Ask question

Ask question

All categories

3 years ago

9

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

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%.

You might be interested in

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