I need to figure out A but i’m not sure how

Two sides of a parallelogram are in the ratio 5 : 3. If its perimeter is 64 cm, find the lengths of its sides.

Xelga [282]

Answer: 32

Explanation:

4 0

2 years ago

A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.

EleoNora [17]

(a) The kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

(b) The work done in firing the projectile is 2,500 J.

<h3>Kinetic energy of the projectile at maximum height</h3>

The kinetic energy of the projectile when it reaches the highest point in its trajectory is calculated as follows;

K.E = ¹/₂mv₀ₓ²

where;

m is mass of the projectile
v₀ₓ is the initial horizontal component of the velocity at maximum height

<u>Note:</u> At maximum height the final vertical velocity is zero and the final horizontal velocity is equal to the initial horizontal velocity.

K.E = (0.5)(2)(30²)

K.E = 900 J

<h3>Work done in firing the projectile</h3>

Based on the principle of conservation of energy, the work done in firing the projectile is equal to the initial kinetic energy of the projectile.

W = K.E(i) = ¹/₂mv²

where;

v is the resultant velocity

v = √(30² + 40²)

v = 50 m/s

W = (0.5)(2)(50²)

W = 2,500 J

Thus, the kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

The work done in firing the projectile is 2,500 J.

Learn more about kinetic energy here: brainly.com/question/25959744

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5 0

11 months ago

You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its l

sdas [7]

This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

a)

We will use the second equation of motion to find out the initial speed of the arrow.

$h= v_it + \frac{1}{2}gt^2\\$

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}$

b)

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

$h= v_it + \frac{1}{2}gt^2\\$

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

$17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0$

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

4 0

1 year ago

An earthquake’s epicenter is _____.

Evgen [1.6K]

The point in which it originates.

4 0

2 years ago

Read 2 more answers

A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A

Sphinxa [80]

Answer:

Density of 127 I = $\rm 1.79\times 10^{14}\ g/cm^3.$

Also, $\rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

Explanation:

Given, the radius of a nucleus is given as

$\rm r=kA^{1/3}$ .

where,

$\rm k = 1.3\times 10^{-13} cm.$

A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

$\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.$

For the nucleus 127 I,

Mass, M = $\rm 2.1\times 10^{-22}\ g.$

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

$\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.$

On comparing with the density of the solid iodine,

$\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

7 0

2 years ago