All categories

2 years ago

An arrow movirg 48.3 m/s has 5.22 kg•m/s of momentum. What is its mass?

Physics

1 answer:

spin [16.1K]2 years ago

6 0

Answer:

0.11 kg

Explanation:

Ft = MV

Ft = momentum 5.22kg m/s

M = mass

V = velocity 48.3m/s

Therefore

5.22 = M x 48.3

Divide both sides by 48.3

5.22/48.3 = M x 48.3/48.3

0.11 = M

M = 0.11kg

You might be interested in

In which of these scenarios would the sound of a drumstick hitting a metal bar get to you in the least amount of time?explain ur

MA_775_DIABLO [31]

Answer:

Scenario 1 is the correct answer.

Explanation:

The sound of the drumstick hitting the metal bar will get to me in a shorter amount of time in Scenario 1 . The sound wave will travel faster in the metal bar than through the air because the speed of sound waves in solid is faster than it is in gases.

6 0

2 years ago

Read 2 more answers

A thin rod of length 0.75 m and mass 0.42 kg is suspended freely from one end. It is pulled to one side and then allowed to swin

Goshia [24]

Answer:

(A) 0.63 J

(B) 0.15 m

Explanation:

length (L) = 0.75 m

mass (m) =0.42 kg

angular speed (ω) = 4 rad/s

To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)

I = Ic + m $h^{2}$

Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis

h is the horizontal distance between the center of mass and the rotational axis of the rod

I = $(\frac{1}{12})(mL^{2} ) + m([tex]\frac{L}{2}$ )^{2}[/tex]

I = $(\frac{1}{12})(0.42 x 0.75^{2} ) + ( 0.42 x ([tex]\frac{0.75}{2}$ )^{2}[/tex])

I = 0.07875 kg.m^{2}

(A) rods kinetic energy = 0.5I $ω^{2}$

= 0.5 x 0.07875 x $4^{2}$ = 0.63 J 0.15 m

(B) from the conservation of energy

initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

Ki + Ui = Kf + Uf

at the maximum height velocity = 0 therefore final kinetic energy = 0

Ki + Ui = Uf

Ki = Uf - Ui

Ki = mg(H-h)

where (H-h) = rise in the center of mass

0.63 = 0.42 x 9.8 x (H-h)

(H-h) = 0.15 m

6 0

2 years ago

On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he

Liula [17]

Answer:

correct is d) a ’= g / 2

Explanation:

For this exercise let's use the kinematics equations

On earth

v = v₀ - a t

a = (v₀- v) / T

On planet X

v = v₀ - a' t’

a ’= (v₀-v) / 2T

Let's substitute the land values in plot X

a’= a / 2

Now let's use Newton's second law

W = ma

m g = m a

a = g

We substitute

a ’= g / 2

So we see that on planet X the acceleration is half the acceleration of Earth's gravity

4 0

2 years ago

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

lianna [129]

<h2>Answer:</h2>

<em><u>Velocity of throwing arrow = 43.13 m/s.</u></em>

<h2>Explanation:</h2>

In the question,

Let us say the height from which the arrow was shot = h

Distance traveled by the arrow in horizontal = 61 m

Angle made by the arrow with the ground = 2°

So,

From the <u>equations of the motion</u>,

$61 =u.t\\t=\frac{61}{u}$

Now,

Also,

Finally, the angle made is 2 degrees with the horizontal.

So,

Final horizontal velocity = v.cos20°

Final vertical velocity = v.sin20°

Now,

u = v.cos20° (No acceleration in horizontal)

Also,

$v=u+at\\vsin20=0+9.8(t)\\t=\frac{v.sin20}{9.8}$

So,

We can say that,

$\frac{v.sin20}{9.8}=\frac{61}{v.cos20}\\v^{2}.sin20.cos20=597.8\\v^{2}=1860.56\\v=43.13\,m/s$

<em><u>Therefore, the velocity with which the arrow was shot by the archer is 43.13 m/s.</u></em>

5 0

2 years ago

At what point is the KINETIC ENERGY at its highest?

igomit [66]

Answer:

An object has the MOST kinetic energy when it's movement is the GREATEST.

Explanation:

8 0

2 years ago