49 = 1/2gt²

PLS HELP!!! A 1200-kg whale swims horizontally to the right at a speed of 6.0 m/s. It suddenly collides directly with a stationa

Fittoniya [83]

Answer:

7200 kg.m/s

Explanation:

According the law of conservation of linear momentum, the sum of momentum before and after collision are equal.

Using this principle, the sum of initial momentum will be given as p=mv where p is momentum, m is mass and v is velocity

Initial momentum

Mass of whale*initial velocity of whale + mass of seal*initial seal velocity

Since the seal is initially stationary, its velocity is zero. By substitution and taking right direction as positive

Initial momentum will be

1200*6+(280*0)=7200 kg.m/s

Since both initial and final momentum should be equal, hence the final momentum will also be 7200 kg.m/s

7 0

3 years ago

[01.02]

julia-pushkina [17]

That information describes the plane's speed.
If he had also told them what direction they were flying, then
they would have been able to put the two pieces of information
together, and they would know the plane's velocity.

6 0

4 years ago

Read 2 more answers

A slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. The stone lands 36.0 meters away.

alisha [4.7K]

A)

It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:

$S=S_{o}+v_{o}t+ \frac{at^2}{2} \\ 20= \frac{10t^2}{2} \\ t=2s$

Through Definition of Velocity, comes:

$\Delta v= \frac{\Delta S}{\Delta t} \\ v_x= \frac{36}{2} \\ \boxed {v_{x}=18m/s}$

B)

Using the Velocity Hourly Equation in vertical direction, we have:

$v_{y}=v_{y_{o}}+gt \\ v_{y}=10\times2 \\ \boxed {v_{y}=20m/s}$

The angle of impact is given by:

$cos(\theta) =\frac{v_{x}}{v_{y}} \\ cos(\theta) = \frac{18}{20} \\ cos(\theta) =0.9 \\ arccos(0.9)=\theta \\ \boxed {\theta \approx 25.84}$

If you notice any mistake in my english, please let me know, because i am not native.

7 0

3 years ago

Read 2 more answers

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

Nostrana [21]

Answer:

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

$E_1 = E_2$

Let the position where net field is zero will lie at distance "r" from q1

$\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}$

now we will have

$\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}$

now square root both sides

$\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}$

now we have

$\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1$

so we have

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

8 0

3 years ago

Help ASAP

EastWind [94]

Answer:

Elastic potential energy, $E=2.35\times 10^{-8}\ J$

Explanation:

Charge, $q=9.4\times 10^{-10}\ C$

Potential, V = 50 V

It is required to find the electric potential energy in a capacitor stored in it. The formula of the electric potential energy in a capacitor is given by :

$E=\dfrac{1}{2}qV\\\\E=\dfrac{1}{2}\times 9.4\times 10^{-10}\times 50\\\\E=2.35\times 10^{-8}\ J$

So, the electric potential energy stored in the capacitor is $2.35\times 10^{-8}\ J$

8 0

3 years ago