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4 years ago

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A teacher performing demonstration finds that a piece of cork displaces 23.5 ml of water. The piece of cork has a mass 5.7 g. Wh

at is the density of the cork ?

1 answer:

Sedbober [7]4 years ago

8 0

Answer: 0.24g/ml

Explanation:

Given that:

Volume of water displaced = 23.5 ml

Mass of cork = 5.7 g

Density of the cork = ?

Recall that density is obtained by dividing the mass of a substance by the volume of water displaced.

i.e Density = Mass/volume

Density = 5.7g /23.5ml

Density = 0.24g/ml

Thus, the density of the piece of cork is 0.24g/ml

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An electron moving with a velocity = 5.0 × 10 7 m/s enters a region of space where perpendicular electric and a magnetic fields

inna [77]

Answer:

The magnetic field is $2 \times 10^{-4}$ T

Explanation:

Given:

Velocity of electron $v = 5 \times 10^{7} \frac{m}{s}$

Electric field $E = 10^{4} \frac{V}{m}$

The force on electron in magnetic field is given by,

$F = qvB \sin \theta$ ......(1)

The force on electron in electric field is given by,

$F = qE$ ......(2)

Compare both equation,

$qE = qvB \sin \theta$

Here $\sin \theta = 1$

$E = vB$

$B= \frac{E}{v}$

$B = \frac{10^{4} }{5 \times 10^{7} }$

$B = 2 \times 10^{-4}$ T

Therefore, the magnetic field is $2 \times 10^{-4}$ T

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3 years ago

Which of the following is an arithmetic sequence? A. 2, 1, 4, 3, 6, 5, …

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Explanation:

it is not a arithmetic sequence

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2 years ago

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myrzilka [38]

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3 years ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
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4 years ago

A bullet with an initial kinetic energy of 400 J strikes a wooden block where a 8000 N resistive force stops the bullet. What is

Sindrei [870]

Answer:

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Explanation:

We must remember the principle of conservation of energy which tells us that energy is transformed from one way to another. For this case, the initial kinetic energy is transformed into useful work that is equal to the product of force by distance.

$E_{k}=F*d\\400 = 8000*d\\d = 0.05 [m] = 50 [mm]$

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3 years ago