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shtirl [24]
2 years ago
5

10) For a horizontally launched projectile, decreasing the velocity of the

Physics
1 answer:
Daniel [21]2 years ago
4 0

Answer:i think the answer is b

Explanation:because if the velocity is decreased then it will not travel as far and wont stay in the air as long as it wont have as much force keeping it in the air

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If a much heavier stone rolled off the same cliff, would it hit the ground more quickly? explain
N76 [4]
No, the rate of gravity remains constant
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3 years ago
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Short Answer Questions:
Otrada [13]

According to Archimede's principle, a physical object experiences an upthrust due to a difference in pressure between upper and lower  surfaces.

<h3>What is an upthrust?</h3>

An upthrust is also referred to as buoyancy and it can be defined as an upward force which is exerted by a fluid (liquid or gas), so as to oppose the weight of a partially or fully immersed physical object that is floating in it.

Based on scientific information, a physical object experiences an upthrust when it is immersed in a fluid due to a difference in height and pressure between upper (top) and lower (bottom) surfaces.

According to Archimede's principle, there is a higher pressure at the bottom of the physical object due to height, and a lower pressure at the top of the physical object.

Read more on upthrust here: brainly.com/question/24389514

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4 0
2 years ago
A train travels at a speed of 30 m/s. The train starts at an initial position of 1000 meters and travels for 30 seconds. What is
pychu [463]
1000 + 30x30 = 1900. Hope that helps
6 0
3 years ago
A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a
makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]              (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°

The distance r is:

r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m

You replace the values of all parameters in the equation (1):

\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0
3 years ago
Why does an energy transfer not always result in phase change?
mash [69]

Answer:

These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules. "If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance

4 0
3 years ago
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