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4 years ago

How can the time period of a pendulum be decreased or increased

Physics

1 answer:

erastovalidia [21]4 years ago

5 0

Explanation:

The period of a simple pendulum is:

T = 2π √(L/g)

where L is the length of the pendulum and g is the acceleration due to gravity.

The period of a pendulum can be decreased by decreasing the length of the pendulum or by moving the pendulum to a place with a higher gravity.

The period of a pendulum can be increased by increasing the length of the pendulum or by moving the pendulum to a place with a lower gravity.

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A rocket has landed on planet x, which has half the radius of earth. An astronaut onboard the rocket weighs twice as much on pla

Nastasia [14]

Answer:

Option (c) u0

Explanation:

The escape velocity has a formula as:

V = √(2gR)

Where V is the escape velocity,

g is the acceleration due to gravity

R is the radius of the earth.

Now, from the question, we were told that the escape velocity for the rocket taking off from earth is u0 i.e

V(earth) = u0

V(earth) = √(2gR)

u0 = √(2gR) => For the earth

Now, let us calculate the escape velocity for the rocket taking off from planet x. This is illustrated below below:

g(planet x) = 2g (earth) => since the weight of the astronaut is twice as much on planet x as on earth

R(planet x) = 1/2 R(earth) => planet x has half the radius of earth

V(planet x) =?

Applying the formula V = √(2gR), the escape velocity on planet x is obtained as follow:

V(planet x) = √(2g(x) x R(x))

V(planet x) = √(2 x 2g x 1/2R)

V(planet x) = √(2 x g x R)

V(planet x) = √(2gR)

The expression obtained for the escape velocity on planet x i.e V(planet x) = √(2gR), is exactly the same as that obtained for the earth i.e V(earth) = √(2gR)

Therefore,

V(planet x) = V(earth) = √(2gR)

But from the question, V(earth) is u0

Therefore,

V(planet x) = V(earth) = √(2gR) = u0

So, the escape velocity on planet x is u0

4 0

3 years ago

What kind of star gives rise to a type i supernova?

larisa86 [58]

a. It occures in binary systems where one of them is a whitedwarf

3 0

3 years ago

Can some answer 7 a and b please

pav-90 [236]

Answer:

a.After $15$ second Mr Comer's speed $=$ $1.7 m/s$

b.Distance travelled by Mr.Comer in $15$ seconds $=$ $18.75\ m$

Explanation:

a. Lets recall our first equation of motion $V_{f} =V_{i} + at$

Now we know that $V_{i} = 0.8m/s$ , $a = 0.06\ m/s^2$ and

$t = 15 \ sec$

Plugging the values we have.

$V_{f} =V_{i} + at$

$V_{f} =0.8 + 0.06 \times 15$

$V_{f} =0.8+ 0.9$

$V_{f} =1.7 m/s$

Then Mr.Comer's speed after $15$ sec $=$ $1.7ms^-1$

Lets find the distance and recall our third equation of motion.

$V_{f} ^2-V{i}^2 = 2as$

So $s =$ distance covered.

Dividing both sides with 2a we have.

$\frac{V_{f} ^2-V{i}^2}{2a} = 2as$

Plugging the values.

$\frac{(1.7)^2-(0.8)^2}{2\times0.06} = 2as$

$s= 18.75\ m$

So Mr.Comer will travel a distance of $s= 18.75\ m$ .

4 0

3 years ago

Write whether true or false...

elixir [45]

Answer:

a --> true, b --> false, c --> true, d -->false

Explanation:

a) since it stays floating the gravity force and the upqards push is the same

b) if it's balanced the rocket won't move from the ground, the force of the rocket, has to exceed the force of gravity

c) since it's going in a diretion the force of gravity is exceeding the force pushing it up

d) since that are speeding up at a rate, meaning growing, the force is unbalanced.

8 0

3 years ago

A car is rounding a 100-m-radius curve at 25 m/s.What is the minimum possible coefficient of static friction between the tires a

Crazy boy [7]

Answer:

The minimum possible coefficient of static friction between the tires and the ground is 0.64.

Explanation:

if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :

Fc = f

m×(v^2)/(R) = μ×m×g

(v^2)/(R) = g×μ

μ = (v^2)/(R×g)

= ((25)^2)/((100)×(9.8))

= 0.64

Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.

4 0

3 years ago