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shutvik [7]
2 years ago
8

The force shown in the attached figure is the net eastward force acting on a ball. The force starts rising at t = 0.012 s, falls

back to zero at t = 0.062 s, and reaches a maximum force of 35 N at the peak. Determine with an error no bigger than 25% (high or low) the magnitude of the impulse (in N-s) delivered to the ball. Hint: Do not use J = FΔt. Look at the figure. Find the area of a nearly equally sized triangle.

Physics
1 answer:
kifflom [539]2 years ago
6 0
Impulse = Integral of F(t) dt from 0.012s to 0.062 s

Given that you do not know the function F(t) you have to make an approximation.

The integral is the area under the curve.

The problem suggest you to approximate the area to a triangle.

In this triangle the base is the time: 0.062 s - 0.012 s = 0.050 s

The height is the peak force: 35 N.

Then, the area is [1/2] (0.05s) (35N) = 0.875 N*s

Answer> 0.875 N*s
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A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe
notsponge [240]

Answer:

Q = \frac{0.068}{E}

where E = electric field intensity

Explanation:

As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical

So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field

while weight of the ball is vertically downwards

so here we have

QE = F_x

mg = F_y

since string makes 30 degree angle with the vertical so we will have

tan\theta = \frac{F_x}{F_y}

tan30 = \frac{QE}{mg}

Q = \frac{mg}{E}tan30

Q = \frac{0.012\times 9.81}{E} tan30

Q = \frac{0.068}{E}

where E = electric field intensity

5 0
2 years ago
Q1: A runner is jogging in a straight line at a steady vr= 6.8 km/hr. When the runner is L= 2.4 km from the finish line, a bird
serious [3.7K]

Answer:

Q1: 3.2km

Q2: 4.8K

Explanation:

Q1:

So db is the distance of bird, and dr is the distance of runner

db = 2vr  and the distance of bird is going to be 2 times greater than the runner.

formulas: db = 2vr & db = 2dr

  1. db = 2dr
  2. L + (L - x) = 2x
  3. 2L - x = 2x
  4. 2L = 3x
  5. x = \frac{2}{3}L

Insert it in x = \frac{2}{3}L

\frac{2}{3}(2.4km) = 1.6km

Now we use formula db = 2dr

  1. db = 2L - x
  2. db = 2(2.4km) - 1.6km
  3. <u>db = 3.2km</u>

Q2:

Formulas: Vr = L /Δt & Vb = db/Δt

  1. Vr = L/ Δt ⇒ Δt = \frac{L}{Vr}
  2. \frac{2.4km}{6.8km/hr}
  3. \frac{6}{17}hr

(Km cancel each other)

  1. Vb = db/Δt ⇒ db = VbΔt
  2. 13.6km/hr(\frac{6}{17}hr )
  3. <u>4.8km</u>

(hr cancel each other)

Hope it helps you :)

6 0
2 years ago
Which of the following equations illustrates the law of conservation of matter?
AlladinOne [14]

Answer: The correct answer is option (B).

Explanation:

Law of conservation of mass: 'In a chemical reaction, mass neither be created nor be destroyed'

In the balance chemical equation,total mass of the reactants is equal to the total mass of the products.

A. 2Na+Cl_2\rightarrow  NaCl

[2\times (23 g/mol)+(35.5 g/mol)\times 2]=[23 g/mol+35.5 g/mol]

117 g/mol  ≠ 58.5 g/mol

Law of Conservation of Mass not followed.

B.2Na+Cl_2\rightarrow 2NaCl

[2\times (23 g/mol)+(35.5 g/mol)\times 2]=2\times [23 g/mol+35.5 g/mol]

117 g/mol  =  117 g/mol

Law of Conservation of Mass is followed.

C.2Na+2Cl_2\rightarrow 2NaCl

[2\times (23 g/mol)+(35.5 g/mol)\times 2]=2\times [23 g/mol+35.5 g/mol]

188 g/mol  ≠ 117 g/mol

Law of Conservation of Mass not followed.

D.Na+Cl_2\rightarrow 2NaCl

[(23 g/mol)+(35.5 g/mol)\times 2]=2\times [23 g/mol+35.5 g/mol]

94 g/mol  ≠ 117 g/mol

Law of Conservation of Mass not followed.

Hence, the correct answer is option (B).

5 0
2 years ago
A ball bearing of radius of 1.5 mm made of iron of density
Serjik [45]

Answer:

\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

Acceleration due to gravity (g) = 980.6 cm/s²

To Find:

Viscosity of glycerine (\sf \eta)

Explanation:

\boxed{ \bold{v =  \frac{2}{9}  \frac{( {r}^{2} ( \rho -  \sigma)g)}{ \eta} }}

\sf \implies \eta =  \frac{2}{9}  \frac{( {r}^{2}( \rho -  \sigma)g )}{v}

Substituting values of r, ρ, σ, v & g in the equation:

\sf \implies \eta =  \frac{2}{9}  \frac{( {(0.15)}^{2}  \times  (7.85 - 1.25) \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \times  \frac{145.6191}{2.25}

\sf \implies \eta =  \frac{2}{9}  \times 64.7196

\sf \implies \eta =  2 \times 7.191

\sf \implies \eta =  14.382 \: poise

6 0
2 years ago
explain why the impact of one heavy stone would produce waves with higher amplitude than the impact of the light stone would
DerKrebs [107]
The heavy stone would produce waves with a higher amplitude, rather than the smaller stone, because since the stone is heaver its going to have a grater impact and displace more water to create a bigger wave.
5 0
2 years ago
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