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Liula [17]
3 years ago
11

Any unit of acceleration MUST have the form of _____.

Physics
1 answer:
KengaRu [80]3 years ago
7 0

Any unit of acceleration must have the dimensions (form) of

(a unit of length) / (a unit of time)²

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______ resources include biodiversity, energy, land and<br> soil, and water resources.
cricket20 [7]

Answer : <u>Natural</u>

Explanation : Natural resources are the resources that are found in the environment and are developed without the intervention of humans

5 0
1 year ago
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A 762 kg car experiences a braking force of 9045 N and skids to a stop in 4.3 seconds. What is the speed of the car just before
PilotLPTM [1.2K]

Answer:

The solution for this problem is:

We will be using the formula for force which is F = ma 

=>10,000 = 2000 * a 

but we need to solve for acceleration so divide both sides by 2000, we will get:

=>a = 5 m/s^2 

Let the initial velocity was u m/s 

=>By v = u - at 

=>0 = u - 5 x 6 

Since acceleration is constant the velocity can be computed by multiplying the acceleration by 6 seconds. 

=>u = 30 m/s

Explanation:

6 0
3 years ago
After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r
mina [271]

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

2 n t = (m + \dfrac{1}{2})\lambda

At m = 0 for the minimum thickness, so

2\times 1.45 \times t = (0 + 0.5)\times 600

2.9 \times t =300

t =\dfrac{300}{2.9}

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

6 0
3 years ago
A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball
Akimi4 [234]

Answer:

Approximately 122.625\; {\rm m} (assuming that g = 9.81\; {\rm m\cdot s^{-2}}, the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let v_{0} denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly (-v_{0}). The velocity of the ball would be changed from v to (-v_{0})\! (such that \Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})) within t = 10\; {\rm s}.

Also because the drag on the ball is negligible, the acceleration of the ball would be a = -g = -9.81\; {\rm m\cdot s^{-2}}. Thus:

\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}.

Since \Delta v = (-2\, v_{0}):

-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}.

\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}.

The ball reaches maximum height when its velocity is v_{1} = 0\; {\rm m\cdot s^{-1}}. Apply the SUVAT equation x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a) to find the displacement x between the original position (ground level, where v_{0} = 49.05\; {\rm m\cdot s^{-1}}) and the max-height position of the ball (where v_{1} = 0\; {\rm m\cdot s^{-1}}.)

\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}.

7 0
2 years ago
How many moles are in 95 g of Na?
nexus9112 [7]
1.8 mol of Na. hope this helps
3 0
3 years ago
Read 2 more answers
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