All categories

3 years ago

Which waves move perpendicular to their energy?

Physics

1 answer:

Step2247 [10]3 years ago

3 0

Transverse waves move perpendicular to their energy

You might be interested in

Help again! Thank you all xoxo

Svetach [21]

Answer:

Amount of charge

Explanation:

Mark me brainliest

6 0

3 years ago

QUICK: A circular loop of radius r is rotated through a magnetic field B, which of the following would increase the magnetic flu

WITCHER [35]

Doubling B and quadrupling r

6 0

3 years ago

Read 2 more answers

if you throw a rock straight up at a speed of 19m/s. How long goes it take the ball to reach its maximum height?

Scrat [10]

Answer:

It requires 1.9 seconds to reach maximum height.

Explanation:

As per given question,

Initial velocity (U) =19 m/s

Final velocity (V) = 0 m/s

$\text { Taking acceleration due to gravity }(a)=10 \mathrm{m} / \mathrm{s}^{2}$

Maximum height = S

Time taken is "t"

Calculating time taken to reach maximum height:

We know that time taken to reach the maximum height is calculated by using the formula V = U + at

Substitute the given values in the above equation.

Final velocity is “0” as there is no velocity at the maximum height.

$0=19+10 \times t$

$-19=10 \times t$

$\frac{-19}{10}=t$

t = 1.9 seconds.

The time taken to reach maximum height is 1.9 seconds.

Calculating maximum height:

$\text { Consider the equation } V^{2}-U^{2}=2 a S$

Solving the equation we will get the value of S

$0-19^{2}=2 \times(-10) \times \mathrm{S} .(-\text { is due to opposite of gravity) }$

-361 = -20S

Negative sign cancel both the sides.

$\mathrm{S}=\frac{361}{20}$

S = 18.05 m

Maximum height is 18.05 m .

3 0

3 years ago

How much force is needed to accelerate an object of mass 90 kg at a rate of 1.2 m/s2? 0.013 N 75 N 108 N 1080 N

lana [24]

F = m*a

F = 90*1,2

F = 108 N

8 0

4 years ago

Read 2 more answers

At the end of a delivery ramp, a skid pad exerts a constant force on a package so that the package comes to rest in a distance d

MatroZZZ [7]

Answer:

increases by a factor of $\sqrt{2}$

Explanation:

First we need to find the initial velocity for it to stop at the distance 2d using the following equation of motion:

$v^2 - v_0^2 = 2a\Delta s$

where v = 0 m/s is the final velocity of the package when it stops, $v_0$ is the initial velocity of the package when it, a is the deceleration, and $\Delta s = d$ is the distance traveled.

So the equation above can be simplified and plug in Δs = d, $v_0 = v_1$ for the 1st case

$-v_1^2 = 2ad$ (1)

For the 2nd scenario where the ramp is changed and distance becomes 2d, $v_0 = v_2$

$-v_2^2 = 4ad$ (2)

let equation (2) divided by (1) we have:

$\left(\frac{v_2}{v_1}\right)^2 = 4ad / 2ad = 2$

$v_2 / v_1 = \sqrt{2}$

$v_2 = \sqrt{2}v_1$

So the initial speed increases by $\sqrt{2}$ . If the deceleration a stays the same and time is the ratio of speed over acceleration a

$t = v / a$

The time would increase by a factor of $\sqrt{2}$

7 0

4 years ago