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3 years ago

A girl runs at a speed of 3.9 m/s off a high dive and hit the water 1.8 s later.

Physics

1 answer:

4vir4ik [10]3 years ago

7 0

Answer:

(a) the height of the diving board is 22.896 m

(b) the horizontal distance traveled by the girl is 7.02 m

Explanation:

Given;

initial speed of the girl, u = 3.9 m/s

time to hit the water, t = 1.8 s

(a) the height of the diving board is calculated as;

h = ut + ¹/₂gt²

h = (3.9 x 1.8) + ¹/₂ x 9.8 x 1.8²

h = 7.02 + 15.876

h = 22.896 m

(b) the horizontal distance traveled by the girl is calculated as;

X = ut

X = 3.9 x 1.8

X = 7.02 m

h = ut + ¹/₂gt²

h = 0 + ¹/₂gt²

2h = gt²

$t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2 \ \times\ 22.896 }{9.8} }\\\\t = 2.16 \ s$

Thus, if she had just dropped off the board, her time to drop to the water would have been longer.

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When will an object sink in water, according to Archimedes' principle?

inna [77]

Answer:

If the buoyant force is greater than the object's weight, the object will rise to the surface and float. If the buoyant force is less than the object's weight, the object will sink. If the buoyant force equals the object's weight, the object will remain suspended at that depth.

Explanation:

Not much explaining to do here!

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3 years ago

An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro

Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

$\sum F = m_b a$

$m_Bg-T_B = m_Ba$

$T_B = m_Bg-m_Ba$

In the case of mass A,

$\sum F = m_A a$

$T_A = m_Ag-m_Aa$

Making summation of Torques in the Pulley we have to

$\sum\tau = I\alpha$

$T_BR_0-T_AR_0=I\alpha$

$T_B-T_A=I\frac{a}{R^2_0}$

Replacing the values previously found,

$(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}$

$(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}$

$a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}$

$a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}$

$a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}$

Replacing with our values

$a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}$

$a=0.7736m/s^2$

PART B) Ignoring the moment of inertia the acceleration would be given by

$a' =\frac{(m_B-m_A)g}{(m_B+m_A)}$

$a' =\frac{(8-6.8)(9.8)}{(8+6.8)}$

$a' = 0.7945$

Therefore the error would be,

$\%error = \frac{a'-a}{a}*100$

$\%error = \frac{0.7945-0.7736}{0.7736}*100$

$\%error = 2.7%$

8 0

3 years ago

A 0.54 kg air hockey puck is initially at rest. What will it's kinetic energy energy be after a net force of 0.56 N acts on it f

Rufina [12.5K]

Answer:

Kf = 470 mJ

Explanation:

According the work-energy theorem, the change in the kinetic energy of one object, is equal to the net work done on it.
Since the puck is initially at rest, the change is kinetic energy is just the final kinetic energy of the puck.
Assuming that the net force is horizontal, and causes a horizontal displacement also, we can find the net work on the puck as follows:

$W_{net} = F_{net} * \Delta X = 0.56 N * 0.84 m = 0.47 J = 470 mJ (1)$

As we have already said, (1) is equal to the final kinetic energy of the puck:
⇒ Kf = 470 mJ (2)

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3 years ago

Which condition is required for Coulomb's law to hold true?

AleksAgata [21]

The correct answer is:
Point charges must be in a vacuum.

In fact, the usual form for of the Coulomb's law is:
 $F= \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}$
where
 $\epsilon_0$ is the permittivity of free space
q1 and q2 are the two charges
q is the separation between the two charges

However, this formula is valid only if the charges are in vacuum. If they are in a material medium, the law is modified as follows:
 $F= \frac{1}{4 \pi \epsilon_0 \epsilon_r} \frac{q_1 q_2}{r^2}$
where $\epsilon_r$ is the relative permittivity, which takes into account the dielectric effects of the material.

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3 years ago

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Fittoniya [83]

Answer:

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Explanation:

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