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3 years ago

A wave has a period of 2 seconds and a wavelength of 4 meters. Calculate its frequency and speed.

Physics

1 answer:

sukhopar [10]3 years ago

4 0

Answer:

frequency = 0.5 /s

speed = 2m/s

Explanation:

frequency = 1/period = 1/2 = 0.5 /s

speed = frequency × wavelength

= 0.5 × 4 = 2m/s

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A2 kg block is accelerating at the rate of 5 m/s² while being acted on by two forces. One of the forces equals 30 N, 0°. What ar

Mademuasel [1]

Answer:

20 [N], in the opposite direction of the first force.

Explanation:

We know that newton's second law stipulates that the sum of forces on a body must be equal to the product of mass by acceleration.

$SumF = m*a\\30 + F = 2*5\\F = 30 - (2*5)\\F = - 20 [N]$

The negative sign means that the other force acting on the body must be in the opposite direction to the force of 30 [N]

6 0

3 years ago

The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p

leva [86]

1) $1.86\cdot 10^6 rad/s^2$

2) 2418 rad/s

3) $27000 m/s^2$

4) 36.3 m/s

Explanation:

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

$\theta=\omega_i t +\frac{1}{2}\alpha t^2$

where:

$\theta$ is the angular displacement

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

$\theta=90^{\circ} = \frac{\pi}{2}rad$ is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

Solving for $\alpha$ , we find:

$\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2$

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

$\omega_f = \omega_i + \alpha t$

where

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

Therefore, the final angular speed is:

$\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s$

The tangential acceleration is related to the angular acceleration by the following formula:

$a_t = \alpha r$

where

$a_t$ is the tangential acceleration

$\alpha$ is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

$a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2$

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

$v=u+at$

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

$a=27900 m/s^2$ (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

$v=0+(27900)(0.0013)=36.3 m/s$

5 0

3 years ago

Who was known for being a pilot and an astronaut that walked on the moon?.

Mkey [24]

Answer:

albert einsteinssssssssssssssssssssssssss

Explanation:

3 0

2 years ago

Read 2 more answers

An insulated container with a divider in the middle contains two separated gases. Gas 1 is initially at a higher temperature tha

12345 [234]

Answer:

Option (D) On average, the molecules of gas 1 lose some of their kinetic energy to the molecules of gas 2 through collisions, resulting in the two gases eventually having the same temperature.

Explanation:

From the question given, Gas 1 was initially at a higher temperature than Gas 2.

As the two gas mixes together, there will be a transfer of heat from Gas 1 molecules to Gas 2 molecules. Now, as this continues over a period of time, the two gas will eventually have the same temperature.

4 0

3 years ago

Read 2 more answers

A 225 kg block is pulled by two horizontal forces. The first force is 178 N at a 41.7-degree angle and the second is 259 N at a

yawa3891 [41]

Answer:

52.9 N, 364.7 N

Explanation:

First of all, we need to resolve both forces along the x- and y- direction. We have:

- Force A (178 N)

$A_x = (178 N)(cos 41.7^{\circ})=132.9 N\\A_y = (178 N)(sin 41.7^{\circ})=118.4 N$

- Force B (259 N)

$B_x = (259 N)(cos 108^{\circ})=-80.0 N\\B_y = (259 N)(sin 108^{\circ})=246.3 N$

So the x- and y- component of the total force acting on the block are:

$R_x = A_x + B_x = 132.9 N - 80.0 N =52.9 N\\R_y = A_y + B_y = 118.4 N +246.3 N = 364.7 N$

7 0

3 years ago