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svet-max [94.6K]

3 years ago

10

Vector A with arrow has a magnitude of 35 units and points in the positive y direction. When vector B with arrow is added to A w

ith arrow, the resultant vector A with arrow + B with arrow points in the negative y direction with a magnitude of 13 units. Find the magnitude and direction of B with arrow.

1 answer:

belka [17]3 years ago

8 0

Answer:

- 43uj

Explanation:

The solution is in the attached file below

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She will use one 2 as a subscript.

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3 years ago

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I have an object that I want to know when it was made. I calculate that the object contains 0.12g of Iodine-131 but started with

Sav [38]

Answer: 71.72 days

Explanation:

This problem can be solved using the <u>Radioactive Half Life Formula: </u>

$A=A_{o}.2^{\frac{-t}{h}}$ (1)

Where:

$A=0.12 g$ is the final amount of Iodine-131

$A_{o}=60 g$ is the initial amount of Iodine-131

$t$ is the time elapsed

$h=8 days$ is the half life of Iodine-131

Knowing this, let's substitute the values and find $t$ from (1):

$0.12 g=(60 g)2^{\frac{-t}{8 days}}$ (2)

$\frac{0.12 g}{60g}=2^{\frac{-t}{8 days}}$ (3)

Applying natural logarithm in both sides:

$ln(\frac{0.12 g}{60g})=ln(2^{\frac{-t}{8 days}})$ (4)

$-6.21=-\frac{t}{8 days}ln(2)$ (5)

Finding $t$ :

$t=71.72 days$

7 0

4 years ago

If the velocity of an object is 9 m/s and its momentum is 72 kgm/s, what is its mass

MAVERICK [17]

An object with a velocity (v) of 9 m/s and a linear momentum (p) of 72 kg.m/s, has a mass (m) of 8 kg.

<h3>What is momentum?</h3>

In Newtonian mechanics, linear momentum, or simply momentum, is the product of the mass and velocity of an object.

It is a vector quantity, possessing a magnitude and a direction.

The mathematical expression for momentum is:

p = m . v

where,

p is the linear momentum of the object.
m is the mass of the object.
v is the velocity of the object.

An object has a velocity (v) of 9 m/s and its linear momentum (p) is 72 kg.m/s. We will use the definition of linear momentum to calculate the mass of the object.

p = m . v

m = p / v

m = (72 kg.m/s) / (9 m/s) = 8 kg

An object with a velocity (v) of 9 m/s and a linear momentum (p) of 72 kg.m/s, has a mass (m) of 8 kg.

Learn more about linear momentum here: brainly.com/question/7538238

#SPJ1

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2 years ago

Find the ratio of the radii of a baseball to the Earth, knowing that the radius of a baseball is .09 m, and that the Earth's rad

Xelga [282]

0.09 / 6.37 x 10⁶ = 1.4129 x 10⁻⁸

The radius of the baseball is 1.4129 x 10⁻⁸ the radius of the Earth.

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3 years ago

The following questions present a twist on the scenario above to test your understanding. Suppose another stone is thrown horizo

Ipatiy [6.2K]

The first part of the text is missing, you can find on google:

"A ball is thrown horizontally from the roof of a building 45 m. If it strikes the ground 56 m away, find the following values."

Let's now solve the different parts.

(a) 3.03 s

The time of flight can be found by analyzing the vertical motion only. The vertical displacement at time t is given by

$y(t) = h -\frac{1}{2}gt^2$

where

h = 45 m is the initial height

g = 9.8 m/s^2 is the acceleration of gravity

When y=0, the ball reaches the ground, so the time taken for this to happen can be found by substituting y=0 and solving for the time:

$0=h-\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(45)}{9.8}}=3.03 s$

(b) 18.5 m/s

For this part, we need to analyze the horizontal motion only, which is a uniform motion at constant speed.

The horizontal position is given by

$x=v_x t$

where

$v_x$ is the horizontal speed, which is constant

t is the time

At t = 3.03 s (time of flight), we know that the horizontal position is x = 56 m. By substituting these numbers and solving for vx, we find the horizontal speed:

$v_x = \frac{x}{t}=\frac{56}{3.03}=18.5 m/s$

The ball was thrown horizontally: this means that its initial vertical speed was zero, so 18.5 m/s was also its initial overall speed.

(c) 35.0 m/s at 58.1 degrees below the horizontal

At the impact, we know that the horizontal speed is still the same:

$v_x = 18.5 m/s$

we need to find the vertical velocity. This can be done by using the equation

$v_y = u_y -gt$

where

$u_y =0$ is the initial vertical velocity

g is the acceleration of gravity

t is the time

Substituting t = 3.03 s, we find the vertical velocity at the time of impact:

$v_y = -(9.8)(3.03)=-29.7 m/s$

So the magnitude of the velocity at the impact (so, the speed at the impact) is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.5^2+(-29.7)^2}=35.0 m/s$

The angle instead can be found as:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-29.7}{18.5})=-58.1^{\circ}$

so, 58.1 degrees below the horizontal.

4 0

3 years ago