All categories

2 years ago

If the mass of an ideal pendulum is doubled how will the oscillation period change?

1 answer:

5 0

Answer: Doubling the mass of the bob will half the period. b) If the length is doubled, the period will increase by a factor of √2 . The mass of the bob will not effect the period

Explanation:

You might be interested in

What is the purpose of the three R’s of resource management?

svlad2 [7]

The three R's<span> – reduce, reuse and recycle – all help to cut down on the amount of waste we throw away. They conserve natural </span>resources<span>, landfill space and energy.

Hope that helped :)</span>

3 0

3 years ago

Read 2 more answers

How does water flowing over a waterfall involve both kinetic energy and potential energy?

Vinvika [58]

Answer:

While the water falls v increases and h decreases, so the kinetic energy increases and the gravitational potential energy decreases, and this happens in a way that the total energy is always the same. (If there is no friction)

Explanation:

7 0

3 years ago

Read 2 more answers

A quarterback claims that he can throw the football a horizontal distance of 167 m. Furthermore, he claims that he can do this b

pychu [463]

Answer:u=42.29 m/s

Explanation:

Given

Horizontal distance=167 m

launch angle $=33.1^{\circ}$

Let u be the initial speed of ball

Range $=\frac{u^2\sin 2\theta }{g}$

$167=\frac{u^2\sin (66.2)}{9.8}$

$u^2=1788.71$

$u=\sqrt{1788.71}$

$u=42.29 m/s$

7 0

3 years ago

Can you walk on the moon

morpeh [17]

You can. But the gravity on the moon is 1/6th the gravity on Earth. This means 300 lbs man would only weigh 50 lbs.

7 0

3 years ago

Read 2 more answers

a train is moving with an initial velocity of 30 m/s, the brakes are applied so as to produce a uniform acceleration of -1.5 m/s

Pepsi [2]

Answer:

$\boxed{\sf Time \ in \ which \ train \ will \ come \ to \ rest = 20 \ sec}$

Given:

Initial velocity (u) = 30 m/s

Final speed (v) = 0 m/s

Acceleration (a) = - 1.5 m/,s²

To Find:

Time in which train will come to rest (t).

Explanation:

$\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ a: \\ \sf \implies 0 = 30 + ( - 1.5)(t) \\ \sf \implies 0 = 30 - 1.5(t) \\ \sf \implies 30 - 1.5(t) = 0 \\ \\ \sf Subtract \: 30 \: from \: both \: sides: \\ \sf \implies (30 - \boxed{ \sf 30}) - 1.5(t) = \boxed{ \sf - 30} \\ \\ \sf 30 - 30 = 0 : \\ \sf \implies - 1.5(t) = - 30 \\ \\ \sf Divide \: both \: sides \: of \: - 1.5(t) = - 30 \: by \: - 1.5 : \\ \sf \implies \frac{ - 1.5(t)}{ \boxed{ \sf - 1.5}} = \frac{ - 30}{ \boxed{ \sf -1.5 }} \\ \\ \sf \frac{ \cancel{ \sf 1.5}}{\cancel{ \sf 1.5}} = 1 : \\ \sf \implies t = \frac{ - 30}{ - 1.5} \\ \\ \sf \frac{ - 30}{ - 1.5} = \frac{\cancel{ \sf 1.5} \times 20}{\cancel{ \sf 1.5}} = 20 : \\ \sf \implies t = 20 \: sec$

So,

Time in which train will come to rest = 20 seconds

4 0

3 years ago