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tangare [24]
3 years ago
11

An object which has a mass 70 kg is sitting on a cliff 10 m high . Calculate the objects gravitational potential energy given G=

10m/s2
Physics
2 answers:
Lina20 [59]3 years ago
6 0

In physics, potential energy is calculated as ‘mass × acceleration due to gravity × height’

So,

P.E. = 70 kg × 10 m/s² × 10 m

P.E. = 7000 Joules

inysia [295]3 years ago
4 0

Answer: 7000 joules

Explanation: The potential energy is the energy "stored" by the object.

For objects that are lifted in the y-axis, the potential energy can be calculated as:

U = g*h*m

where g is the gravity's acceleration, h is the height and m is the mass of the object.

Here, g = 10m/s^2, m=70kg and h = 10m

U = 10m/s^2*70kg*10m = 7000 (m/s)^2*kg = 7000 joules

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Two identical small charged spheres hang in equilibrium with equal masses (0.02kg). The length of the strings is equal (0.18m) a
Yuliya22 [10]

Answer:

The value is q = 3.4 *10^{-6} \ C

Explanation:

From the question we are told that

    The mass of each sphere is m_1 = m_2  = m  =  0.020 \ kg

     The length of the string is  l = 0.18 \  m

     The angle of with the vertical is \theta  =  7^o

      The acceleration due to gravity is g = 9.8 \ m/s^2

Generally the force acting between the forces is mathematically represented as

       F  =  T cos \theta =  \frac{k*  q^2}{ r^2}

=>     T cos \theta =  \frac{k*  q^2}{ r^2}

Generally from Pythagoras theorem the radius of the circular curve created by the force is

         r = 2 L sin (\theta )

=>      r = 2* 0.180 sin (7)

=>      r = 0.043 \ m  

=>     q = tan \theta * \frac{m * g * r^2 }{k}

=>      q = tan(7)* \frac{ 0.02 * 9.8 * 0.043^2 }{9*10^{9}}

=>      q = 3.4 *10^{-6} \ C

7 0
3 years ago
One of the largest barometers ever built was an oil-filled barometer constructed in Leicester, England in 1991. The oil had a he
ZanzabumX [31]

Answer:

ρ = 830.32 kg/m³

Explanation:

Given that

Oil head = 12.2 m

h= 12.2 m

Pressure P = 1.013 x 10⁵ Pa

Lets take density of the liquid =ρ

The pressure due to liquid P given as

P = ρ g h

Now by putting the all values in the above equation

1.013 x 10⁵ Pa = ρ x 10 x 12.2                 ( take g =10 m/s²)

ρ = 830.32 kg/m³

Therefore the density of oil is 830.32  kg/m³

5 0
3 years ago
The graph to the right shows the change in Canada‘s harvest of Atlantic cod from 1950-2004 what year shows the clearest evidence
Ann [662]

The correct answer is C. 1995

Explanation:

The graph shows the changes in the harvest of Atlantic cod. In general, this graph illustrates how the peak occurred in the 1980s but then there was a sudden and sharp decline in 1995. Indeed, 1995 is the year with the lowest number of harvested cod as in this year there were approximately least than 10 thousand metric tonnes of cods. Also, this year shows the collapse of fishing stocks or that the population of this fish collapsed, which made it impossible to harvest as many fish as in previous years. According to this, the year that shows the collapse of fishing stocks is 1995.

3 0
3 years ago
What happens to the open circuit if a small fan is connected at point f and the circuit is closed
liraira [26]
There's no way to tell.  Without seeing a diagram of the circuit,
I'll need to know much more about it than you've told me.
I don't know anything about the components or power supply
that are in the circuit, and I don't know where point ' f ' is in it.

Right now, even with the copious volume of all the available
information, no answer to your question is possible. 
4 0
3 years ago
A wheel rotates about a fixed axis with an initial angular velocity of 13 rad/s. During a 8-s interval the angular velocity incr
____ [38]

Answer:

The number of revolutions is 44.6.

Explanation:

We can find the revolutions of the wheel with the following equation:

\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}

Where:

\omega_{0}: is the initial angular velocity = 13 rad/s              

t: is the time = 8 s

α: is the angular acceleration

We can find the angular acceleration with the initial and final angular velocities:

\omega_{f} = \omega_{0} + \alpha t

Where:

\omega_{f}: is the final angular velocity = 57 rad/s

\alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2}

Hence, the number of revolutions is:

\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev

Therefore, the number of revolutions is 44.6.

       

I hope it helps you!

4 0
3 years ago
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