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son4ous [18]
3 years ago
9

Please help me**** i need some answers now

Physics
1 answer:
sweet-ann [11.9K]3 years ago
4 0
Output can not be greater than input because the conversion of energy can not be greater than 100%.
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a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone​
Dafna1 [17]

Answer:

The centripetal acceleration of the stone is 5 m/s²

Explanation:

The length of the string to which the stone is attached, r = 1 m

The speed with which the string is rotated, v = 5 m/s

The centripetal acceleration, a_c, is given as follows;

a_c = \dfrac{v^2}{r}

Therefore, the centripetal acceleration of the stone found as follows;

a_c = \dfrac{(5 \ m/s)^2}{1 \ m} = 5 \ m/s^2

The centripetal acceleration of the stone, a_c = 5 m/s².

5 0
3 years ago
HELP ASAP <br> Describe when contact metamorphism occurs?
diamong [38]

Contact metamorphism occurs adjacent to igneous intrusions and results from high temperatures associated with the igneous intrusion. Since only a small area surrounding the intrusion is heated by the magma, metamorphism is restricted to the zone surrounding the intrusion, called a metamorphic or contact aureole

5 0
3 years ago
Read 2 more answers
This time particle A starts from rest and accelerates to the right at 65.5 cm/s
FrozenT [24]

Answer:

t = 4 s

Explanation:

As we know that the particle A starts from Rest with constant acceleration

So the distance moved by the particle in given time "t"

d = v_i t + \frac{1}{2}at^2

d = 0 + \frac{1}{2}(65.5)t^2

d_1 = 32.75 t^2 cm

Now we know that B moves with constant speed so in the same time B will move to another distance

d_2 = 44 \times t

now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

d_1 = 349 + d_2

32.75 t^2 = 349 + 44 t

on solving above kinematics equation we have

t = 4 s

4 0
3 years ago
At its widest point, the diameter of a bottlenose dolphin is 0.50 m. Bottlenose dolphins are particularly sleek, having a drag c
fiasKO [112]

Answer:

497.00977 N

3742514.97005

Explanation:

\rho = Density of water = 1000 kg/m³

C = Drag coefficient = 0.09

v = Velocity of dolphin = 7.5 m/s

r = Radius of bottlenose dolphin = 0.5/2 = 0.25 m

A = Area

Drag force

F_d=\frac{1}{2}\rho CAv^2\\\Rightarrow F_d=\frac{1}{2}\times 1000 \times 0.09(\pi 0.25^2)7.5^2\\\Rightarrow F_d=497.00977\ N

The drag force on the dolphin's nose is 497.00977 N

at 20°C

\mu = Dynamic viscosity = 1.002\times 10^{-3}\ Pas

Reynold's Number

Re=\frac{\rho vd}{\mu}\\\Rightarrow Re=\frac{1000\times 7.5\times 0.5}{1.002\times 10^{-3}}\\\Rightarrow Re=3742514.97005

The Reynolds number is 3742514.97005

8 0
3 years ago
A high power line carries a current of 1.0 kA. What is the strength of the magnetic field this line produces at the ground, 10 m
solmaris [256]

Answer:

The strength of the magnetic field that the line produces is 2x10^{-5} Tesla.

Explanation:

From Biot-Savart law, the equation to determine the strength of the magnetic field for any straight wire can be deduced:

           

B = \frac{\mu_{0}I}{2\pi r} (1)      

                                     

Where \mu_{0} is the permiability constant, I is the current and r is the distance from the wire.    

             

Notice that it is necessary to express the current, I, from kiloampere to ampere.

I = 1.0kA \cdot \frac{1000A}{1kA} ⇒ 1000A

Finally, equation 1 can be used:

B = \frac{(4\pi x10^{-7}T.m/A)(1000A)}{2\pi (10m)}    

           

B = 2x10^{-5}T    

Hence, the strength of the magnetic field that the line produces is 2x10^{-5} Tesla.

         

8 0
3 years ago
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