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3 years ago

If the weight of a submerged object is equal to the buoyant force, what net force acts upon the object?

Physics

1 answer:

Y_Kistochka [10]3 years ago

7 0

Answer:

Explanation:

The weight is acting downwards where as the buoyant force acting upwards (opposite) direction with equal amount of force. so the opposite forces cancel out each other (because of the force amount being equal) and no net force is acting on the object.

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4 uses of plane mirror

faust18 [17]

Answer:

they are used in periscopes,for signalling,in kaleidoscopes,to see round dangerous bends

Explanation:

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3 years ago

What is the unit for height

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The unit of height is:
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3 years ago

Read 2 more answers

A 2.5 m -long wire carries a current of 8.0 A and is immersed within a uniform magnetic field B⃗ . When this wire lies along the

leva [86]

Answer:

Explanation:

Let the magnetic field be B = B₁i + B₂j + B₃k

Force = I ( L x B ) , I is current , L is length and B is magnetic field .

In the first case

force = - 2.3 j N

L = 2.5 i

puting the values in the equation above

- 2.3 j = 8 [ 2.5 i x ( B₁i + B₂j + B₃k )]

= - 20 B₃ j + 20 B₂ k

comparing LHS and RHS ,

20B₃ = 2.3

B₃ = .115

B₂ = 0

In the second case

L = 2.5 j

Force = I ( L x B )

2.3i−5.6k = 8 ( 2.5 j x (B₁i + B₂j + B₃k )

= - 20 B₁ k + 20B₃ i

2.3i−5.6k = - 20 B₁ k + 20B₃ i

B₃ = .115

B₁ = .28

So magnetic field B = .28 i + .115 B₃

Part A

x component of B = .28 T

Part B

y component of B = 0

Part C

z component of B = .115 T .

8 0

3 years ago

A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T

miv72 [106K]

Answer:

C. In the first collision has twice the momentum as when it stays still ( second colllions)

Explanation:

To see which statement is correct, it is best to solve the problem, the momentum is equal to the variation of the moment

I = Δp = m vf - m v₀

I = m (vf -v₀)

Case 1. In car bounces, the initial speed is 0.3 m / s, say that this direction is positive, when the magnitude of the speed bounces it remains constant, but its direction is reversed (vf = -0.3 m / s)

I₁ = m (-0.3 - 0.3)

I₁ = -0.6 m

Case 2. The expensive one that still after the crash so its speed is zero (vf = 0)

I₂ = m (0 - 0.3)

I₂ = -0.3 m

Let's calculate the relationship between the two impulses

I₁ / I₂ = -0.6m / -0.3m

I₂ / I₂ = 2

When it bounces it has twice the momentum as when it stays still

Now let's analyze the answers:

A. False The momentum changes

B. False. The momentum is less in the second collision

C. True. The momentum is double in this collision

D. False. Can be calculated, because the mass is the same throughout the exercise and is eliminated in the equations

E. False. When they say bounces it implies the same speed with the opposite direction

4 0

3 years ago

Block A of mass M is at rest and attached to the top of a spring. The block compresses the spring a distance d from its uncompre

Anni [7]

Answer:

a) k = Mg / d , b) v = √2gh , c) v_{f} = $\frac{2}{3} \ \sqrt{2gh}$ , d) x² + 6d x - $\frac{8}{3}$ dh = 0

e)the spring must compress a greater distance.

Explanation:

a) when the block of mass M is placed on the spring, we have an equilibrium condition,

∑ F = 0

$F_{e}$ - W = 0

k d = Mg

k = Mg / d

b) let's use the concepts of energy to find the velocity of the block just before the collision

starting point. Position when released

Em₀ = U = m g h

lowest point. Right at the point of shock

$Em_{f}$ = K = ½ m v²2

as there is no friction, energy is conserved

Em₀ = Em_{f}

mg h = ½ m v²

v = √2gh

c) The velocity of the two blocks after the collision, we define a system formed by the two blocks, in such a way that the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

p₀ = 2M v + M 0

final instant. Just after the shock, before the spring compression begins

p_{f} = (2M + M) v_{f}

the moment is preserved

p₀ = p_{f}

2M v = 3M v_{f}

v_{f} = ⅔ v

v_{f} = $\frac{2}{3} \ \sqrt{2gh}$

d) now we work with the joined system after the collision, let's use the concepts of energy

starting point. After shock, before beginning spring compression

Em₀ = K = ½ (3M) $v_{f}^2$

Em₀ = 3/2 M (\frac{2}{3} \ \sqrt{2gh})²

Em₀ = 4/3 M gh

final point. With the spring fully compressed

Em_f = K_e + U = ½ k x² + (3M) g x

in this case we have taken the zero of gravitational potential energy at the point where the blocks collide, as there is no friction, the energy is conserved

Em₀ = Em_f

4/3 M g h = ½ k x² + 3M g x

½ k x² + 3Mg x - 4/3 Mgh = 0

we substitute the expression for k

$\frac{1}{2}$ ( $\frac{Mg}{d}$ ) x² + 3Mg x - $\frac{4}{3}$ Mgh = 0

$\frac{x^{2} }{2d}$ + 3 x - $\frac{4}{3}$ h = 0

to find the value of the spring compression, the second degree equation must be solved

x² + 6d x - $\frac{8}{3}$ dh = 0

x = [-6d ± $\sqrt{(36 d^{2} - 4 \frac{8}{3} dh) }$ ] / 2

x = [-6d ± 6d $\sqrt{ 1 - \frac{32}{3 \ 36} \ \frac{h}{d} }$ ]/2

x = 3d ( -1± $\sqrt{ 1 - 0.296 \frac{h}{d} }$ )

e) If the collision elastic force would not lose any part of the kinetic energy during the collision, therefore the speed of the block of mass M would be much higher and therefore the spring must compress a greater distance.

8 0

3 years ago