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Dvinal [7]
1 year ago
6

All forces on the bullets cancel so that the net force on a bullet is zero, which means the bullet has zero acceleration and is

in a state known as _____.
Physics
1 answer:
andre [41]1 year ago
7 0
The acceleration is defined as the rate of change of velocity.
So, if the acceleration is zero, this means that the rate of change of velocity is zero, which also means that the body is moving with constant velocity.
Since we are given that the net forces acting on the body is zero, this means that the body is at equilibrium

Based on this:
<span>All forces on the bullets cancel so that the net force on a bullet is zero, which means the bullet has zero acceleration and is in a state known as equilibrium.

Note that if this constant velocity is equal to zero, then the body would be at rest (not moving)</span>
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A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'
lbvjy [14]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3  

                                              = 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

Y =\frac{mgL}{A\Delta L}

\Delta L =\frac{mgL}{A\times Y}

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

5 0
1 year ago
Suppose Galileo dropped a lead ball (100 kilograms) and a glass ball (1 kilogram) from the Leaning Tower of Pisa. Which one hit
Reika [66]

The Answer is C Both at the same time

3 0
1 year ago
A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s
Leona [35]

Answer:

0.6

Explanation:

The volume of a sphere = \frac{4}{3} \pi (\frac{D}{2})^3

Therefore \pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3

r of the disc = 1.15(\frac{ D}{2} )

Using conservation of angular momentum;

The M_i of the sphere = \frac{2}{5} m \frac{D}{2}^2

M_i of the disc = m*\frac{   \frac{1.15*D}{2}^2 }{2}

\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{  m * \frac{(\frac{`.`5*D}{2})^2 }{2} }

= 0.6

5 0
2 years ago
3. A 4.1 x 10-15 C charge is able to pick up a bit of paper when it is initially 1.0 cm above the paper. Assume an induced charg
Anni [7]

Answer:

\mathbf{1.51\times10^{-15}N}

Explanation:

The computation of the weight of the paper in newtons is shown below:

On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.

Therefore the paper charge is

q_{paper}=-4.1\times10^{-15}C

Now the distance from the charge is

r=1cm=0.01m

Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.

mg=\frac{k_{e}q_{1}q_{2}}{r^{2}}

\Rightarrow W=mg

=\frac{9\times10^{9}\times(4.1\times10^{-15})^{2}}{0.01^{2}}

=\mathbf{1.51\times10^{-15}N}

6 0
1 year ago
1. Each year at a college, there is a tradition of having a hoop rolling competition. Alex rolls his 0.350 kg hoop down the cour
grigory [225]

Question 1:

Answer:

The moment of inertia of Alex's rolling hoop is 0.197 kg \cdot cm^2

Explanation:

<u>Given</u>:

Mass of the hoop = 0.350 g

Radius of the hoop = 75.0 cm

<u>To Find:</u>

The moment of inertia of Alex's rolling hoop = ?

<u>Solution</u><u>:</u>

The moment of inertia  = mr^2

where

m is the mass

r is the radius

Converting cm to m, we get

75.0 cm = 0.75 m

Now substituting the values,

=> moment of inertia  = (0.350)(0.75)^2

=> moment of inertia  = (0.350)(0.5625)

=> moment of inertia  = (0.197)

Question 2:

Answer:

The combined angular momentum of the masses is 1.76 kg m^2 s^{-1}

If she pulls her arms in to 0.12 m, her new linear speed  is  18.33 m/s^2

Explanation:

Given:

Mass  = 2.0 kg

Radius = 0.8 m

Velocity =  1.2 m/s

a.The combined angular momentum of the masses:

L = r \cdot m \cdot v_1

Substituting the values,

L = 0.8 \cdot 2.0 \cdot 1.1

L= 1.76 kg m^2 s^{-1}

b. If she pulls her arms in to 0.12 m, what is her new linear speed

0.12 \cdot 0.8 \cdot v_2 = 1.76

0.096 cdot v_2 = 1.76

v_2 = \frac{1.76}{0.096}

v_2 = 18.33 m/s^2

6 0
1 year ago
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