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3 months ago

6

Two projectiles are in flight at the same time. The acceleration of one relative to the other: A. is always 9.8 m/s2 B. can be a

s large as 19.8 m/s2 C. can be horizontal D. is zero E. none of these

1 answer:

Keith_Richards [3.2K]3 months ago

5 0

Answer:

D. Is Zero

Explanation:

The acceleration experienced by projectiles moving through the atmosphere solely results from gravity, which measures 9.81 m/s².

Apart from gravity, there is no other factor causing acceleration. Air resistance only affects projectiles differently based on their surface areas, leading to varied rates of slowing.

As friction isn't accounted for here (no details about projectile surface areas are provided), we do not need to factor it in, hence ONLY GRAVITY COUNTS.

Both objects are influenced equally by gravity, resulting in no relative acceleration between them.

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1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

Softa [3030]

Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle $\theta$ to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

where $y_o$ represents the initial height from ground level, $v_{oy}$ is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

$v_y=v_{oy}-gt$

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of $v_o$ .

The peak height is attained when $v_y=0$ , which allows us to compute the time to reach that height.

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum height reached is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is equal to 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Continuing with the calculations for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Substituting known values yields

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

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3 months ago

A transverse wave on a string has an amplitude A. A tiny spot on the string is colored red. As one cycle of the wave passes by,

Sav [3153]

Since it's classified as a transverse wave, the particle on the string moves horizontally as the wave progresses, without actual forward or backward travel. Consequently, the red dot shifts 'A' to the left, returns 'A' to the center, moves 'A' to the right, and goes back 'A' to the center once again. Thus, the red dot collectively travels a distance totaling 4A.

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2 months ago

A metal, M, forms an oxide having the formula M2O3 containing 52.92% metal by mass. Determine the atomic weight in g/mole of the

ValentinkaMS [3465]

Answer:

The molar mass of the metal in grams per mole is calculated to be 8.87.

Explanation:

Initially, we can consider a sample of the compound weighing 100 g. This results in:

52.92% metal: 52.92 g M
47.80% oxygen: 47.80 g O

By utilizing the molar mass of oxygen, which is 16 g / mol, we can determine the quantity of moles of oxygen in the sample via the rule of three:

$moles of oxygen=\frac{47.8g*1mol}{16g}$

moles of oxygen=2.9875

The formula for the metal oxide indicates that:

2 M⁺³ + 3 O²⁻ ⇒ M₂O₃

From the previous equation, it is evident that 3 oxygen ions are necessary to react with 2 metal ions. Hence:

$2.9875 moles of oxygen*\frac{2 moles of metal M}{1 mol of oxygen} = 5.975 moles of metal M$

Given 52.92 g of metal in the sample, the molar mass of the metal is:

$molar mass=\frac{52.92 g}{5.975 mol}$

molar mass≅ 8.87 g/mol

The molar mass of the metal in grams per mole is 8.87.

The value that most closely corresponds to this is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.

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2 months ago