What are three properties of components of the universe that can be determined using electromagnetic radiation ?

A 5 kilograms bowling ball is dropped out a window. It hits the ground, and bounces upward. The velocity change of the ball is n

ioda

Answer:

13.5

Explanation:

Mass: 5kg

Initial Velocity: -15

Final Velocity: 12

Force: 10

We can use the equation: Vf = Vi + at

We need to find acceleration, and we can use the equation, F=ma,

We have mass and the force so it would look like this, 10=5a, and 5 times 2 would equal 10, so acceleration would be 2.

Now we have all the variables to find time.

Back to Vf = Vi + at, plug the numbers in, 12 = -15 + 2(t)

Plugging them in into desmos gives 13.5 for time.

4 0

2 years ago

An airplane is moving at 350 km/hr. If a bomb is

bogdanovich [222]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

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

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final height

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb's initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

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

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's final velocity

Knowing this, let's begin with the answers:

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity </h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

3 years ago

a uniform rule balance on a knife edge at the 55cm mark when a mass of 40g is hung from the 95cm mark. find the weight of the ru

BigorU [14]

Answer:

W = 3.1 N

Explanation:

moments about any convenient point will sum to zero.

I choose summing about the knife edge mark and will assume the ruler of weight W is of uniform construction.

I will assume the ruler weight makes a positive moment

W[55 - 50) - 0.040(9.8)[ 95 - 55] = 0

5W = 15.68

W = 3.136

7 0

3 years ago

Which of the following is NOT a good example of a short-term goal?

Molodets [167]

Loose 10 pounds im pretty sure

5 0

4 years ago

A rescue pilot drops a survival kit while her plane is flying at an altitude of 1.5 km with a forward velocity of 100.0 m/s. If

tresset_1 [31]

1750 meters.
First, determine how long it takes for the kit to hit the ground. Distance over constant acceleration is:
d = 1/2 A T^2
where
d = distance
A = acceleration
T = time
Solving for T, gives
d = 1/2 A T^2
2d = A T^2
2d/A = T^2
sqrt(2d/A) = T
Substitute the known values and calculate.
sqrt(2d/A) = T
sqrt(2* 1500m / 9.8 m/s^2) = T
sqrt(3000m / 9.8 m/s^2) = T
sqrt(306.122449 s^2) = T
17.49635531 s = T
Rounding to 4 significant figures gives 17.50 seconds. Since it will take
17.50 seconds for the kit to hit the ground, the kit needs to be dropped 17.50
seconds before the plane goes overhead. So just simply multiply by the velocity.
17.50 s * 100 m/s = 1750 m

3 0

3 years ago

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