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3 years ago

9

What are the two main types of defense that teams employ?

1 answer:

Gnesinka [82]3 years ago

7 0

Answer:

they employ front and Back defenses

Explanation:

there are the most positions in these types

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Which property increases as an electromgnetic waze's decreases

kirza4 [7]

Answer:

the frequency decreases

Explanation:

As a wave's wavelength increases, the frequency decreases

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3 years ago

According to ohm's law if you don't change the value of the resistor & you double the voltage in a circuit the amount of cur

Nana76 [90]

They double!! Hope im not too late!!

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3 years ago

How does a dynamo Work?

marin [14]

The generator/dynamo<span> is made up of stationary magnets (stator) which create a powerful magnetic field, and a rotating magnet (rotor) which distorts and cuts through the magnetic lines of flux of the stator. When the rotor cuts through lines of magnetic flux it makes electricity.</span>

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3 years ago

It is interesting to speculate on the properties of a universe with different values for the fundamental constants.

qwelly [4]

Answer:

Part a)

$\lambda = 0.345 m$

Part b)

$\Delta x = 0.274 m$

Part c)

$r = 2.8 \times 10^{11} m$

Explanation:

Part a)

De broglie wavelength is given as

$\lambda = \frac{h}{mv}$

$\lambda = \frac{1}{(0.145)(20)}$

$\lambda = 0.345 m$

Part b)

By principle of uncertainty we know that

$\Delta x \times \Delta P = \frac{h}{4\pi}$

$\Delta x \times (0.145)(21 - 19) = \frac{1}{4\pi}$

$\Delta x = 0.274 m$

Part c)

As we know that

$\frac{kq_1q_2}{r^2} = \frac{mv^2}{r}$

also we know

$mvr = \frac{nh}{2\pi}$

$v = \frac{h}{2\pi mr}$

now we have

$\frac{ke^2}{r} = \frac{mh^2}{4\pi^2m^2 r^2}$

$r = \frac{h^2}{4\pi^2mke^2}$

$r = 2.8 \times 10^{11} m$

5 0

3 years ago

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0° above the horizo

Norma-Jean [14]

Answer:

0.528m

Explanation:

a)58.7 cm = 0.587 m

Let g = 9.8m/s2. When the frog jumps from ground to the highest point its kinetic energy is converted to potential energy:

$E_p = E_k$

$mgh = mv^2/2$

where m is the frog mass and h is the vertical distance traveled, v is the frog velocity at take-off

$v^2 = 2gh = 2*9.8*0.587 = 11.5$

$v = \sqrt{11.5} = 3.4 m/s$

b) Vertical and horizontal components of the velocity are

$v_v = vsin(\alpha) = 3.4sin(58^0) = 2.877 m/s$

$v_h = vcos(\alpha) = 3.4cos(58^0) = 1.8 m/s$

The time it takes for the vertical speed to reach 0 (highest point) under gravitational acceleration g = -9.8m/s2 is

$\Delta t = \Delta v / g = \frac{0 - 2.877}{-9.8} = 0.293s$

This is also the time it takes to travel horizontally, we can multiply this with the horizontal speed to get the horizontal distance it travels

$s_h = v_ht = 1.8*0.293 = 0.528 m$

3 0

3 years ago