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3 years ago

A boat floats in water a. Archimedes b. Bernoulli c. Density d. Pascal e. Pressure

Physics

2 answers:

Ivenika [448]3 years ago

7 0

The answer is Density !, Do you also need an example ?

Rate this the brainliest answer , Thank youuu !

liubo4ka [24]3 years ago

4 0

Hey there,

Your question states: <span>A boat floats in water . . .

Your correct answer would be "density" basically because the density of the water would make the boat float.

Hope this helps you!

~Jurgen</span>

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A body decelerates uniformly to a constant speed and after some time it accelerates uniformly Draw the shape of speed time graph

Alex73 [517]

Answer:

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3 years ago

A sled and rider with a combined mass of 70 kg are at the top of a hill that rises 9 m above the level ground below. The sled is

kvv77 [185]

Answer:

(A) 6174J, 0J.

(B) 7374J

Explanation:

See the attachment below for the calculation.

The princi6of conservation of energy has been used.

E = K1 + U1 = K2 + U2

Where E = total mechanical energy.

Taking the bottom as reference, and h =0 and as a result U2 = mg(0) = 0J.

The complete solution to the problem can be found in the attachment below.

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2 years ago

A small 12.00g plastic ball is suspended by a string in a uniform, horizontal electric field with a magnitude of 10^3 N/C. If th

tensa zangetsu [6.8K]

Answer:

The net charge is $67.89 \mu C$

Solution:

As per the question:

Mass of the plastic bag, m = 12.0 g = $12.0\times 10^{-3}\ kg$

Magnitude of electric field, E = $10^{3}\ N/C$

Angle made by the string, $\theta = 30^{\circ}$

Now,

To calculate the net charge, Q on the ball:

Vertical component of the tension in the string, $T = Tcos\theta$

Horizontal component of the tension in the string, $T = Tsin\theta$

Now,

Balancing the forces in the x-direction:

$Tsin\theta = QE$

$Q = {Tsin\theta}{E}$ (1)

Balancing the forces in the y-direction:

$Tcos\theta = mg$

where

g = acceleration due to gravity = $9.8\ m/s^{2}$

Thus

$T = \frac{mg}{cos\theta }$

$T = \frac{12.0\times 10^{-3}\times 9.8}{cos30^{\circ}} = 0.1357\ N$

Use T = 0.1357 N in eqn (1):

$Q = {0.1357\times sin30^{\circ}}{10^{3}} = 6.789\times 10^{- 5}\ C$

$Q = 67.89\times 10^{- 5}\ C = 67.89\mu C$

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3 years ago

The drawings show three examples of the force with which someone pushes against a vertical wall. In each case the magnitude of t

Monica [59]

Complete Question

The complete question is shown on the

Answer:

The ascending order would be 2nd < 1st < 3rd

Explanation:

Generally the the Normal force is mathematically represented as

$Normal \ Force = Fsin \theta$

=> $Normal \ Force \ \alpha \ sin \theta$

For the first drawing the value $\theta$ is between that of the the second and the third drawing so the Normal force would also be between the normal forces of the second and the third drawing

For the second drawing whose value of $\theta$ is less than that of the first and the third the normal force would also be less than that of the first and third

For the the third drawing whose value is (90°) which is higher than the first and the second the normal force would also be higher than the first and the second

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2 years ago

Read 2 more answers

A 2.0 kg particle moves in a circle of radius 3.1 m. As you look down on the plane of its orbit, the particle is initially movin

Ghella [55]

Answer

given,

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mass of particle = 2 Kg

radius of the circle = 3.1 m

a) torque

τ = $\dfrac{dL}{dt}$