All categories

3 years ago

I need help on god!!!

1 answer:

6 0

Solubility - is how much solute will dissolve in the solvent (so that would be a measure of how much salt (solute) would dissolve in the water (solvent) making up a solution)

You might be interested in

How are science and technology related?

aalyn [17]

One way you could think about it is that you used science of electronics to develop the technology of a computer

4 0

3 years ago

Read 2 more answers

Please help! If the instantaneous speed of an object remains constant, can its instantaneous velocity change? If the instantaneo

kondaur [170]

I think the the answer would be no. If the instantaneous speed of an object remains constant, then its instantaneous velocity would not change. Since v = c for c is some constant, then |v| = |c|. This is regardless of the sign of c. Hope this answers the question.

4 0

4 years ago

A fuzzy bunny sees a butterfly and chases it right off of a 20 m high cliff at 7m/s

bogdanovich [222]

A) 2.02 s

To find the time it takes for the bunny to reach the rocks below the cliff, we just need to analyze its vertical motion. This motion is a free fall motion, which is a uniformly accelerated motion. So we can use the suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where

s is the vertical displacement

u is the initial vertical velocity

a is the acceleration

t is the time

For the bunny here, choosing downward as positive direction,

$u_y = 0$ (initial vertical velocity is zero)

s = 20 m

$a=g=9.8 m/s^2$ (acceleration of gravity)

And solving for t, we find the time of flight:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s$

B) 14.1 m

For this part, we need to consider the horizontal motion of the bunny.

The horizontal motion of the bunny is a uniform motion with constant velocity, which is the initial velocity of the bunny:

$v_x = 7 m/s$

Therefore the distance covered after time t is given by

$d=v_x t$

And substituting the time at which the bunny hits the ground,

t = 2.02 s

We find how far the bunny went from the cliff:

$d=(7)(2.02)=14.1 m$

C) 21.0 m/s at $70.5^{\circ}$ below the horizontal

The horizontal component of the velocity is constant during the entire motion, so it will still be the same when the bunny hits the ground:

$v_x = 7 m/s$

Instead, the vertical velocity is given by

$v_y = u_y +at$

And substituting t = 2.02 s, we find the vertical velocity at the moment of impact:

$v_y = 0+(9.8)(2.02)=19.8 m/s$

So, the magnitude of the final velocity is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{7^2+(19.8)^2}=21.0 m/s$

And the angle is given by

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{19.8}{7})=70.5^{\circ}$

below the horizontal

5 0

3 years ago

When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6

Stella [2.4K]

Answer:

it is separated by 80 cm distance

Explanation:

As per Coulombs law we know that force between two point charges is given by

$F = \frac{kq_1q_2}{r^2}$

here we know that

$q_1 = +8.4\mu C$

$q_2 = +5.6 \mu C$

force between two charges is given as

$F = 0.66 N$

now we have

$F = \frac{kq_1q_2}{r^2}$

$0.66 = \frac{(9\times 10^9)(8.4 \mu C)(5.6 \mu C)}{r^2}$

$r = 0.8 m$

so it is separated by 80 cm distance

3 0

3 years ago

Read 2 more answers

A box of books weighing 305 N is shoved

Anna [14]

Answer:

3.1 s

Explanation:

from the question we are given the the following:

Weight of books (Fb) = 305 N

Push force (Fp) = 516 N

distance (s) = 5.82 m

Angle of force exerted = 32 degrees

acceleration due to gravity (g) = 9.81 m/s^{2}

coefficient of friction Uk= 0.59

time (t) = ?

mass of the book (m) = weight / (g) = 305 / 9.8 = 31 kg

lets first get the net force from the summation of vertical forces

Fnet = Fpush + Fgravity

F net = 516 + 305sin32

F net = 677.6 N

now lets get the acceleration from the summation of the horizontal forces

Fpcos32 - friction force = m x a

Fpcos32 - (Uk x F net) = m x a

516cos32 - (0.59 x 677.6) = 31 x a

37.8 = 31a

a = 1.22m/s^{2}

now that we have our acceleration we can get the time from the equation of motion

s = ut + o.5at^{2}

u ( initial velocity ) = 0 because the box was initially at rest

5.82 = (0 x t ) + (0.5 x 1.22 x t^{2})

5.82 = 0.61t^{2}

t = 3.1 s

5 0

3 years ago