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umka2103 [35]
3 years ago
7

Matthew throws a ball straight up into the air. It rises for a period of time and then begins to drop. At which points in the ba

ll’s journey will gravity be the greatest force acting on the ball?
Physics
1 answer:
PolarNik [594]3 years ago
3 0

Answer:

If air resistance is taken as negligible, then the ball is in freefall the moment it is thrown so gravity is the only force acting on the object. If air resistance is not negligible then gravity will be the greatest force acting on the ball while it is going up and coming down, because Fair has to be less than gravity at all times otherwise the atmosphere would wither away.

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Which statement describes the magnetic field inside a bar magnet?
kakasveta [241]

The statement that describes the magnetic field inside a bar magnet is as follows: it points from north to south.

<h3>What is a bar magnet?</h3>

A bar magnet is a permanent magnet of rectangular shape.

A magnet generally possess a magnetic field, which is a condition in the space around a magnet which there is a detectable magnetic force and the presence of two magnetic poles.

A bar magnet like every other magnet possesses a magnetic field that points from the north pole to the south pole.

Learn more about magnets at: brainly.com/question/13026686

#SPJ1

5 0
2 years ago
At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction
lbvjy [14]

Answer:

F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

|F_T|=2\sqrt{34}k\frac{Q^2}{L}

\theta=tan^{-1}(\frac{5}{3})=59.03\°

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

and the magnitude is:

|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}

the direction is:

\theta=tan^{-1}(\frac{5}{3})=59.03\°

4 0
3 years ago
In a fireworks display, a rocket is launched from the ground with a speed of 18.0 m/s and a direction of 51.0° above the horizon
Masja [62]

Answer:38.66 m

Explanation:

Given

launch angle \theta =51^{\circ}

launch velocity u=18 m/s

center of mass continue to travel its original Path so it center of mass will be at a distance of

R=\frac{u^2\sin 2\theta }{g}

R=\frac{18^2\sin 102}{9.8}

R=32.33 m

Center of mass will be at x=32.33 m

(b)if one of the piece will be at x=26  m then other will be at

x_{com}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

x_{com}=\frac{\frac{m}{2}\cdot 26+\frac{m}{2}\cdot x_0}{\frac{m}{2}+\frac{m}{2}}

32.33=\frac{26+x_0}{2}

x_0=38.66 m

8 0
3 years ago
A liquid of density 830 kg/m3 flows through a horizontal pipe that has a cross-sectional area of 1.20 x 10-2 m2 in region A and
kicyunya [14]

Answer:

A) volume flow rate = 0.047 m3/s

B) mass flow rate = 39.01 kg/s

Explanation:

Detailed explanation and calculation is shown in the image below

3 0
3 years ago
A. 24.89<br> B. 25.89<br> C. 17.74<br> D. 19.73
Veronika [31]

Answer: D

Explanation:

Just did it got an 100

5 0
2 years ago
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