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3 years ago

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A 1.2-kg mass is projected from ground level with a velocity of 31.3 m/s at some unknown angle above the horizontal. A short tim

e after being projected, the mass barely clears a 16-m tall fence. Disregard air resistance and assume the ground is level. What is the kinetic energy of the mass as it clears the fence

1 answer:

My name is Ann [436]3 years ago

8 0

Answer:

The kinetic energy will be "399.65 J".

Explanation:

Given:

Mass,

m = 1.2 kg

Velocity,

v = 31.3 m/s

The total energy of mass will be:

⇒ $E=K+U$

or,

⇒ $E=.5mv^2+mgh$

By putting the values, we get

$=.5(1.2)(31.3)^2+0$

$=0.6\times 979.69+0$

$=587.81 \ J$

Since,

The system's total energy is unchanged, then

⇒ $E=K+U$

or,

⇒ $E=K+mgh$

$587.81=K+1.2(9.8)(16)$

$587.81=K+188.16$

$K=587.81-188.16$

$=399.65 \ J$

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A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal di

Lyrx [107]

Answer:

D

Explanation:

From the information given:

The angular speed for the block $\omega = 50 \ rad/s$

Disk radius (r) = 0.2 m

The block Initial velocity is:

$v = r \omega \\ \\ v = (0.2 \times 50) \\ \\ v= 10 \ m/s$

Change in the block's angular speed is:

$\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s$

However, on the disk, moment of inertIa is:

$I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2$

The time t = 10s

∴

Frictional torques by the wall on the disk is:

$T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10}) \\ \\ =0.6 \ N.m$

Finally, the frictional force is calculated as:

$F = \dfrac{T}r{}$

$F= \dfrac{0.6}{0.2} \\ \\ F = 3N$

8 0

3 years ago

HELLO , PLZ HELP .

sergey [27]

Answer:

the pressure due to the water on the diver is 200,000 pascal

pressure = height × density × acceleration due to gravity

p = 20×1000×10

p=200,000 pascal

6 0

3 years ago

A thin layer of liquid methylene iodide (n = 1.756) is sandwiched between two flat, parallel plates of glass (n = 1.50). What mu

Harrizon [31]

Answer:

t = 96.1 nm

Explanation:

For strong reflection through liquid layer we know that the path difference between two reflected light rays must be integral multiple of wavelength

now we know that the path difference of two reflected light from thin liquid layer is given as

$2\mu t - \frac{\lambda}{2} = N\lambda$

here we know that

$\mu = 1.756$

t = thickness of layer

N = 0 (for minimum thickness of layer)

$\lambda = 675 nm$

now we have

$2(1.756) t = \frac{675 nm}{2}$

$t = 96.1 nm$

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3 years ago

At constant volume, the heat of combustion of a particular compound is − 3550.0 kJ / mol. When 1.075 g of this compound ( molar

swat32

Answer:

$C=1,25\cdot 10^{5} kJ/^{\circ}C$

Explanation:

First of all let's define the specific molar heat capacity.

$C = \frac{-Q}{n\cdot \Delta T}$ (1)

Where:

Q is the released heat by the system

n is the number of moles

ΔT is the difference of temperature of the system

Now, we can find n with the molar mass (M) the mass of the compound (m).

$n=\frac{m}{M}=6.95\cdot 10^{-3} moles$

Using (1) we have:

$C=\frac{-3550}{6.95\cdot 10^{-3} 4.073}$

$C=1,25\cdot 10^{5} kJ/^{\circ}C$

I hope it helps!

6 0

3 years ago

The density of a block of wood is 694 kg/m3. Its mass is 689 g. We tie the block to the bottom of a swimming pool using a single

Serhud [2]

Answer:

<em>The tension in the string = 2.065 N</em>

Explanation:

From Archimedes principle,

R.d = density of the wood block/density of water = weight of the wood block/Upthrust of the wood block in water.

R.d = D₁/D₂ = W/U

W/U =D₁/D₂.................................. Equation 1

Where W = weight of the wood block, U = upthrust of the wood block in water, D₁ = Density of the wood block, D₂ = Density of water.

Making U the subject of the equation,

U = WD₂/D₁........................... Equation 2

Given: W = mg = (689/1000)9.8 = 6.75 N, D₁ = 694 kg/m³, D₂ = 1000 kg/m³.

Substituting these values into equation 2,

U = 6.75(694)/1000

U = 4684.5/1000

U = 4.685 N.

Note: Three forces act on the wood block in the pool. and they are

(i) The weight(W) acting downs

(ii) The upthrust (U) acting upwards,

(iii) The Tension (T) in the string, acting upwards.

Thus,

W = U+T

T = W - U ................................. Equation 3

Where W = 6.75 N, U = 4.685 N

T = 6.75 - 4.685

T = 2.065 N.

T = 2.065 N

<em>Thus the tension in the string = 2.065 N</em>

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3 years ago