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SashulF [63]
2 years ago
6

Can you help me with this physics homework? please

Physics
1 answer:
KengaRu [80]2 years ago
3 0

Answer:

Area under a curve can be estimated via block counting, calculated with simple geometry if the velocity has constant accelerations, or calculated by integration if the velocity can be described by a mathematical equations.

The area under a velocity time curve will be the product of velocity and time which equals displacement.

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A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) How much time elapses before the b
daser333 [38]

Answer:

<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b) The bullet travels horizontally 110.6 m</h2>

Explanation:

a)  Consider the vertical motion of bullet

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Displacement, s = 1.5 m      

     Substituting

                      s = ut + 0.5 at²

                      1.5 = 0 x t + 0.5 x 9.81 xt²

                      t = 0.553 s

      Time elapsed before the bullet hits the ground is 0.553 seconds.

b) Consider the horizontal motion of bullet

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 200 m/s

        Acceleration, a = 0 m/s²  

        Time, t = 0.553 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 200 x 0.553 + 0.5 x 0 x 0.553²

                      s = 110.6 m

      The bullet travels horizontally 110.6 m

6 0
3 years ago
Two astronauts are taking a spacewalk outside the International Space
Tems11 [23]

Answer:

im not sure

Explanation:

5 0
3 years ago
A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it w
Bumek [7]

Answer:

Explanation:

Given that,

Mass attached m = 0.95kg

Spring constant k = 16N/m

Instantaneous speed v = 36cm/s = 0.36m/s

Amplitude A=?

When x = 0.7A

Using conservation of energy

∆K.E + ∆P.E = 0

K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0

At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.

Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0

So, the equation becomes

— K.E(initial) + P.E(final) = 0

K.E(initial) = P.E(final)

½mv² = ½kA²

mv² = kA²

0.95 × 0.36² = 16×A²

0.12312 = 16•A²

A² = 0.12312/16

A² = 0.007695

A = √0.007695

A = 0.088 m

A = 8.8cm

B. Speed at x = 0.7A

Using the same principle above

K.E(initial) = P.E(final)

½mv² = ½kA²

Where A = 0.7A = 0.7 × 0.088 = 0.0614m

Then,

½× 0.95 × v² = ½ × 16 × 0.0614²

0.475v² = 0.0310644

v² = 0.0310644/0.475

v² = 0.0635

v = √0.0635

v = 0.252 m/s

v = 25.2 cm/s

8 0
3 years ago
4 facts about relative dating?
sasho [114]
Relative dating is used to arrange geological events….

Relative dating puts geologic events in chronological order without requiring that a specific numerical age be assigned to each event….

Relative Dating uses the half life of isotopes to get the exact age of a rock or mineral.
6 0
3 years ago
A gold sphere of radius R=100 μm and density 19g/cm^3 falls through water. Given the viscosity of water is about 10^-3​ Pa s and
icang [17]

The terminal velocity of gold sphere is 39.2 cm/s

<h3>What is terminal velocity?</h3>

Terminal velocity is the maximum velocity attainable for an object as it falls through a fluid.

<h3>How to calculate the terminal velocity of the gold sphere?</h3>

The terminal velocity of the gold sphere is given by v = 2gr²(ρ - σ)/9η where

  • g = acceleration due to gravity = 9.8 m/s²,
  • r = radius of sphere = 100 μm = 100 × 10⁻⁶ m = 10⁻⁴ m = 10⁻² cm,
  • ρ = density of sphere = 19 g/cm³,
  • σ = density of water = 1.0 g/cm³ and
  • η = viscosity of water = 10⁻³ Pa-s

So, susbtituting the values of the variables into the equation, we have that

v = 2gr²(ρ - σ)/9η

v = 2 × 9.8m/s²× (10⁻² cm)²(19 g/cm³ - 1.0 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 9.8 m/s² × 10⁻⁴ cm² × (18 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 980 cm/s² × 10⁻⁴ cm² × 2 g/cm³/(1 × 10⁻³ Pa-s)

v = 3920 g/s² × 10⁻⁴/(1 × 10⁻³ Pa-s)

v = 392 cm/s × 10³ × 10⁻⁴

v = 392 × 10⁻¹ cm/s

v = 39.2 cm/s

So, the terminal velocity is 39.2 cm/s

Learn more about terminal velocity of sphere here:

brainly.com/question/21684177

#SPJ1

4 0
1 year ago
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