Is this statement true or false? The littoral zone is exposed to air.

Help me complete it. differentiate the types of collision

kolbaska11 [484]

Elastic
☑ occur when the colliding objects bounce off of each other
☑ hard or bouncy objects
☑ kinetic energy is conserved

Inelastic
☑ occur when two objects collide
☑ two objects colliding stick together
☑ kinetic energy is not conserved — some of the initial kinetic energy is converted into other types of energy (heat, sound, etc.)

4 0

3 years ago

A 7450 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci

bagirrra123 [75]

Answer:

a) 520m

b) 10.30 s

c) 100,95 m/s

Explanation:

a) According the given information, the rocket suddenly stops when it reach the height of 520m, because the engines fail, and then it begins the free fall.

This means the maximum height this rocket reached before falling was 520 m.

b) As we are dealing with constant acceleration (due gravity) $g=9.8 \frac{m}{s^{2}}$ we can use the following formula:

$y=y_{o}+V_{o} t-\frac{gt^{2}}{2}$ (1)

Where:

$y_{o}=520 m$ is the initial height of the rocket (at the exact moment in which it stops due engines fail)

$y=0$ is the final height of the rocket (when it finally hits the launch pad)

$V_{o}=0$ is the initial velocity of the rocket (at the exact moment in which it stops the velocity is zero and then it begins to fall)

$g=9.8m/s^{2}$ is the acceleration due gravity

$t$ is the time it takes to the rocket to hit the launch pad

Clearing $t$ :

$0=520 m+0-\frac{9.8m/s^{2} t^{2}}{2}$ (2)

$t^{2}=\frac{-520 m}{-4.9 m/s^{2}}$ (3)

$t=\sqrt{106.12 s^{2}$ (4)

$t=10.30 s$ (5) This is the time

c) Now we need to find the final velocity $V_{f}$ for this rocket, and the following equation will be perfect to find it:

$V_{f}=V_{o}-gt$ (6)

$V_{f}=0-(9.8 m/s^{2})(10.30 s)$ (7)

$V_{f}=-100.95 m/s$ (8) This is the final velocity of the rocket. Note the negative sign indicates its direction is downwards (to the launch pad)

7 0

3 years ago

At a point between the earth and the moon, the magnitude of the force exerted on an object by the earth's gravity equals the mag

S_A_V [24]

1.62 m/s is the correct answer to the problem

8 0

4 years ago

A car tire rotates at a constant angular velocity of 3.5 rotations during a time interval of 0.75 seconds. What is the angular s

Andru [333]

Answer : $\theta = 2.625\ radians$

Explanation :

Given that,

Angular velocity $\omega = 3.5\ rad/s$

Time $t = 0.75 s$

We know , the formula of angular velocity

$\omega = \dfrac{\theta}{t}$

$\theta = \omega \times t$

$\theta = 3.5\ rad/s\times0.75\ s$

The angular speed of the tire is

$\theta = 2.625\ radians$

Hence, this is the required solution.

4 0

3 years ago

A rubber duck sat is floating on the surface

lana66690 [7]

The other force that must be acting on the duck is upthrust of 10 N.

The given parameters;

weight of the rubber duck, W = 10 N

When the weight of the object is equal to the upthrust, the apparent weight will be zero and the object will float in the liquid. This is known as principle of floatation.

Since the given weight of the rubber duck is 10 N, then the upthrust on the duck will be 10 N to enable the duck float on the water.

Thus, the other force acting on the duck is upthrust of 10 N.

Learn more here:brainly.com/question/24777505

7 0

2 years ago