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Dafna1 [17]
3 years ago
8

Perpetual Endless Relentless Stubborn Immovable Steady Tireless Enduring Never-ending Constant Ever-lasting

Physics
1 answer:
klasskru [66]3 years ago
5 0

Eleven truly inspiring adjectives !
You've made my day.

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a man can swim in still water with speed of 2kmhr-1 wants to cross a river 200m wide flowing with 1kmhr-1 . Calculate the time t
tangare [24]

Answer:

d savage

Explanation:

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3 years ago
A chemical equation lists NaHCO3(s) as a reactant. What does (s) indicate?
Ostrovityanka [42]

Answer:

The (s) indicates that the state of matter for NaHCO3 is solid.

Explanation:

When a chemical reaction is written, the state of matter for each components of the reactants and products are mentioned in brackets along with their names or formulas.

For example, NaHCO3 has (s) mentioned in the brackets. The s shows that the state of matter for NaHCO3. (l) represents liquid format. (g) represents that the state of matter is gas.

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3 years ago
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Answer rain gauge measures rain shadow units millimetres

7 0
2 years ago
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A ball is thrown vertically upwards from the edge of the cliff and hits the ground at the base of the cliff with a speed of 30 m
olya-2409 [2.1K]

To solve this problem we will apply the linear motion kinematic equations. From the definition of the final velocity, as the sum between the initial velocity and the product between the acceleration (gravity) by time, we will find the final velocity. From the second law of kinematics, we will find the vertical position traveled.

v = v_0 -gt

Here,

v = Final velocity

v_0 = Initial velocity

g = Acceleration due to gravity

t = Time

At t = 4s, v = -30m/s (Downward)

Therefore the initial velocity will be

-30 = v_0 -9.8(4)

v_0 = 9.2m/s

Now the position can be calculated as,

y = h +v_0t -\frac{1}{2}gt^2

When it has the ground, y=0 and the time is t=4s,

0 = h+(9.2)(4)-\frac{1}{2} (9.8)(4)^2

h = 41.6m

Therefore the cliff was initially to 41.6m from the ground

7 0
3 years ago
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
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