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3 years ago

Calculate the force needed to accelerate a baseball with a mass of 0.45

Physics

1 answer:

RideAnS [48]3 years ago

8 0

Mass=0.45kg
Acceleration=175m/s^2

$\\ \bull\tt\dashrightarrow F=ma$

$\\ \bull\tt\dashrightarrow F=0.45(175)$

$\\ \bull\tt\dashrightarrow F=78.75N$

$\\ \bull\tt\dashrightarrow F=78.8N$

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If it takes 50n of force to lift a 450n what is the ma of the machine

otez555 [7]

If the machine is 100% efficient, then its
Mechanical Advantage is (450/50) = 9 .

If the machine is less than 100% efficient,
then the MA is more than 9 .

7 0

4 years ago

If two micro coulomb of charge is flowing in a circuit for 5 minutes. What is the amount of current in the circuit?

Murrr4er [49]

Answer:

6.67×10¯⁹ A

Explanation:

From the question given above, the following data were obtained:

Quantity of electricity (Q) = 2 μC

Time (t) = 5 mins

Current (I) =?

Next, we shall convert 2 μC to C. This can be obtained as follow:

1 μC = 1×10¯⁶ C

Therefore,

2 μC = 2 μC × 1×10¯⁶ C / 1 μC

2 μC = 2×10¯⁶ C

Next, we shall convert 5 mins to seconds. This can be obtained as follow:

1 min = 60 secs

Therefore,

5 min = 5 min × 60 sec / 1 min

5 mins = 300 s

Finally, we shall determine the current in the circuit. This can be obtained as follow:

Quantity of electricity (Q) = 2×10¯⁶ C

Time (t) = 300 s

Current (I) =?

Q = It

2×10¯⁶ = I × 300

Divide both side by 300

I = 2×10¯⁶ / 300

I = 6.67×10¯⁹ A

Thus, the current in the circuit is 6.67×10¯⁹ A

4 0

3 years ago

slader A transcontinental flight of 4890 km is scheduled to take 40 min longer westward than eastward. The airspeed of the airpl

11111nata11111 [884]

Answer:

$v_{s}=65.2km/h$

Explanation:

Given data

Flight distance S=4890 km

Time difference Δt=t₂-t₁=40 min

Air speed of plane=980 km/h

To find

Speed of jet stream

Solution

When moving in the same direction as the jet stream time taken as t₁=d/(v+vs),v is velocity of plane and vs is velocity of plane

While moving in opposite direction t₂=d/(v+vs)

$t_{2}-t_{1}=\frac{d}{(v-v_{s}) } - \frac{d}{(v+v_{s}) }\\t_{2}-t_{1}=\frac{d(v+v_{s})-d(v-v_{s})}{(v-v_{s})(v+v_{s})} \\t_{2}-t_{1}=\frac{2dv_{s}}{(v)^{2} -(v_{s})^{2} }\\0.666667h=\frac{2(4890km)v_{s}}{(980km/h)^{2} -(v_{s})^{2} }\\0.666667((980km/h)^{2} -(v_{s})^{2})=9780v_{s}\\640267-0.666667(v_{s})^{2}-9780v_{s}=0\\0.666667(v_{s})^{2}+9780v_{s}-640267=0$

Apply quadratic formula to solve for vs

$v_{s}=65.2km/h$

4 0

3 years ago

Which example describes constant acceleration due ONLY to a change in direction?

Brut [27]

'Traveling around a circular track' can be a description of constant
acceleration due only to changes in direction. But if it is, then the
progress around the circular track must be at constant speed.

5 0

4 years ago

Read 2 more answers

The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cyl

Ulleksa [173]

Hi there!

We can begin by using the equation for energy density.

$U = \frac{1}{2}\epsilon_0 E^2$

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
$\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}$

Creating a Gaussian surface being the lateral surface area of a cylinder:
$A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}$

Now, we can calculate the energy density using the equation:
$U = \frac{1}{2} \epsilon_0 E^2$

Plug in the expression for the electric field and solve.

$U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}$

Now, we can integrate over the volume with respect to the radius.

Recall:
$V = \pi r^2L \\\\dV = 2\pi rLdr$

Now, we can take the integral of the above expression. Let:
$r_i$ = inner cylinder radius

$r_o$ = outer cylindrical shell inner radius

Total energy-field energy:

$U = \int\limits^{r_o}_{r_i} {U_D} \, dV = \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr$

Plug in the equation for the electric field energy density and solve.

$U = \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\$

Bring constants in front and integrate. Recall the following integration rule:
$\int {\frac{1}{x}} \, dx = ln(x) + C$

Now, we can solve!

$U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})$

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

$\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}$

And here's our equation!

3 0

2 years ago