a toy airplane is flying at a speed of 8 m/s with an acceleration of 0.9 m/s^2. How fast is it flying after 2 seconds?

A 1232 kg car moving north at 25.6 m/s collides with a 2028 kg car moving north at 17.5 m/s . They stick together. In what direc

Citrus2011 [14]

Answer:

I. Angle = 41.7° Northeast.

II. Vr = 7.08m/s

Explanation:

Let the two cars be denoted by A and B

Given the following data;

Mass of car A = 1232 Kg

Velocity of car A = 25.6 m/s

Mass of car B = 2028 Kg

Velocity of car B = 17.5m/s

First of all, we would solve for momentum;

Momentum = mass × velocity

Momentum, M1 = 1232 × 25.6

Momentum, M1 = 31539.2 Kgm/s

Momentum, M2 = 2028 × 17.5

Momentum, M2 = 35490 Kgm/s

Now, let's find the resultant momentum using the Pythagoras theorem;

R² = M1² + M2²

R² = 31539.2² + 35490²

R² = 994721136.6 + 1259540100

R² = 2254261237

Taking the square root of both sides, we have

Resultant momentum, R = 47479.06 Kgm/s

To find the direction;

Angle = tan¯¹(M1/M2)

Angle = tan¯¹(31539.2/35490)

Angle = tan¯¹(0.89)

Angle = 41.7° Northeast.

To find the speed;

R = (M1 + M2)Vr

47479.06 = (31539.2 + 35490)Vr

47479.06 = 67029.2Vr

Vr = 47479.06/67029.2

Vr = 7.08m/s

6 0

2 years ago

The symbol for the charge on a proton is:

timofeeve [1]

$p +$
I think is that

4 0

3 years ago

A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread nonuniformly throug

Vaselesa [24]

Answer:

$1.57 * 10^{3} Q$

Explanation:

The volume charge density is defined by ρ = $\frac{Q}{V}$ (Equation A), where Q is the charge and V, the volume.

The units in the S.I. are $\frac{Coulombs}{m^{3} }$ , so we have to express the radius in meters:

inner radius = $4 cm * \frac{1 m}{100 cm} = 0.04m$

outer radius = $6 cm * \frac{1m}{100cm} = 0.06m$

Now, we know that the volume of the sphere is calculated by the formula:

$V = \frac{4}{3}\pi r^{3}$ , and as we have an spherical shell, the volume is calculated by the difference between the outher and inner spheres:

V = $\frac{4}{3}\pi (r_{o} ^{3} - r_{i} ^{3})$ , where $r_{o}$ is the outer radius and $r_{i}$ is the inner radius.

Replacing the volume formula in the Equation A:

ρ = $\frac{Q}{\frac{4}{3}\pi(r^{3} _{o}-r_{i} ^{3})}$

ρ = $\frac{3Q}{4\pi (r_{o} ^{3}-r_{i} ^{3} ) }$

Replacing the values of the outer and inner radius whe have:

ρ = $\frac{3Q}{4\pi (1.52 * 10^{-4})}$

ρ = $1.57 * 10^{3} Q$

4 0

3 years ago

Question 19 (Multiple Choice Worth 3 points) A mirror will reflect

Jobisdone [24]

So far, all we have is a very short true statement. If it's a multiple choice question, then where are they ?

4 0

3 years ago

The bar graph shows the population of male and female deer in a forest, measured in 10-year intervals. According to the graph, w

saul85 [17]

Answer: option D. the ratio of the population of male deer is not constant.

Explanation:

The bar graph permits to compare the results for two different populations: male and female deer in a very easy visual way.

These features are remarkable:

The polulation of male deer (blue bars) decrease from 1961 to 1971, then increase in the next 10 year, decrease in the next decade, and increase for the next two decades. So, its trend is erratic, with ups and downs.

This discards the option A, which states that the population of male deer increases each decade from 1961 to 2011.

The population of female deer (purple or brown bars) decreases every decade.

This discards the option B. which states that when the polulation of male deer increases, the poluplation of female deer also increases.

The populations never are equal, hence this discards the option C.

Since, one popultion increases and decreases, while the other population only decreases, you conclude that the ratio of the population of male deer to female deer is not constant, which is the option D.

4 0

3 years ago

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