What is the kinetic energy of a 3-kilogram ball that is rolling at 2 meters per second

A cannon fired horizontally at 20 m/s from the top of a cliff lands 80m away. how tall is the cliff

Scorpion4ik [409]

Answer:

The height of the cliff is, h = 78.4 m

Explanation:

Given,

The horizontal velocity of the projectile, Vx = 20 m/s

The range of the projectile, s = 80 m

The projectile projected from a height is given by the formula

S = Vx [Vy + √(Vy² + 2gh)] / g

Therefore,

h = S²g/2Vx²

Substituting the values

h = 80² x 9.8/ (2 x 20²)

= 78.4 m

Hence, the height of the cliff is, h = 78.4 m

8 0

3 years ago

Please help me to solve the problem ?cm=1m

Oxana [17]

1m is equivalent to 100cm

1m = 100cm

4 0

3 years ago

Saki is ice-staking to the left with velocity of 5 m/s. A steady breeze starts blowing. Causing saki to accelerate leftward at a

Pavlova-9 [17]

Answer:

Final velocity will be equal to 14 m/sec

Explanation:

We have given initial velocity u = 5 m/sec

Constant acceleration is given $a=1.5m/sec^2$

Time t = 6 sec

We have to find the final velocity

From first equation of motion $v=u+at$ , here v is final velocity, u is initial velocity , a is acceleration and t is time

So $v=5+1.5\times 6=14m/sec$

So equal final velocity will be equal to 14 m/sec

4 0

2 years ago

Scientist often use ______________ which are programs that combine what is known about atmospheric circulation

Serggg [28]

Answer:

General circulation model.

Explanation:

A general circulation model (GCM) is a type of climate model that employs a mathematical model of the general circulation of a planetary atmosphere or ocean. GCM uses the Navier–Stokes equations on a rotating sphere with thermodynamic terms for various energy sources (radiation, latent heat). These equations are the basic equations for computer programs used to simulate the Earth's atmosphere or oceans.

7 0

3 years ago

Read 2 more answers

A small mirror is attached to a vertical wall, and it hangs a distance of 1.86 m above the floor. The mirror is facing due east,

Katyanochek1 [597]

Answer:

$t=2.044\ hr$

Explanation:

From the schematic we can visualize the situation and the position of the rays falling on the floor.

Considering the given data from the lowest edge of the mirror.

We get a triangle with height of 1.86 meters.

In the first instance the base of the triangle is 3.8 meters.

While in the second instance the base is 1.22 meters.
Speed of rotation of earth, $\omega=15^{\circ}\ hr^{-1}$

Now applying the trigonometric ratio to known sides in the first instance:

$tan\ \theta_1=\frac{1.86}{3.8}$

$\theta_1=26.08^{\circ}$

Applying the trigonometric ratio to known sides in the second instance:

$tan\ \theta_2=\frac{1.86}{1.22}$

$\theta_2=56.74^{\circ}$

Now by the law of reflection we know that the angle of incidence is equal to the angle of reflection. So the sun would have been at the same angle on the opposite side of the normal.

Hence the change in angle of the sun with respect to the mirror (also the earth)

$\Delta \theta=\theta_2-\theta_1$