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3 years ago

I need to know the answer to the question

Physics

1 answer:

galben [10]3 years ago

4 0

I would go with A. An Observation. Because he's looking at visual differences, for example: "based on smell, color change, and release of bubbles.

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Which one do I press guys?

taurus [48]

Answer:

The Nucleus...

Explanation:

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3 years ago

A=i+j-k<br>b=j+k<br><img src="https://tex.z-dn.net/?f=a%20%3D%20i%20%2B%20j%20-%20k%20%20%20%5C%5C%20b%20%3D%20j%20%2B%20k" id="

olga nikolaevna [1]

Answer:

Explanation:

.--------------------.

4 0

3 years ago

A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of

Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

$h=(u\sin60)\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2$

$h=41.6\ m$

(b). We need to calculate the maximum height of the ball

Using formula of height

$h_{max}=\dfrac{(u\sin\theta)^2}{2g}$

Put the value into the formula

$h=\dfrac{(33\sin60)^2}{2\times 9.8}$

$h=41.67\ m$

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

$v=u-gt$

$v=u\sin\theta-gt$

Put the value into the formula

$v_{y}=33\times\sin 60-9.8\times3.0$

$v_{y}=-0.82\ m/s$

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=33\times\cos60$

$v_{x}=16.5\ m/s$

We need to calculate the ball's impact speed

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(16.5)^2+(-0.82)^2}$

$v=16.52\ m/s$

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

3 0

3 years ago

¿Cuál es la magnitud y dirección de la aceleración de los cuerpos en caída libre en el planeta Tierra?

Tatiana [17]

La magnitud es de 9.8 m/s² ... la aceleración de la gravedad en o cerca de la superficie de la Tierra.

La dirección es hacia el centro de la Tierra. (Llamamos a esa dirección "abajo").

6 0

3 years ago

A proton moving at 3.90 106 m/s through a magnetic field of magnitude 1.80 T experiences a magnetic force of magnitude 7.20 10-1

VARVARA [1.3K]

Answer:

$\theta=40^0$

Explanation:

The magnitude of the magnetic force is

$F_m=evB\sin\theta$

To find the angle, we make $\sin\theta$ subject of the formula

$\implies \sin\theta=\frac{F_m}{evB}=\frac{7.20\times 10^{-13}}{1.6\times 10^{-19}\times 3.90\times 10^6\times 1.80}$

$\implies \sin\theta=0.641025641$

$\therefore \theta=\sin^{-1}=39.8683^0\\\implies \theta\approxeq 40^0$

8 0

3 years ago