Help me with this please

John flies directly east for 20km, then turns to the north and flies for another 10 km before dodging a flock of geese. what’s t

KIM [24]

The distance is 30 km and the displacement is 22.4 km North East

7 0

3 years ago

You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the

Sphinxa [80]

Answer:

Option (c) : 20°C

Explanation:

$t(final) = \frac{w1 \times t1 + w2 \times t2}{w1 + w2}$

T(final) = 500* 10 + 100*70/600 = 20°C

4 0

3 years ago

A plane is landing at an airport. The plane has a massive amount of kinetic energy due to it's motion. When the plane lands, it

marshall27 [118]

Answer:

A. The brakes used a coil system to convert the kinetic energy into potential energy stored in the brakes

Explanation:

Based on the law of conservation of energy, the brakes used a coil system to convert the kinetic energy into potential energy stored in the brakes.

The law of conservation of energy states that energy is neither created nor destroyed in a system but it is transformed from one form to another.

As the airplane slows down, the kinetic energy which is presented in the motion of the plane is gradually converted to potential energy.

The potential energy is the energy due to the position of a body.

8 0

3 years ago

A block of mass m = 4.8 kg slides from left to right across a frictionless surface with a speed Vi=7.3m/s. It collides with a bl

nirvana33 [79]

Answer:

The speed of the 11.5kg block after the collision is V≅4.1 m/s

Explanation:

ma= 4.8 kg

va1= 7.3 m/s

va2= - 2.5 m/s

mb= 11.5 kg

vb1= 0 m/s

vb2= ?

vb2= ( ma*va1 - ma*va2) / mb

vb2= 4.09 m/s ≅ 4.1 m/s

7 0

3 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago