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2 years ago

Please Help! A ball is thrown straight up from the ground. What way does its acceleration point at the top?

Physics

1 answer:

QveST [7]2 years ago

3 0

Answer: A. Vertical

Explanation:

If the ball is thrown straight up from the ground (asuming the ground as height 0), this means its initial velocity is greater than zero.

While the ball rises and gains height, its velocity decreases until it reaches its maximum height where it stops (velocity equal to zero) and then it begins to fall to the ground.

Now, <u>during all the movement, the ball has an acceleration</u>, which is the acceleration due gravity (9.8 m/s^{2} on Earth), even at the top or maximum height (when the ball stops just for fraction of time), and this acceleration points vertical and downward to the Earth's center.

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What distance does electromagnetic radiation travel in 55.0 μs ?

Komok [63]

Electromagnetic radiation is an energy that is known as light. so electromagnetic radiation will have the same speed as the speed of light which is 3 x 10^8 m/s. so the distance it travel at 55 x 10^-6 s is:
D = ( 3 x 10^8 m/s ) ( 55 x 10^-6 s )
D = 16500 m

4 0

3 years ago

Calculate the molecular weight of Aluminium hydroxide

Vedmedyk [2.9K]

Al(OH)3 = 26.98 + [(16×3) + (1.01×3)] = 26.98 + 51.03 = 78.01 and the unit will be g/mol

5 0

3 years ago

In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1

asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, $f=5\times 10^{14}\ Hz$

Separation between slits, $d=2.2\times 10^{-5}\ m$

(a) The condition for maxima is given by :

$d\ sin\theta=n\lambda$

For third maxima,

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{nc}{fd})$

$\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})$

$\theta=4.69^{\circ}$

(b) For second dark fringe, n = 2

$d\ sin\theta=(n+1/2)\lambda$

$\theta=sin^{-1}(\dfrac{5\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{5c}{2df})$

$\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})$

$\theta=3.90^{\circ}$

Hence, this is the required solution.

8 0

3 years ago

A gymnast is swinging on a high bar. The distance between his waist and the bar is 0.905 m, as the drawing shows. At the top of

Citrus2011 [14]

Answer:

5.959 m/s

Explanation:

m = Mass of gymnast

u = Initial velocity

v = Final velocity

$h_i$ = Initial height

$h_f$ = Final height

From conservation of Energy

$\frac{1}{2}mv^2+mgh_f=\frac{1}{2}mu^2+mgh_i\\\Rightarrow\frac{1}{2}mv^2+mg0=\frac{1}{2}m0^2+mgh_i\\\Rightarrow \frac{1}{2}mv^2=mgh_i\\\Rightarrow v=\sqrt{2gh_i}$

$h_i=2r$

$v=\sqrt{4gr}\\\Rightarrow v=\sqrt{4\times 9.81\times 0.905}\\\Rightarrow v=5.959\ m/s$

Velocity of gymnast at bottom of swing is 5.959 m/s

5 0

3 years ago

Write the realtionship between the density of a liquid and its upthrust? clarify

kipiarov [429]

Answer:

B = ρ g V_liquid

the thrust is proportional to the density of the liquid

Explanation:

The density of a liquid is defined as the relationship between the mass and the volume of the liquid

ρ = m / V

The upward push of the liquid is given by the principle of Archimedes Archimedes establishes that the push is equal to the weight of the dislodged liquid

B = W_liquid

B = m _liquid g

we substitute mass for density

B = ρ g V_liquid

therefore we see that the thrust is proportional to the density of the liquid

8 0

2 years ago