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hammer [34]
3 years ago
8

a container of water is knocked off a 10.0 meter high ledge with a horizontal velocity of 1.00 meters/second. calculate the time

it takes for the container to reach the ground
Physics
1 answer:
Evgen [1.6K]3 years ago
3 0

Answer:

1.43 s

Explanation:

The time it takes for the container to reach the ground is determined only by the vertical motion of the container, which is a free-fall motion, so a uniformly accelerated motion with a constant acceleration of g=9.8 m/s^2 towards the ground.

The vertical distance covered by an object in free fall is given by

S=ut + \frac{1}{2}at^2

where

u = 0 is the initial vertical speed

t is the time

a= g = 9.8 m/s^2 is the acceleration

since u=0, it can be rewritten as

S=\frac{1}{2}gt^2

And substituting S=10.0 m, we can solve for t, to find the duration of the fall:

t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(10.0 m)}{9.8 m/s^2}}=1.43 s

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What amount of heat is required to raise the temperature of 200 g of water by 15°C (the specific heat of water is 1 cal/g°C)
Sergeeva-Olga [200]

Answer:

Heat energy required (Q) = 3,000 J

Explanation:

Find:

Mass of water (M) = 200 g

Change in temperature (ΔT) = 15°C

Specific heat of water (C) = 1 cal/g°C

Find:

Heat energy required (Q) = ?

Computation:

Q = M × ΔT × C

Heat energy required (Q) = Mass of water (M) × Change in temperature (ΔT) × Specific heat of water (C)

Heat energy required (Q) = 200 g × 15°C × 1 cal/g°C

Heat energy required (Q) = 3,000 J

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3 years ago
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Yuliya22 [10]

Answer:

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Pick a point on the graph for Ian and plug in values.

For example, 20 minutes -> 4km

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7 0
3 years ago
Read 2 more answers
Mike is standing on the roof of a building looking at the roof of the neighboring building that is 15 meters away and 10 meters
amid [387]

Answer:

Part a)

t = 1.65 s

Part b)

x = 40.4 m

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

v = 27.3 m/s

direction of velocity is given as

[tex]\theta = 26.35 degree

Explanation:

Part a)

acceleration due to gravity on this planet is 3/4 times the gravity on earth

So the acceleration due to gravity on this new planet is given as

a = \frac{3}{4}(9.81)

a = 7.36 m/s^2

now the vertical displacement covered by the canister is given as

y = 10 m

now by kinematics we have

y = \frac{1}{2}gt^2

10 = \frac{1}{2}(7.36)t^2

t = 1.65 s

Part b)

Horizontal speed of the canister is given as

v_x = 24.5 m/s

now the distance moved by it

x = v_x t

x = 24.5 (1.65)

x = 40.4 m

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

Final velocity in X direction will remains the same

v_x = 24.5 m/s

final velocity in Y direction

v_y = v_i + at

v_y = 0 + (7.36)(1.65)

v_y = 12.14 m/s

now magnitude of velocity is given as

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{24.5^2 + 12.14^2}

v = 27.3 m/s

direction of velocity is given as

\theta = tan^{-1}\frac{v_y}{v_x}

\theta = tan^{-1}\frac{12.14}{24.5}

[tex]\theta = 26.35 degree

6 0
3 years ago
Which type of map would you most likely use to locate mineral deposits? A. topographic B. geologic C. satellite D. hazard
qwelly [4]
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2 years ago
Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45 m diamete
Readme [11.4K]

Answer:

In m/s^2:

a=11.3778 m/s^2

In units of g:

a=1.161 g

Explanation:

Since the racing greyhounds are capable of rounding corners at very high speed so we are going use the following formula of acceleration for circular paths.

a=\frac{v^2}{r}

where:

v is the speed

r is the radius

Now,

a=\frac{16^2}{45/2}\\ a=11.3778 m/s^2

In g units:

a=\frac{11.3778\ g}{9.8}\\ a=1.161\ g

7 0
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