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3 years ago

A diver is on a board 1.80 m above

Physics

2 answers:

Virty [35]3 years ago

8 0

Answer:

v = 6.95 m/s

Explanation:

Given that,

A diver is on a board 1.80 m above the water, s = 1.8 m

The initial speed of the diver, u = 3.62 m/s

Let v is the speed with which she hit the water. It will move under the action of gravity. Using the equation of motion as follows :

$v^2-u^2=2gs\\\\v=\sqrt{u^2+2gs} \\\\v=\sqrt{(3.62)^2+2(9.8)(1.8)} \\\\v=6.95\ m/s$

So, she will hit the water with a speed of 6.95 m/s.

love history [14]3 years ago

4 0

Answer:

6.95

Explanation:

Acellus

Correct

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A charge q1 of -5.00 x 10^-9 C and a charge q2 of -2.00x 10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium po

aleksandrvk [35]

Answer:

Explanation:

Let the equilibrium position of third charge be x distance from q₁.

Force on third charge due to q₁

= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²

Force on third charge due to q₂

= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

Both the force will act in opposite direction and for balancing , they should be equal.

9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

5 / x² = 2 / ( .4 - x )²

Taking square root on both sides

2.236 / x = 1.414 / .4 - x

2.236 ( .4 - x ) = 1.414 x

.8944 - 2.236 x = 1.414 x

.8944 = 3.65 x

x = .245 m

24.5 cm

So the third charge should be at a distance of 24.5 cm from q₁ .

4 0

2 years ago

A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance.

Nookie1986 [14]

Answer:

a. 0.6 m b. 0.385 m c. 3.6 m/s at 287.78° to the horizontal

Explanation:

a. Using s = ut - 1/2gt² for motion under gravity where s = vertical distance = height of table, u = initial vertical velocity of book = 0 m/s, t = time of flight = 0.350 s and g = acceleration due to gravity = 9.8 m/s².

Substituting these these values into s and taking the top of the table as position 0 m, we have.

0 - s = 0t - 1/2gt²

-s = -1/2gt²

s = 1/2gt²

s = 1/2 × 9.8 m/s² × (0.350 s)²

s = 0.6 m

b. Using d = v't where d = horizontal distance from table, v' = horizontal velocity of book = 1.10 m/s and t = time of flight = 0.350 s

d = v't = 1.10 m/s × 0.350 s = 0.385 m

c. Using v² = u² - 2gs where u = initial vertical velocity of book = 0 m/s and g = 9.8 m/s², s = -0.6 m (negative since we are at the bottom and 0 m is at the top)and v = final vertical velocity of book

v² = u² - 2gs

= 0 - 2 × 9.8 m/s² × (-0.6 m)

= 11.76 m²/s²

v = √11.76 m/s

= 3.43 m/s

So, the magnitude of the resultant velocity is V = √(v² + v'²)

= √((3.43 m/s)² + (1.10 m/s)'²)

= √(11.76 m²/s² + 1.21 m²/s²)

= √12.97 m²/s²

= 3.6 m/s

Its direction Ф = tan⁻¹(-v/v') since v is in the negative y direction

= tan⁻¹(-3.43 m/s/1.10 m/s)

= tan⁻¹(-3.1182)

= -72.22°

Ф = -72.22°+ 360 = 287.78° since it is in the third quadrant

7 0

3 years ago

A single point charge is placed at the center of an imaginary cube that has 10 cm long edges. The electric flux out of one of th

12345 [234]

Answer:

Explanation:

Given:

Length of each side of the cube, $L=10\ cm$

The Elecric flux through one of the side of the cube is, $\phi =-1 kNm^2/C.$

The net flux through a closed surface is defined as the total charge that lie inside the closed surface divided by $\epsilon_0$

Since Flux is a scalar quantity. It can added to get total flux through the surface.

$\phi_{total}=\dfrac{Q_{in}}{\epsilon_0}\\6\times {-1}\times 10^3=\dfrac{Q_{in}}{\epsilon_0}\\\\Q_{in}=-6}\times 10^3\epsilon_0\\Q_{in}=-6}\times 10^3\times8.85\times 10^{-12}\\Q_{in}=-53.1\times10^{-9}\ C$

So the the charge at the centre is calculated.

5 0

3 years ago

When the temperature of a metal cylinder is raised from 0.0°C to 110°C, its length increases by 0.22%. Find the percent change i

monitta

Volume of cylinder = pi*r*2*L

As, from the above formula,volume is directly proportional to length,
So, if we increase in length also increases in volume by 0.22%

we know

density=mass/volume

As, density is inversely proportional to volume it means increasing in volume decreases the density by 50.22%

8 0

3 years ago

Read 2 more answers

NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low- mass sail and the ene

RUDIKE [14]

Answer:

d. The sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.

Explanation:

Let us take the momentum of a photon unit as u

we know that the rate of change of momentum is proportional to the force exerted.

For a absorbing surface, the photon is absorbed, therefore the final momentum is zero. From this we can say that

F = (u - 0)/t = u/t

for a unit time, the force is proportional to the momentum of the wave due to its energy density. Therefore,

F = u

For a reflecting surface, the momentum of the wave strikes the sail and changes direction. Since we know that the speed of light does not change, then the force is proportional to

F = (u - (-u))/t = 2u/t

just as the we did above, it becomes

F = 2u.

From this we can see that the force for a reflective sail is twice of that for an absorbing sail, and we know that the pressure is proportional to the force for a given area. From these, we conclude that the sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.

3 0

3 years ago