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Ne4ueva [31]
3 years ago
10

Why would the time of flight depend on the angle of the launch

Physics
1 answer:
DIA [1.3K]3 years ago
3 0
The time of flight depends on the vertical speed. The angle of the launch determines how much of the total launch velocity is horizontal speed and how much of it is vertical speed.
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It’s Newton’s third law
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Pressure, V is volume , T is temperature of a gas and R is gas constant.
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2 years ago
A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it
Veseljchak [2.6K]

Incomplete question as the unit of volume is not written correctly.So the complete question is here:

A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it in a graduated cylinder. What is the density of a 240-g rock that displaces 89.0 cm³?

Answer:

d_{Density}=2.7g/cm^{3}

Explanation:

Given data

Mass m=240g

Volume V=89.0 cm³

To find

Density d

Solution

If rock displaces 89.0 cm³ of water means volume of rock is also 89cm³

So

d_{Density}=\frac{mass}{volume}\\d_{Density}=\frac{240g}{89.0cm^{3} } \\d_{Density}=2.7g/cm^{3}

5 0
3 years ago
A closed, uninsulated system fitted with movable piston, so no matter is exchanged with the surroundings, was assembled. Introdu
xeze [42]

Answer: You do not specify what is being asked for. ∆E? ∆H?

∆E = (430 - 238) J = 192 J

∆H = 430 J

Explanation:

If asked for the value of ∆H the answer is simply the change in heat, and in the question, it states introduction of 430 J of heat is causing the system to expand.

Therefore ∆H = 430 J

If asked for ∆E, we know that ∆E = ±q (heat) + work (-P∆V) = ±q + w

The question states that 238 J of work are done AND the system expanded

(work is negative because expansion means work is done BY the system, releasing energy/heat... Conversely, if the system were compressed, work is done ON the system, absorbing heat/energy)

Therefore, ∆E = (430 - 238) J = 192 J

8 0
2 years ago
Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at
dedylja [7]

Answer:

50.91 \mu C

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for q_{1} and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted byq_{1} on q. Then we can know the magnitude of the force exerted by q_{2} about q, finally this will allow us to know the magnitude of q_{2}

q_{1} exerts a force on q in +y direction, and q_{2} exerts a force on q in -y direction.

F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\

The net force on q is:

F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}

Rewriting for q_{2}:

q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C

8 0
3 years ago
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