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3 years ago

77. The first law of motion applies to

1 answer:

8 0

In the first law, an object will not change its motion unless a force acts on it. So I think the answer is A. Only objects that are moving.

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A car with a mass of 1,324 kilograms, traveling at a speed of 20 meters/second, crashes into a wall and stops. What is the kinet

Valentin [98]

I think the correct answer from the choices listed above is option A. The kinetic energy after the perfectly inelastic collision would be zero Joules. <span>A </span>perfectly inelastic collision<span> occurs when the maximum amount of kinetic energy of a system is lost. Hope this answers the question.</span>

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4 years ago

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At what angle of projectile range of maximum

Agata [3.3K]

Answer:

$45°$

Explanation:

The textbooks say that the maximum range for projectile motion (with no air resistance) is 45 degrees.

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4 years ago

Una tortuga se desplaza en línea recta 100cm al norte y luego 80cm al este, determine su desplazamiento

valentinak56 [21]

Answer:

is this the full question?

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3 years ago

What medical technique uses sound wave pulses and their reflections to map structures and organs inside the body?

garri49 [273]

The correct answer is “C” ultrasound. Hope this helps!

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3 years ago

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A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a

makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

$\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]$ (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

$\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°$

The distance r is:

$r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m$

You replace the values of all parameters in the equation (1):

$\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}$

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0

3 years ago