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3 years ago

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2. An object is moving with an initial velocity of 12 m/s. It accelerates at a rate of 1.5 m/s^2 over a distance of 40 m. What i

s its new velocity?

1 answer:

nikklg [1K]3 years ago

4 0

Answer:

16.2 m/s

Explanation:

2ad=Vf^2-Vi^2

2 (1.5) (40) = Vf^2 -(12)^2

Vf= 16.2 m/s

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A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap

viktelen [127]

Answer:

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

$V_r + V_L + V_c = V_{net}$

now we will have

$iR + L\frac{di}{dt} + \frac{q}{C} = 12 V$

now we have

$1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12$

So we will have

$q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}$

at t = 0 we have

q = 0

$0 = 3\times 10^{-6} + c_1 + c_2$

also we know that

at t = 0 i = 0

$0 = -4000 c_1 - 1000c_2$

$c_2 = -4c_1$

$c_1 = 1 \times 10^{-6}$

$c_2 = -4 \times 10^{-6}$

so we have

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

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3 years ago

Two identical 9.10-g metal spheres (small enough to be treated as particles) are hung from separate 300-mm strings attached to t

Musya8 [376]

Answer:

$n = 1.266\times 10^{12}$

Explanation:

Given data:

mass of sphere is 10 g

Angle between string and vertical axis is $\theta = 13 degree$

thickness of string 300 mm = 0.3 m

$sin\theta =\frac{2}{0.3 m}$

r =0.3 sin 13 = 0.067 m

$Fe = \frac{ kq_1 q-2}{d^2}$

$Fe = \frac{kq^2}{(2r)^2} = mg tan\theta$

$q^2 = mg tan\theta \frac{(2r)^2}{k}$

$= 0.0091 \times 9.8 tan13 \times \frac{(2\times 0.067)^2}{9\times 10^9}$

$q^2 = 4.10\times 10^{-14}$

$q = 2.026 \times 10^{-7} C$

q = ne

$n = \frac{1.6\times 10^{-19}}{2.02\times 10^{-7}}$

$n = 1.266\times 10^{12}$

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3 years ago

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4 years ago

A baseball of mass 300-g is hit at a velocity of 40 m/s. If the ball is caught by the third baseman and the ball penetrates 2.0

OlgaM077 [116]

The force is -12,000 N

Explanation:

First of all, we calculate the acceleration of the ball, by using the following suvat equation:

$v^2-u^2=2as$

where:

v = 0 is the final velocity of the baseball (it comes to rest)

u = 40 m/s is the initial velocity

a is the acceleration

s = 2.0 cm = 0.02 m is the displacement of the ball

Solving for a,

$a=\frac{v^2-u^2}{2s}=\frac{0-40^2}{2(0.02)}=-40,000 m/s^2$

Now we can calculate the average force exerted on the ball, by using Newton's second law:

$F=ma$

where

m = 300 g = 0.3 kg is the mass of the ball

$a=-40,000 m/s^2$ is the acceleration

Substituting,

$F=(0.3)(-40,000)=-12,000 N$

where the negative sign indicates that the direction of the force is opposite to the direction of motion of the ball.

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NARA [144]

Answer:

Humus

Explanation:

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3 years ago