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jenyasd209 [6]
3 years ago
10

Consider the following statement: The magnitude of the buoyant force is equal to the weight of fluid displaced by the object. Un

der what circumstances is this statement true?
Physics
1 answer:
kenny6666 [7]3 years ago
3 0

Answer:

Buoyant force enables the objects to float. So if the object is completely or partially immersed in the fluid then the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object.

Explanation:

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If an object is propelled upward from a height of 128 feet at an initial velocity of 112 feet per​ second, then its height h aft
Marina CMI [18]

Explanation:

The equation of motion of an object is given by :

h(t)=-16t^2+112t+128

Where

t is the time in seconds

We need to find the time when the object hits the ground. When the object hits the ground, h(t) = 0

So,

-16t^2+112t+128=0

-t^2+7t+8=0

On solving above equation using online calculator, t = 8 seconds. So, the object hit the ground after 8 seconds. Hence, this is the required solution.

8 0
3 years ago
What effect does time have on the speed of a moving object
sergejj [24]
Velocity is d/t distance over time. Increase velocity (speed) decrease. Increase d velocity increases.
6 0
3 years ago
A bug splats against the windshield of a car traveling at high speeds down a backcountry road. Which statement correctly compare
zvonat [6]

Answer:

C. The bug's change in momentum is equal to the car's change in momentum.

Explanation:

As we know by Newton's 2nd law

F = \frac{\Delta P}{\Delta t}

here we have also know that when car hits the bug then force applied by wind shield on the bug is same as the force applied by the bug on the car's wind shield as per Newton's III law

F_{12} = F_{21}

so we know that

\frac{\Delta P_{12}}{\Delta t} = \frac{\Delta P_{21}}{\Delta t}

so we have

\Delta P_{12} = \Delta P_{21}

so correct answer will be

C. The bug's change in momentum is equal to the car's change in momentum.

6 0
4 years ago
The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 1.10 ✕ 105 kg train mov
Mrac [35]

Answer:

(a) k = 1684.38 N/m = 1.684 KN/m

(b) Vi = 0.105 m/s

(c) F = 1010.62 N = 1.01 KN

Explanation:

(a)

First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = ?

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = 0.35 m/s

s = distance covered by train before stopping = 2 m

Therefore,

a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)

a = 0.0306 m/s²

Now, we calculate the force applied on spring by train:

F = ma

F = (1.1 x 10⁵ kg)(0.0306 m/s²)

F = 3368.75 N

Now, for force constant, we use Hooke's Law:

F = kΔx

where,

k = Force Constant = ?

Δx = Compression = 2 m

Therefore.

3368.75 N = k(2 m)

k = (3368.75 N)/(2 m)

<u>k = 1684.38 N/m = 1.684 KN/m</u>

<u></u>

<u>(</u>c<u>)</u>

Applying Hooke's Law with:

Δx  = 0.6 m

F = (1684.38 N/m)(0.6 m)

<u>F = 1010.62 N = 1.01 KN</u>

<u></u>

(b)

Now, the acceleration required for this force is:

F = ma

1010.62 N = (1.1 kg)a

a = 1010.62 N/1.1 x 10⁵ kg

a = 0.0092 m/s²

Now, we find initial velocity of train by using 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = -0.0092 m/s² (negative sign due to deceleration)

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = ?

s = distance covered by train before stopping = 0.6 m

Therefore,

-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)

Vi = √(0.0092 m/s²)(1.2 m)

<u>Vi = 0.105 m/s</u>

4 0
4 years ago
What holds the positively charged metal ions together within a metallic bond?
enot [183]

Answer:

The electron cloud

Explanation:

Metallic bonds result from interaction of positively charged metal ions with free valence electrons which now forms an electron cloud around the metal ions. Electrostatic interaction between the metal ions and the electron cloud holds the metal ions together in the metallic bond.

4 0
3 years ago
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