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Ganezh [65]
3 years ago
13

True or false The sound of a police siren would be an example of the Doppler effect

Physics
1 answer:
Stella [2.4K]3 years ago
3 0
The answer to the question is True.
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Which group within the Senate has the most power over legislation?
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Work out
Simora [160]

the father has to sit 0.5meter away from the kid because he is a 3/4 heavier that the kid

6 0
3 years ago
Una niña está empujando un baúl. El PESO del baúl es de 230 N y el roce es de 50 N, la niña sólo logra ejercer una fuerza de 30
Gre4nikov [31]

Answer:

Su padre necesita aplicar una fuerza de 20 newtons en la misma dirección que la fuerza aplicada por su hija.

Explanation:

Asúmase que el baúl se mueve en una superficie horizontal. La fuerza de rozamiento dada por el problema es la fuerza de rozamiento estático máximo, se requiere una fuerza externa antiparalela a la fuerza de rozamiento estático máximo para que el baúl se empiece a mover. La ecuación de equilibrio de fuerzas horizontales sobre el baúl es:

\Sigma F = P - f = 0

Donde:

P - Fuerza externa aplicada sobre el baúl, medida en newtons.

f - Fuerza de rozamiento estático máximo, medida en newtons.

Se despeja la fuerza externa:

P = f

Si f = 50\,N, entonces:

P = 50\,N

Si la niña solo logra ejercer una fuerza de 30 newtons, su padre necesita aplicar una fuerza de 20 newtons paralela a aquella fuerza.

5 0
3 years ago
QUESTION 1
lina2011 [118]

Solution :

Magnetic field at the centre due to $I_P$ :

$B_1 = \frac{\mu_0 I_P}{2 \pi d}$

$B_1 = \frac{4 \pi \times 10^{-7} \times 0.2}{2 \pi d}$

$B_1 =10^{-7} \ T$

Its direction will be downwards in the plane of the paper.

Magnetic field at the centre due to $I_Q$ :

$B_2 = \frac{\mu_0 I_Q}{2 \pi d}$

$B_2 = \frac{4 \pi \times 10^{-7} \times 0.6}{2 \pi \times 0.8}$

$B_2 =1.5 \times 10^{-7} \ T$

Its direction will be upwards in the plane of the paper

Magnetic field due to the coil:

$B_3 = \frac{\mu_0 I_{coil}}{2 r}$

$B_3 = \frac{4 \pi \times 10^{-7} \times 0.5}{2 \times 0.2}$

$B_3 =3.14 \times 10^{-7} \ T$

Its direction will be rightwards.

Now the resultant of the magnetic field at the centre.

$B_{net} = \sqrt{(B_2 -B_1)^2+B^2_3}$

$B_{net} = \sqrt{(0.5 \times 10^{-7})^2+(3.14 \times 10^{-7})^2}$

$B_{net} = 3.18 \times 10^{-7} \ T$

Now the direction ,

$\tan \theta = \frac{B_2-B_1}{B_3}$

$\tan \theta = \frac{0.5 \times 10^{-7}}{3.14 \times 10^{-7}}$

Therefore $\theta = 9^\circ$ (from right to upward)

3 0
3 years ago
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