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2 years ago

A girl on a bridge throws a rock

Physics

1 answer:

Anon25 [30]2 years ago

6 0

t = 1.3 s.

To solve this exercise we have to use the freefall equation t = (v-v₀)/g. v = 0 due the final velocity when the rock reach the ground is 0.

With v₀ = -12.5m/s and g = 9.8m/s²

t = -(-12.5m/s)/9.8m/s²

t = 1.3 s

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What is the mathematical relationship between voltage, resistance, and current?

Alex73 [517]

The formula for Voltage is V = IR or in words
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2 years ago

What is the kinetic energy of a 0.5 kg puppy that is running 1.5m/s

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2 years ago

A rectangular loop with dimensions 4.20 cm by 9.50 cm carries current I. The current in the loop produces a magnetic field at th

tiny-mole [99]

Answer:

I=1.48 A

Explanation:

Given that

B=3.1 x 10⁻5 T

b= 4.2 cm

l= 9.5 cm

The relationship for magnetic field and current given as

$B=\dfrac{2\mu _oI}{\pi}D$

Where

$D=\dfrac{\sqrt{l^2+b^2}}{lb}$

By putting the values

$D=\dfrac{\sqrt{l^2+b^2}}{lb}$

$D=\dfrac{\sqrt{0.042^2+0.095^2}}{0.042\times 0.095}$

D=26.03 m⁻¹

$B=\dfrac{2\mu _oI}{\pi}D$

$3.1\times 10^{-5}=\dfrac{2\times 4\times \pi \times 10^{-7} I}{\pi}\times 26.03$

$I=\dfrac{3.1\times 10^{-5}}{{2\times 4\times 10^{-7} }\times 26.03}$

I=1.48 A

8 0

2 years ago

Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and fi

steposvetlana [31]

Answer:

$W=K_f-K_i$

Explanation:

The work done on a particle by external forces is defined as:

$W=\int\limits^{r_f}_{r_i} {F\cdot dr} \,$

According to Newton's second law $F=ma$ . Thus:

$W=\int\limits^{r_f}_{r_i}{ma\cdot dr} \,\\$

Acceleration is defined as the derivative of the speed with respect to time:

$W=m\int\limits^{r_f}_{r_i}{\frac{dv}{dt}\cdot dr} \,\\\\W=m\int\limits^{r_f}_{r_i}{dv \cdot \frac{dr}{dt}} \,$

Speed is defined as the derivative of the position with respect to time:

$W=m\int\limits^{v_f}_{v_i} v \cdot dv \,$

Kinetic energy is defined as $K=\frac{mv^2}{2}$ :

$W=m\frac{v_f^2}{2}-m\frac{v_i^2}{2}\\W=K_f-K_i$

3 0

2 years ago

A wire, 1.0 m long, with a mass of 90 g, is under tension. A transverse wave is propagated on the wire, for which the frequency

Mice21 [21]

Answer:

T = 712.9 N

Explanation:

First, we will find the speed of the wave:

v = fλ

where,

v = speed of the wave = ?

f = frequency = 890 Hz

λ = wavelength = 0.1 m

Therefore,

v = (890 Hz)(0.1 m)

v = 89 m/s

Now, we will find the linear mass density of the wire:

$\mu = \frac{m}{L}$

where,

μ = linear mass density of wie = ?

m = mass of wire = 90 g = 0.09 kg

L = length of wire = 1 m

Therefore,

$\mu = \frac{0.09\ kg}{1\ m}$

μ = 0.09 kg/m

Now, the tension in wire (T) will be:

T = μv² = (0.09 kg/m)(89 m/s)²

7 0

2 years ago