If ∠G=30degrees, g=3, k=4 what is the approximate value of h?2.2 or 4.7

4.7

5.7

Answers

Answer 1

Answer: Without a picture of the situation, we don't know the relationship
between angle-G, g, k, and 'h'. We don't even know for sure
whether 'h' is on the same planet as the other things. No answer
is possible without seeing the picture.

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After a radioactive atom decays, it is the same element that it was before with no measurable change in mass. Which kind of decay has occurred, and how do you know? alpha decay because alpha particles have no mass beta decay because this kind of decay cannot change one element into another alpha decay because it creates a new isotope of the same element gamma decay because photons have no mass.

Answers

Gammadecay is the production of extremely high-frequency electromagnetic radiation . Gamma decay is the right answer from all types of decay.

What is gamma-decay?

Gammadecay is the production of extremely high-frequency electromagnetic radiation in order to stabilize the unstablenucleus. You must be well-versed in the various energylevels found in an atom.

The Nucleus has its own amount of energy. Gamma decay is the nucleus' method of transitioning from a higher energy level to a lower energy level by emitting high energy photons.

Hence gamma decay is the right answer for all types of decay. Because photons have no mass.

To learn more about gamma decay refer to the link;

brainly.com/question/25455333?referrer=searchResults

Answer:

D. gamma decay because photons have no mass

Explanation:

Waves are created when a source of _________ (force) causes a vibration.

Answers

The answer is "energy"

The density of seawater is 1.025 g/ml. what is the mass of 1000 liters of seawater in g amd kg?

Answers

You are giventhe density of seawater which is 1.025 grams per milliliters and its volume at1000 liters. You are required to find the mass of the seawater in grams andkilograms. You need to know that density is equal to mass over volume. Let usdenote D as our density, V our volume and M as our mass. So,

D = M/V
M = D*V
M = (1.025g/mL)(1000 L)(1000mL/1L)
M =1,025,000 g
To changeinto kilograms, simply divide it by 1000
M = (1,025,000g)(1kg/1000g) = 1,025kg

An object with a mass of 1.5 kg changes its velocity from +15 m/s to +22 m/s during a time interval of 3.5 seconds. What impulse was delivered to the object? 3.0 N • s 11 N • s 25 N • s 56 N • s

Answers

Answer: The impulse delivered to the object is 33 N • s.

Explanation: Impulse = Change in momentum

The momentum of an object is calculated by multiplying its mass by its velocity:

Momentum = Mass x Velocity

Given:

Mass of the object = 1.5 kg

Initial velocity = +15 m/s

Final velocity = +22 m/s

Time interval = 3.5 seconds

First, we need to calculate the change in momentum:

Change in momentum = Final momentum - Initial momentum

To find the initial momentum, we multiply the mass by the initial velocity:

Initial momentum = Mass x Initial velocity

Initial momentum = 1.5 kg x 15 m/s

To find the final momentum, we multiply the mass by the final velocity:

Final momentum = Mass x Final velocity

Final momentum = 1.5 kg x 22 m/s

Now, we can calculate the change in momentum:

Change in momentum = Final momentum - Initial momentum

Change in momentum = (1.5 kg x 22 m/s) - (1.5 kg x 15 m/s)

Simplifying the equation:

Change in momentum = 33 kg m/s

Finally, we have found the change in momentum, which is equal to the impulse delivered to the object. Therefore, the impulse delivered to the object is 33 N • s.

Answer:

11 Ns

Explanation:

Impulse = change in momentum

J = Δp

J = mΔv

J = (1.5 kg) (22 m/s − 15 m/s)

J = 10.5 kg m/s

Rounded to two significant figures, the impulse is 11 Ns.

What is the density of the block? Assume that the block is 40.0% submerged.First Sketches says block weigh 50g
Express the density numerically in grams per cubic centimeter.

Answers

There are forces are acting on the block: 1. Fs = upward force from scale (up = + Fs) 2. W = downward weight force (down = - W) 3. Fb = upward buoyancy force (up = + Fb)
Since the block is at rest, (net force = 0), the force equation is:
Fs + Fb - W = 0 ~(eq1)
Substituting Fs, Fb and W into ~(eq1): ==> {m (on scale) x g} + {ρ x V x g} - (m (block) x g} = 0
Dividing both sides of equation by g: m (on scale) + (ρ x V) - m (block) = 0 ~(eq2)
Given: V = volume of water displaced = 40% volume of the block = 0.40 x V (block)
Substituting 0.40 x V (block) into ~(eq2): m (on scale) + (ρ x 0.40 x V (block)) - m (block) = 0 ~(eq3) where, m (on scale) = 5.6 g ρ = density of water at 20ºC = 0.99821 g/cm^3 m (block) = 50 g
Substituting m (on scale), ρ and m (block) into ~(eq3): 5.6 g + (0.99821 g/cm^3 x 0.40 x V (block)) - 50 g = 0 ~(eq4)
Simplifying ~(eq4) and solving for V (block): ==> 5.6 g + 0.399284 g/cm^3 x V (block) - 50 g = 0 ==> 0.399284 g/cm^3 x V (block) - 44.4 g = 0 ==> 0.399284 g/cm^3 x V (block) = 44.4 g ==> V (block) = 44.4 g / (0.399284 g/cm^3) ==> V (block) = 111.199 cm^3
ρ (block) = density of block = m (block) / V (block) = 50 g / 111.199 cm^3 = 0.449644 g / cm^3

If you had thrown the pennies down from the observation deck 321m above street level at an initial velocity of 1.99m/s, determine their speed as they reach the ground.

Answers

Answer: 79.3 m/s

Explanation: Use the equation V^2-Vo^2=2a(ΔY)

We know that the initial velocity is 1.99 m/s, so that is our Vo. We also are given that the height of the deck is 321, so therefore, our delta y is 321. Finally, our acceleration is 9.81 m/s^2 since that is the acceleration of gravity during free fall. We are solving for the final velocity (V)

So, our equation is V^2-(1.99^2)=2(9.8)(321)

Next, V^2-(3.96)=6291.6

V^2=6295.56

Take the square root of both sides, and V=79.3 m/s

Answer:79.3 m/s

Explanation:

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