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UkoKoshka [18]
1 year ago
13

A grandfather clock depends on the period of a pendulum to keep correct time.(ii) Suppose a grandfather clock is calibrated corr

ectly at sea level and is then taken to the top of a very tall mountain. Does the grandfather clock now run (a) slow, (b) fast, or (c) correctly?
Physics
1 answer:
ANEK [815]1 year ago
6 0

The grandfather clock will now run slow (Option A).

<h3>What is Time Period of an oscillation?</h3>
  • The time period of an oscillation refers to the time taken by an object to complete one oscillation.
  • It is the inverse of frequency of oscillation; denoted by "T".

Now,

  • \sqrt{\frac{L}{g}}, where L is the length and g is the gravitational constant, is the formula for a pendulum's period.
  • The period will increase as one climbs a very tall mountain because g will slightly decrease.
  • Due to this and the previous issue, the clock runs slowly and it seems that one second is longer than it actually is.

Hence, the grandfather clock will now run slow (Option A).

To learn more about the time period of an oscillation, refer to the link: brainly.com/question/26449711

#SPJ4

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A 50 kg pitcher throws a baseball with a mass of 0.15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is
Andreas93 [3]

The velocity of the pitcher is <u>0.105 m/s</u> in a direction opposite to the velocity of the ball.

When no external force acts on a system, the total momentum of the system is conserved. The total initial momentum of the system is equal to the total final momentum of the system.

The pitcher and the ball are initially at rest, therefore, the total initial momentum of the system is zero.

Since no external forces act on the system comprising of pitcher and the ball, the total final momentum of the system is also equal to zero.

If the mass of the pitcher is mp and its speed is vp, the mass of the ball is mb and the ball's speed is vb, then the final momentum of the system of pitcher and the ball is given by,

p=m_pv_p+m_bv_b=0

Therefore,

v_p=-\frac{m_b}{m_p} v_p

Substituet 0.15 kg for mb, 50 kg for mp and 35 m/s for vb.

v_p=-\frac{m_b}{m_p} v_p=-\frac{0.15 kg}{50 kg} (35m/s)=-0.105 m/s

The pitcher has a velocity <u> 0.105 m/s</u> opposite to the direction of the velocity of the ball.

8 0
3 years ago
What are non examples of light years?
abruzzese [7]
Hi , here are some examples: An astronomical unit A parsec A meter
5 0
3 years ago
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The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m
Julli [10]

Answer:

v=(6ti+6k)\ m/s

Explanation:

Given that,

The position of a particle is given by :

r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m

Let us assume we need to find its velocity.

We know that,

v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s

So, the velocity of the particle is (6ti+6k)\ m/s.

5 0
3 years ago
A 10cm long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire
soldi70 [24.7K]

Answer:

a)  v=4.0m/s

b)  B=2.958T

Explanation:

From the question we are told that:

Wire Length l=10cm=>0.10m

Resistance R=0.35

Force F=1.0N

Power P=4.0W

a)

Generally the equation for Power is mathematically given by

P=Fv

Therefore

v=\frac{P}{F}

v=\frac{4.0}{1.0}

v=4.0m/s

b)

Generally the equation for Magnetic Field is mathematically given by

B=\frac{\sqrt{PR}}{vl}

B=\frac{\sqrt{4*0.35}}{4*0.10}

B=2.958T

5 0
3 years ago
Acceleration toward the center of a curved path is called
Serggg [28]

Answer:

Centripetal acceleration.

Explanation:

Centripetal acceleration is a property of a body moving in a uniform circular path and it is directed radially towards the center of the circle in which body is rotating.

The force which causes this acceleration is centripetal force which is also directed towards the center of the circle and pulls the body towards its center.

It is calculated through following formula

a=v^2/r

where v is velocity and r is the radius of the circle.

7 0
3 years ago
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