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3 years ago

A cannonball is fired in such a way

Physics

1 answer:

Goryan [66]3 years ago

5 0

Answer:

θ = 53.13° above horizontal

Explanation:

Ignore air resistance.

The time the ball takes to reach its max height will equal the time to travel half the max range R.

√(2h/g) = (R/2)/vcosθ

vsinθ/g = R/2vcosθ

2v²sinθcosθ = Rg

2v²sinθcosθ = 3hg

v² = u² + 2gh

0² = v²sin²θ + 2gh

v²sin²θ = 2gh

v² = 2gh/sin²θ

2v²sinθcosθ = 3hg

(2gh/sin²θ)2sinθcosθ = 3hg

(2/sinθ)2cosθ = 3

cotθ = 3/4

θ = 53.130

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calculate the energy spent on spraying a drop of mercury of 1 cm radius into 10^6 droplets all of same radius. surface tension o

WARRIOR [948]

Answer:

ANS : .Energy spent on spraying = $4.3542*10^{-4}J$

Explanation:

Given:

Radius of mercury = 1cm initially ;
split into $10^{6}$ drops ;

Thus, volume is conserved.

i.e ,

$\frac{4}{3} \pi R_{o}^{3} = 10^{6}*\frac{4}{3} \pi R_{n}^{3}\\R_{n}=\frac{R_{o}}{10^{2}} = \frac{1cm}{100} = 0.01 cm$

Energy of a droplet = $T$ Δ $A$

Where ,

T is the surface tension
ΔA is the change in area

Initial energy $E_{i} = T*A_{i}\\= 0.0035 * 4 *\pi *0.01^{2}\\=4.398*10^{-6}J$

Final energy $E_{f}=10^{6}*T*A_{f}\\=10^{6}*0.0035*4*\pi *(0.0001)^2\\=4.39823*10^{-4}$

∴ .Energy spent on spraying = $=E_{f}-E{i}\\=(439.823-4.39823)*10^{-6}\\=4.3542*10^{-4}J$

ANS : .Energy spent on spraying = $4.3542*10^{-4}J$

6 0

3 years ago

A parallel-plate air capacitor with a capacitance of 260 pF has a charge of magnitude 0.155 Î¼C on each plate. The plates have a

butalik [34]

Answer:

The potential difference between the plates is 596.2 volts.

Explanation:

Given that,

Capacitance $C=260\ pF$

Charge $q=0.155\ \mu\ C$

Separation of plates = 0.313 mm

We need to calculate the potential difference between the plates

Using formula of potential difference

$V= \dfrac{Q}{C}$

Where, Q = charge

C = capacitance

Put the value into the formula

$V=\dfrac{0.155\times10^{-6}}{260\times10^{-12}}$

$V=596.2\ volts$

Hence,The potential difference between the plates is 596.2 volts.

7 0

4 years ago

During the warm-up and your scheduled physical activity, what was the weather like? Did the

yulyashka [42]

Yes, if weather was hot you need more fluids in your body due to the sweat pores needing to cool you down.

5 0

3 years ago

When is the velocity of a mass on a spring at its maximum value?

ehidna [41]

Answer:

A. when the mass has a displacement of zero

Explanation:

The velocity of a mass on a spring can be calculated by using the law of conservation of energy. In fact, the total energy of the mass-spring system is equal to the sum of the elastic potential energy (U) of the spring and the kinetic energy (K) of the mass:

$E=U+K=\frac{1}{2}kx^2 + \frac{1}{2}mv^2$

where

k is the spring constant

x is the displacement of the mass with respect to the equilibrium position of the spring

m is the mass

v is the velocity of the mass

Since the total energy E must remain constant, we can notice the following:

- When the displacement is zero (x=0), the velocity must be maximum, because U=0 so K is maximum

- When the displacement is maximum, the velocity must be minimum (zero), because U is maximum and K=0

Based on these observations, we can conclude that the velocity of the mass is at its maximum value when the displacement is zero, so the correct option is A.

8 0

4 years ago

Read 2 more answers

An accelerating voltage of 2.47 x 10^3 V is applied to an electron gun, producing a beam of electrons originally traveling horiz

Dmitry [639]

Answer:

6.3445×10⁻¹⁶ m

Explanation:

E = Accelerating voltage = 2.47×10³ V

m = Mass of electron

Distance electron travels = 33.5 cm = 0.335 cm

$E=\frac{mv^2}{2}\\\Rightarrow v=\sqrt{\frac{2E}{m}}\\\Rightarrow v=\sqrt{\frac{2\times 2470\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}}\\\Rightarrow v=29455356.08671\ m/s$

Deflection by Earth's Gravity

$\Delta =\frac {gt^2}{2}$

Now, Time = Distance/Velocity

$\Delta =\frac {g\frac{s^2}{v^2}}{2}\\\Rightarrow \Delta =\frac{9.81\frac{0.335^2}{29455356.08671^2}}{2}\\\Rightarrow \Delta =6.3445\times 10^{-16}\ m$

∴ Magnitude of the deflection on the screen caused by the Earth's gravitational field is 6.3445×10⁻¹⁶ m

3 0

3 years ago