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Elena-2011 [213]

4 years ago

11

Which of these substances would rise to the top if placed in a container with the rest?

1 answer:

Vladimir [108]4 years ago

4 0

What substances? Depends on their density, the lower density floats on top. For example, oil floats on top of water

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erastova [34]

Answer:

Playing hockey, driving a car, and even simply taking a walk are all everyday examples of Newton's laws of motion.

7 0

3 years ago

As an acapulco cliff diver drops to the water from a height of 42.0 m, his gravitational potential energy decreases by 27,500 j.

seraphim [82]

We know that potential energy is simply energy due to position which has the formula:

PE = mass * gravity * height

We know that mass * gravity is weight, therefore:

PE = weight * height

27500 J = weight * 42 m

<span>weight = 654.76 Newtons</span>

3 0

4 years ago

A 5.7 kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t = 0s, the bl

Genrish500 [490]

Answer:

Explanation:

Given

mass of block $m=5.7\ kg$

at $t=0 s$

displacement is $x=-0.7\ m$

velocity $v=-0.8\ m/s$

acceleration $a=2.7\ m/s^2$

suppose $x=A\cos (\omega t+\phi )$ is the general equation of SHM

where A=amplitude

$\omega$ =natural frequency of oscillation

therefore velocity and acceleration is given by

$v=-A\omega \sin (\omega t+\phi )$

$a=A\omega ^2\cos (\omega t+\phi )$

for t=0

$-0.7=A\cos (\phi )---1$

$v=-0.8=-A\omega \sin(\phi)---2$

$a=2.7=-A\omega ^2\cos(\phi )----3$

divide 1 and 3 we get

$\omega ^2=\frac{27}{7}$

$\omega =\sqrt{\frac{27}{7}}$

Now square and 1 and 2 we get

$(0.7)^2+(\frac{0.8}{\omega })^2=A^2$

$A^2=0.49+0.166$

$A=0.81\ m$

3 0

3 years ago

A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find

sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge $q_{1}=9\times10^{-6}\ C$ at origin

Second charge $q_{2}=6\times10^{-6}\ C$

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

$E=\dfrac{kq}{r^2}$

$0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}$

$\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}$

$\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1$

$\dfrac{1}{d}=1.82$

$d=\dfrac{1}{1.82}$

$d=0.55\ m$

Hence, The coordinates of the point is (0,0.55).

3 0

3 years ago

3. A pendulum with a 1.0-kg weight is set in motion from a position 0.04 m above the lowest point on the path of the weight.

gavmur [86]

Answer: K.E = 0.4 J

Explanation:

Given that:

M = 1.0 kg

h = 0.04 m

K.E = ?

According to conservative of energy

K.E = P.E

K.E = mgh

K.E = 1 × 9.81 × 0.04

K.E = 0.3924 Joule

The kinetic energy of the pendulum at the lowest point is 0.39 Joule

6 0

3 years ago