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3 years ago

1. What is work done in holding a 15kg suitcase while waiting for a bus for 15 minutes?

1 answer:

7 0

The man is holding the suitcase at the same height above the surface of earth. So the gravitation potential energy remains the same.

<span>work done is force * displacement = weight * 0 = 0</span>

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The period of a pendulum is the time it takes for the pendulum to make one full back-and-forth swing. The period of a pendulum

mixer [17]

Answer: 0.04 s

Explanation:

The given equation for the period $T$ of a pendulum when the acceleration due gravity is given in feet per squared second $g=32 ft/s^{2}$ is:

$T=2 \pi \sqrt{\frac{l}{32 ft/s^{2}}}$

Where $l=\frac{1}{18} ft$ is the length of the pendulum

$T=2 \pi \sqrt{\frac{\frac{1}{18} ft}{32 ft/s^{2}}}$

$T=0.04 s$

3 0

4 years ago

Throw a ball upward from point 0 with an

Mariana [72]

Answer:

69m

Explanation:

4 0

3 years ago

A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05

7nadin3 [17]

Answer:

$W_{drag} = 4.223\,J$

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

$U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}$

The work done on the ball due to drag is:

$W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})$

$W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})$

$W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]$

$W_{drag} = 4.223\,J$

7 0

3 years ago

Read 2 more answers

A ball is dropped off a very tall canyon ledge.

oksano4ka [1.4K]

Answer:

V = 49.05 [m/s]

Explanation:

We can easily find the result using kinematics equations, first, we will find the distance traveled during the 5 seconds.

$y =y_{o}+(v_{o}*t)+(\frac{1}{2}*g*t^{2} )$

where:

Yo = initial position = 0

y = final position [m]

Vo = initial velocity = 0

t = time = 5 [s]

g = gravity aceleration = 9.81 [m/s^2]

The initial speed is zero, as the body drops without imparting an initial speed. Therefore:

y = 0 + (0*5) + (0.5*9.81*5^2)

y = 122.625[m]

Now using the following equation we can find the speed it reaches during the 5 seconds.

$v_{f} ^{2}= v_{i} ^{2}+(2*g*y)\\v_{f}=\sqrt{2*9.81*122.625} \\v_{f}=49.05 [m/s]$

6 0

4 years ago

a particle of mass m and charge q moving with velocity v in a magnetic field B. The velocity of the particle is perpendicular to

Lerok [7]

Explanation:

When a charged particle describes a circular path in a uniform magnetic field, the charged particle experiences a magnetic force towards the center of circular path, according to Fleming's left hand rule. Therefore the magnetic force and velocity (tangent to circular path) are perpendicular to each other during the circular motion. As the direction of displacement, is the direction of velocity, hence force and displacement are perpendicular to each other.

Therefore work done by magnetic force,

W=Fscos90=0

3 0

3 years ago