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1 year ago

When air resistance is ignored, _____ of the projectile affect(s) the range and maximum height of the projectile.

1 answer:

8 0

When air resistance is ignored, initial velocity of the projectile affect the range and maximum height of the projectile.

Projectile is a missile designed to be fired from a rocket or gun.

A projectile is the object that is propelled by the application of an external force and then moves freely under the influence of gravity and air resistance.

The range is defined as the distance between the launch point and the point where the projectile hits the ground.

The height from the ground at the top most position of projectile is referred to as maximum height.

When air resistance is ignored, initial velocity of the projectile affect the range and maximum height of the projectile.

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Answer:

Mg will replace Ag in a compound

Explanation:

A single replacement reaction is driven by the position of ions on the activity series.

As a rule of thumb, the position of metal ions on the activity series determines their reactivity.

Metal ions that are above another are more reactive and they will displace those that are lower.

Generally, activity increases as we go up the group.

Mg ions are higher than Ag ions on the series so, Mg will displace Ag from a solution.

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3 years ago

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2 years ago

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Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the sp

kotegsom [21]

Answer:

m2 = 83.3 g

Explanation:

by conservation of momentum principle we have

$m_1v_{i1} + m_2v_{i2} = m_2v_{f2}$

as both sphere has same speed so $v_{i2} = v_{i1}$

$m_2 = \frac{m_1}[\frac{v_f2}{v_{f1}}+1}$

from conservation of kinetic energy principle we have

$\frac{1}{2}m_1v^{2}_{i1} + \frac{1}{2}m_2v^{2}_{i2} = \frac{1}{2}m_2v^{2}_f2$

$v_{f1} = \sqrt {\frac{(m_1+m_2) v^2_i1}{m_2}$

$v_{f1} = v_{i2}\sqrt {\frac{(m_1+m_2)}{m_2}$

$\frac{v_{f1}}{v_{i2}} =\sqrt {\frac{(m_1+m_2)}{m_2}$

substituting this value in above equation to get m2 value

$m_2 = \frac{m_1}{\sqrt {\frac{(m_1+m_2)}{m_2}+1}}$

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2 years ago

Sixty identical drippers, each with a hole of diameter 1.00 mm, are used to water a yard. If the water in the main pipe of diame

VashaNatasha [74]

Answer:

$V = 5.30 *10^{-2} \ m^3$

$v_1 = 0.3127 \ m/s$

Explanation:

From the question we are told that

The number of identical drippers is n = 60

The diameter of each hole in each dripper is $d = 1.0 \ mm = 0.001 \ m$

The diameter of the main pipe is $d_m = 2.5 \ cm = 0.025 \ m$

The speed at which the water is flowing is $v = 3.00 \ cm/s = 0.03 \ m/s$

Generally the amount of water used in one hour = 3600 seconds is mathematically represented as

$V = A * v * 3600$

Here A is the area of the main pipe with value

$A = \pi * \frac{d^2}{4}$

=> $A = 3.142 * \frac{0.025^2}{4}$

=> $A = 0.0004909 \ m^2$

=> $V = 0.0004909 * 0.03 * 3600$

=> $V = 5.30 *10^{-2} \ m^3$

Generally the area of the drippers is mathematically represented as

$A_1= n * \pi \frac{d^2}{4}$

=> $A_1 = 60 * 3.142 * \frac{0.001 ^2}{4}$

=> $A_1 = 4.713 *10^{-5} \ m^2$

Generally from continuity equation we have that

$Av = A_1 v_1$

=> $0.0004909 * 0.03 = 4.713 *10^{-5} * v_1$

=> $v_1 = 0.3127 \ m/s$

3 0

3 years ago