You want to create a device to warm food that is safe. Which type of electromagnetic wave would be most useful to investigate?

Differences between work against friction and work against gravity.

Goshia [24]

Friction is the force that is resisting the motion of an object so it will always point in the opposite direction of that of movement. ... The force of gravity points downwards . So when you do work against gravity it means that the force acting on that object points in the upward direction .

4 0

3 years ago

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Colin knows that as water moves through the water cycle, condensation also takes place. Condensation occurs when water vapor coo

Alborosie

Answer:

Monday

Explanation:

3 0

3 years ago

A rocket is launched from a height of 3 m with an initial velocity of 15 m/s What is the maximum height of the rocket? When will

Fudgin [204]

If no extra acceleration is added to the rocket, then its velocity at time t is

v = 15 m/s - g t

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity.

Also, recall that

v² - u² = 2 a ∆x

where u is initial speed, v is final speed, a is acceleration, and ∆x is net displacement.

At the rocket's maximum height ∆x, the velocity is 0. So, the maximum height is

0² - (15 m/s)² = 2 (-g) ∆x

∆x = (15 m/s)² / (2 * (9.80 m/s²)) ≈ 11.48 m

But this assumes the rocket is launched from the ground. We're given that the rocket is launced from 3 m above the ground, so we need to add this to the height above. So the maximum height is closer to 14.48 m.

As mentioned before, this happens when vertical velocity is 0:

0 = 15 m/s - g t

t = (15 m/s) / (9.80 m/s²) ≈ 1.53 s

5 0

3 years ago

A boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. The ball is 20 m above the ground as he rele

kirill115 [55]

Explanation:

It is given that,

Speed of the ball, v = 10 m/s

Initial position of ball above ground, h = 20 m

(a) Let H is the maximum height reached by the ball. It can be calculated using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh'$

$h'=\dfrac{v^2}{2g}$

$h'=\dfrac{(10)^2}{2\times 9.8}$

h' = 5.1 m

The maximum height above ground,

H = 5.1 + 20

H = 25.1 meters

So, the maximum height reached by the ball is 25.1 meters.

(b) The ball's speed as it passes the window on its way down is same as the initial speed i.e. 10 m/s.

Hence, this is the required solution.

6 0

3 years ago

An airplane is flying in a horizontal circle at a speed of 480 km/h. If its wings are tilted 40° to the horizontal, what is the

guajiro [1.7K]

Answer:

r = 2161.9 m

Explanation:

Aerodynamic lift(L) is perpendicular to the wing, which is tilted 40 degrees to the horizontal.

Since the plane is moving in a horizontal circle, the vertical component of the lift must cancel the weight W of the airplane, but the horizontal component is the centripetal force that keeps it in a circle.

L is perpendicular to wing at angle θ with respect to horizontal

Thus,

Vertical component of lift is:

L cosθ = W = mg

Thus, m = L cosθ / g - - - - (eq1)

Horizontal component of lift is:

L sinθ = centripetal force = mv² / r - - - - (eq2)

Combining equations 1 and 2,we have;

L sinθ = (L cosθ / g)(v² / r)

L cancels out on both sides to give;

tanθ = v²/ rg

r = v² / (g tanθ)

We are given;

velocity; v = 480 km/hr = 480 x 10/36 = 133.33 m/s

r = 133.33²/[(9.8) tan(40)] = 2161.9 m

3 0

3 years ago