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lesya [120]
2 years ago
15

A 50-kg platform diver hits the water below with a kinetic energy of 5000 Joules. The height (relative to the water) from which

the diver dove was approximately ____ meters.
Physics
1 answer:
artcher [175]2 years ago
6 0

Answer:

h = 10 m

Explanation:

given,

mass of platform = 50 Kg

Kinetic energy = 5000 J

height from which the diver dove = ?

taking acceleration due to gravity = 10 m/s²

using conservation of energy

Kinetic energy is converted into mechanical energy

K.E = P.E

K.E =  m g h

5000 = 50 x 10 x h

500 h = 5000

 h = \dfrac{5000}{500}

    h = 10 m

The height from which the diver dove is equal to h = 10 m

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An airplane is delivering food to a small island. It flies 100 m above the ground at a speed of 150 m/s .
miss Akunina [59]

Answer:

The airplane should release the parcel 6.7*10^2 m before reaching the island

Explanation:

The height of the plane is y_0=100m, and its speed is v=150 m/s

When an object moves horizontally in free air (no friction), the equation for the y measured with respect to ground is

y=y_0 - \frac{gt^2}{2}    [1]

And the distance X is

x = V.t     [2]

Being t the time elapsed since the release of the parcel

If we isolate t from the equation [1] and replace it in equation [2] we get

X = V . \sqrt{\frac{2y_0}{g}}

Using the given values:

x = 150 m/s  \sqrt{\frac{2\times 100m}{9.8 m/sec^2}}

x = 6.7*10^2 m

4 0
2 years ago
A circular coil 17.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this locati
sergeinik [125]

Answer:

The torque in the coil is  4.9 × 10⁻⁵ N.m  

Explanation:

T = NIABsinθ

Where;

T is the  torque on the coil

N is the number of loops = 9

I is the current = 7.8 A

A is the area of the circular coil = ?

B is the Earth's magnetic field = 5.5 × 10⁻⁵ T

θ is the angle of inclination = 90 - 56 = 34°

Area of the circular coil is calculated as follows;

A = \frac{\pi d^2}{4} \\\\A = \frac{\pi 0.17^2}{4} =0.0227 m^2

T = 9 × 7.8 × 0.0227 × 5.5×10⁻⁵ × sin34°

T = 4.9 × 10⁻⁵ N.m

Therefore, the torque in the coil is  4.9 × 10⁻⁵ N.m

5 0
2 years ago
A gas has an initial volume of 168 cm3 at a temperature of 255 K and a pressure of 1.6 atm. The pressure of the gas decreases to
erastovalidia [21]
Would presume you are asked to find the volume, since there is no second volume.

By General Gas Law:

P₁V₁/T₁ = P₂V₂/T₂

1.6 * 168 /255 = 1.3*V₂/285

V₂ = 1.6 * 168 * 285 / (1.3*255)

V₂ = 231.095

Final volume ≈ 231 cm³
7 0
2 years ago
Read 2 more answers
The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17°C and
mina [271]

Answer:

\frac{dQ}{dt}= 4312 W

Explanation:

As we know that base of the slab is given as

A = 11 \times 8

A = 88 m^2

now we know that rate of heat transfer is given as

\frac{dQ}{dt} = \frac{kA}{x} (T_2 - T_1)

here we know that

k = 1.4 W/m k

Also we have

x =0.20

\frac{dQ}{dt} = \frac{1.4(88)}{0.20}(17 - 10)

\frac{dQ}{dt}= 4312 W

7 0
2 years ago
A 5.00 kg crate is on a 21.0 degree hill. Using X-Y axes tilted down the plane, what is the y-component of the normal force?
Rama09 [41]

Answer:

The y-component of the normal force is 45.74 N.

Explanation:

Given that,

Mass of the crate, m = 5 kg

Angle with hill, \theta=21^{\circ}

We need to find the y component of the normal force. We know that the y component of the normal force is given by :

F_y=F\ \cos\theta\\\\F_y=mg\ \cos\theta\\\\F_y=5\times 9.8\ \cos(21)\\\\F_y=45.74\ N

So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.

7 0
1 year ago
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