity (v) = 4.6 m/s

Kinetic energy (KE)

= 1/2 * m * v²

= 1/2 * 65 * 4.6²

= 687.7 J

hope it helps :)

7 0

3 years ago

Bro why cant I post this T-T

Leni [432]

Answer: 12.0 m/s^2

Explanation:

Let $\alpha$ be the angular acceleration of the end of the rod

Taking torque about the link, we have:

$\tau = W \times OM\\ or\\ \tau = mg\times \left(\dfrac{L}{2}\right)\sin 55^\circ ....(i)$

Torque is also given in terms of moment of inertia of the rod and its angular acceleration i.e.

$\tau = I_{rod}\ \alpha......(ii)$

From equations (i) and (ii) we have:

$mg\times \left(\frac{L}{2}\right)\sin 55^\circ = \left(\frac{mL^2}{3}\right)\alpha\ \ \ \ \ \ \ (\because I_{rod} = \dfrac{mL^2}{3}) \\ \\ \alpha = \left(\frac{3g}{2L}\right)\sin 55^\circ\\ \\ \alpha = \left(\frac{3\times 9.8}{2\times 2.4}\right)\sin 55^\circ\\ \\ \boxed{\alpha = 5.02\ rad/s^2} }$

The acceleration of the end of the rod farthest from the link is given by:

$a = L\alpha\\ \\ a = (2.4\ m)(5.02\ rad/s^2)\\ \\ \boxed{a = 12.0\ m/s^2}$

7 0

2 years ago

Two point charges are brought closer together, increasing the force between them by a factor of 20. By what factor was their sep

olga55 [171]

The separation between them is $\frac{r}{\sqrt{20} }$

Concept :

If the force increases, distance between charges must decrease. Force is indirectly proportional to the distance squared.

Given,

Two point charges are brought closer together, increasing the force between them by a factor of 20.

Original force is

F = $\frac{kq_{1} q_{2} }{r^{2} }$ -------- ( 1 )

Here, $q_{1} , q_{2}$ are charges and r is the distance between them

New force F' = $\frac{kq_{1q_{2} } }{r'^{2} }$ ----------- (2 )

Divide ( 1 ) and ( 2 )

$\frac{F'}{F}$ = $\frac{\frac{kq_{1}q_{2} }{r'^{2} } }{\frac{kq_{1}_{2} }{r^{2} } }$

20 = $\frac{r^{2} }{r'^{2} }$

r' = $\frac{r}{\sqrt{20} }$

Given that force between them are increasing and therefore distance between them decrease by $\frac{r}{\sqrt{20} }$

Learn more about two point charges here : brainly.com/question/24206363

#SPJ4

8 0

2 years ago

At a certain moment, a car travelling at 40mph begins braking for a railroad crossing. Assume thatthe acceleration is constant (

egoroff_w [7]

Answer:

Part a)

$v_f = 17.88 - (0.15) t$

Part b)

$d = 1072.8 m$

Explanation:

As we know that initial speed of the car is

$v = 40 mph$

$v = 40 \times (\frac{1609}{3600})$

$v = 17.88 m/s$

now we have

$v_f = v_i + at$

$0 = 17.88 + a(120)$

$a = -0.15 m/s^2$

Part a)

As we know that acceleration is constant here

so we have

$v_f = v_i + at$

$v_f = 17.88 - (0.15) t$

Part b)

distance moved by the car is given as

$d = \frac{v_F + v_i}{2} t$

now we have

$d = \frac{0 + 17.88}{2}(120)$

$d = 1072.8 m$

3 0

3 years ago